3130005 cvpde gtu_study_material_e-notes_all_18072019070728_am

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I N D E X
UNIT 1 – COMPLEX FUNCTION AND CONFORMAL MAPPING............................1
1). METHOD – 1: BASIC EXAMPLES...........................................................................................................3
2). METHOD – 2: SQUARE ROOT OF COMPLEX NUMBER.................................................................7
3). METHOD – 3: NTH ROOT OF COMPLEX NUMBER...........................................................................9
4). METHOD – 4: TRIGONOMETRIC FUNCTION OF COMPLEX NUMBER ................................12
5). METHOD – 5: LOGARITHM OF COMPLEX NUMBER..................................................................13
6). METHOD – 6: DIFFERENTIBILITY OF COMPLEX FUNCTION.................................................15
7). METHOD – 7: ANALITICITY OF COMPLEX NUMBER.................................................................17
8). METHOD – 8: TO FIND HARMONIC FUNCTION ..........................................................................20
9). METHOD – 9: FIXED POINT, CRITICAL POINT, ORDINARY POINT .....................................22
10). METHOD – 10: ELEMENTARY TRANSFORMATION ..................................................................23
11). METHOD – 11: BILINEAR TRANSFORMATION ...........................................................................25
UNIT-2 » COMPLEX INTEGRAL, SEQUENCE AND SERIES...............................27
12). METHOD – 1: LINE INTEGRAL...........................................................................................................28
13). METHOD – 2: MAXIMUM MODULUS THEOREM.........................................................................30
14). METHOD – 3: CAUCHY INTEGRAL THEOREM.............................................................................31
15). METHOD – 4: CAUCHY INTEGRAL FORMULA..............................................................................32
16). METHOD – 5: CONVERGENCE OF A SEQUENCE..........................................................................35
17). METHOD – 6: CONVERGENCE OF SERIES......................................................................................37
18). METHOD – 7: REDIUS OF CONVERGENCE ....................................................................................38
19). METHOD – 8: TAYLOR’S SERIES AND MACLAURIN’S SERIES...............................................40
UNIT-3 » LAURENT’S SERIES AND RESIDUES................................................42
20). METHOD – 1: LAURENT’S SERIES ....................................................................................................42
21). METHOD – 2: RESIDUES OF FUNCTION.........................................................................................45
22). METHOD – 3: CAUCHY’S RESIDUE THEOREM.............................................................................46

I N D E X
23). METHOD – 4: CONTOUR INTEGRATION BY USING RESIDUE THEOREM .........................49
UNIT-4 » FIRST ORDER PARTIAL DIFFERENTIAL EQUATION.........................51
24). METHOD – 1: EXAMPLE ON FORMATION OF PARTIAL DIFFERENTIAL EQUATION...52
25). METHOD – 2: EXAMPLE ON LAGRANGE’S DIFFERENTIAL EQUATION.............................55
26). METHOD – 3: EXAMPLE ON NON-LINEAR PDE...........................................................................56
27). METHOD – 4: EXAMPLE ON NON-LINEAR PDE...........................................................................57
UNIT-5 » HIGHER ORDER PARTIAL DIFFERENTIAL EQUATION .....................59
28). METHOD – 1: EXAMPLE ON SOLUTION OF HIGHER ORDERED PDE..................................60
29). METHOD – 2: EXAMPLE ON SOLUTION OF HIGHER ORDERED PDE..................................62
30). METHOD – 3: EXAMPLE ON SEPARATION OF VARIABLES.....................................................64
31). METHOD – 4: EXAMPLE ON CLASSIFICATION OF 2ND ORDER PDE ....................................65
32). METHOD – 5: EXAMPLE ON WAVE, HEAT AND LAPLACE EQUATION...............................73
SYLLABUS OF CVPDE – 3130005..…..……………..…………..………..……………….***

U N I T - 1 » C o m p l e x F u n c t i o n a n d C o n f o r m a l M a p p i n g [ 1 ]
UNIT 1 – COMPLEX FUNCTION AND CONFORMAL MAPPING
 INTRODUCTION:
 A number z = x + iy is called a complex number, where x, y ∈ ℝ and i = √−1.
 x is called the real part of z and is denoted by Re(z).
 y is called the imaginary part of z and is denoted by Im(z).
 Conjugate of a complex number z = x + iy is denoted by z
̅ and is defined by z
̅ = x − iy.
 Two complex number x + iy and x − iy are said to be complex conjugate of each other.
 GEOMETRICAL REPRESENTATION OF COMPLEX NUMBER:
 Let XOY to be a complex plane, where
OX
⃡ and OY
⃡ are called Real axis and
Imaginary axis respectively.
 The ordered pair P(x, y) represents the
complex number z = x + iy. The xy − plane is
now known as Argand plane or Complex
plane or Gaussian Plane.
 OP
̅̅̅̅ Represents the distance between complex numbers P and O, it is called modulus of
z and denoted by |z|.
𝐢. 𝐞. |𝐳| = 𝐫 = √𝐱𝟐 + 𝐲𝟐 = √𝐳. 𝐳
̅
 Let OP
̅̅̅̅ makes an angle θ with positive real axis, it is called argument of z.
𝐢. 𝐞. 𝛉 = 𝐭𝐚𝐧−𝟏
(
𝐲
𝐱
)
 RULES TO DETERMINE ARGUMENT OF A NON-ZERO COMPLEX NUMBER:
 If x > 0 & y > 0 , θ = tan−1
(
y
x
)  If x > 0 & y < 0 , θ = − tan−1
|
y
x
|
 If x < 0 & y > 0 , θ = π − tan−1
|
y
x
|  If x < 0 & y < 0 , θ = −π + tan−1
|
y
x
|
 Notes:
(1). If −π < θ ≤ π, then argument of z is called “PRINCIPAL ARGUMENT” of z. It is
denoted by 𝐀𝐫𝐠(𝐳).
𝜃
𝑃(𝑥, 𝑦)
|
𝒛
|
=
𝒓
Real axis
Imaginary
axis
𝑶
𝑿
𝒀

(2). 𝐢. 𝐞. 𝐀𝐫𝐠(𝐳) = 𝐭𝐚𝐧−𝟏
(
𝐲
𝐱
)
(3). Arg(z) is a Single-Valued Function.
(4). The “GENERAL ARGUMENT” of argument of z is denoted by “arg(z)”.
(5). Relation between “arg(z)” and “Arg(z)”. 𝐚𝐫𝐠(𝐳) = 𝐀𝐫𝐠(𝐳) + 𝟐𝐤𝛑 ; 𝐤 = 𝟎, ±𝟏, ±𝟐, …
(6). arg(z) is a Multi-Valued Function.
(7). For z = 0 = 0 + i0, argument is not defined.
 ARITHMETIC OPERATIONS OF COMPLEX NUMBERS:
 Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers then
 Addition
z1 + z2 = (x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2)
 Subtraction
z1 − z2 = (x1 + iy1) − (x2 + iy2) = (x1 − x2) + i(y1 − y2)
 Multiplication
z1 ∙ z2 = (x1 + iy1) ∙ (x2 + iy2) = (x1x2 − y1y2) + i(x1y2 + x2y1)
 Division
z1
z2
=
x1 + iy1
x2 + iy2
×
x2 − iy2
x2 − iy2
=
(x1x2 + y1y2) + i(x2y1 − x1y2)
x2
2
+ y2
2
 PROPERTIES:
 Let z1and z2be two complex numbers then
(z1
̅ )
̅̅̅̅̅ = z1 (
z1
z2
)
̅̅̅̅̅
=
z1
̅̅̅
z2
̅̅̅
; z2 ≠ 0
|z1| = |z1
̅ |
z1+z1
̅̅̅
2
= Re(z1)
z1 ± z2
̅̅̅̅̅̅̅̅̅ = z1
̅ ± z2
̅
z1−z1
̅̅̅
2i
= Im(z1)
z1 ⋅ z2
̅̅̅̅̅̅̅̅ = z1
̅ ⋅ z2
̅ z ⋅ z
̅ = x2
+ y2
= |z|2
|z1 + z2| ≤ |z1| + |z2| |z1 − z2| ≥ ||z1| − |z2||
|z1 ∙ z2| = |z1| ∙ |z2| |
z1
z2
| =
|z1|
|z2|

 POLAR REPRESENTATION OF A COMPLEX NUMBER:
 Let z = x + iy be a complex number. Let x = r cos θ and y = r sin θ ; θ ∈ (−π, π].
Now, z = x + iy = rcos θ + i rsin θ = r(cos θ + i sin θ) = 𝐫 𝐜𝐢𝐬 𝛉
Thus, z = r(cos θ + i sin θ) is called Polar representation of a complex number.
 By Euler Formula, eiθ
= cos θ + i sin θ
Then, z = reiθ
is called “Exponential representation”.
METHOD – 1: BASIC EXAMPLES
C 1 Find the Real & Imaginary part of f(z) = z2
+ 3z.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑𝐞(𝐟(𝐳)) = 𝐱𝟐
+ 𝟑𝐱 − 𝐲𝟐
, 𝐈𝐦(𝐟(𝐳)) = 𝟐𝐱𝐲 + 𝟑𝐲
H 2 Separate real and imaginary parts of f(z) = z2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑𝐞(𝐟(𝐳)) = 𝐱𝟐
− 𝐲𝟐
; 𝐈𝐦(𝐟(𝐳)) = −𝟐𝐱𝐲
H 3 Find the value of Re(f(z)) and Im(f(z)) at the indicated point
where f(z) =
1
1−z
at 7 + 2i.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑𝐞(𝐟(𝐳)) = −
𝟑
𝟐𝟎
; 𝐈𝐦(𝐟(𝐳)) =
𝟏
𝟐𝟎
C 4 Find the real part of f(z) =
1
z−2 i
at point 1 + i.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑𝐞(𝐟(𝐳)) =
𝟏
𝟐
H 5 Separate real and imaginary parts of sinh z.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑𝐞(𝐬𝐢𝐧𝐡 𝐳) = 𝐬𝐢𝐧𝐡 𝐱 𝐜𝐨𝐬 𝐲 ; 𝐈𝐦(𝐬𝐢𝐧𝐡 𝐳) = 𝐜𝐨𝐬𝐡 𝐱 𝐬𝐢𝐧 𝐲
C 6 Determine the modulus of following complex number.
z = 3 + 4i ; z =
1−2i
i−1
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟓 ; √
𝟓
𝟐

H 7 Determine the modulus of following complex number.
z = (4 + 2i)(−3 + √2i); z =
1 − 7i
(2 + i)2
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐√𝟓𝟓 ; √𝟐
C 8 Find the Principal Value of argument (Principal Argument).
z = − √3 + i ; z = − √3 − i ; z = √3 − i
𝐀𝐧𝐬𝐰𝐞𝐫:
𝟓𝛑
𝟔
; −
𝟓𝛑
𝟔
; −
𝛑
𝟔
H 9 Find the Principal Value of argument (Principal Argument).
z =
2 + 6√3i
5 + √3i
; z =
1 + 2i
1 − 3i
; z = −
2
1 + √3 i
𝛑
𝟑
;
𝟑𝛑
𝟒
;
𝟐𝛑
𝟑
T 10 Find z if arg(z + 1) =
π
6
& arg(z − 1) =
2π
3
.
𝟏
𝟐
+
√𝟑 𝐢
𝟐
.
T 11 Is Arg(z1 z2) = Arg(z1) + Arg(z2) ? Justify.
Answer: NO
C 12 Express √3 − i into polar form.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐳) = 𝟐 (𝐜𝐨𝐬
𝛑
𝟔
− 𝐢 𝐬𝐢𝐧
𝛑
𝟔
)
H 13 Express the following into polar form.
z = (
2 + i
3 − i
)
2
; z =
1 + i
1 − i
; z = 3 + 4i
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐳) =
𝟏
𝟐
𝐜𝐢𝐬 (
𝛑
𝟐
) ; 𝐟(𝐳) = 𝐜𝐢𝐬 (
𝛑
𝟐
) ; 𝐟(𝐳) = 𝟓 𝐜𝐢𝐬 (𝐭𝐚𝐧−𝟏
𝟒
𝟑
)
T 14
Prove that |e−2z| < 1 if and only if Re(z) > 0.

 BASIC DEFINITIONS:
 Distance:
Let z = a + ib and w = c + id be complex numbers. Distance between z & w is defined as
below.
i. e. |z − w| = √(a − c)2 + (b − d)2
So, Modulus of a complex number z, |z| = √a2 + b2 is distance form origin.
 Circle:
If z′ is a complex number and r is a positive number, then equation of circle is |z − z′| = r.
It gives the set of all those z whose distance from z′
is r. [Points on the boundary] [Fig A]
 Open Circular Disk:
The equation |z − z′| < r means set of all points inside the disk of radius r about z′
.
Here, “OPEN” means that points on the boundary of circle are not in the set. [See Fig B ]
 Closed Circular Disk:
The equation |z − z′| ≤ r means set of all points on the boundary and inside the disk of
radius r about z′
. It is union of circle and open circular disk.
Here, “CLOSED” means that points on the boundary of circle are in the set. [See Fig C ]
𝑶 𝑿
𝒀
𝒓
𝒛′
𝑭𝒊𝒈 𝑨
𝑶 𝑿
𝒀
𝒓
𝒛′
𝑭𝒊𝒈 𝑩
𝑶 𝑿
𝒀
𝒓
𝒛′
𝑭𝒊𝒈 𝑪

 Neighborhood:
The neighborhood of a point z0 is set of points inside
the circle centered at z0 and radius ϵ.
i. e. |z − z0| < ϵ
Neighborhood is nothing but a open circular disk with
center z0 and radius ϵ.
 Deleted Neighborhood:
The deleted neighborhood of a point z0 is set of points
inside the circle centered at z0 and radius ϵ except the
center z0.
i. e. 0 < |z − z0| < ϵ
A deleted neighborhood is also known as “Punctured
Disk”.
 Annulus or Annular Region:
The region between two concentric circle of radii r1 &
r2 can be represented as
r1 < |z − z0| < r2
 Interior , Exterior and Boundary Points:
A point z0 is said to be interior point of a set S
whenever there is some neighborhood of z0 that
contains only points of S.
A point z1 is said to be exterior point of a set S
whenever there is no neighborhood of z1 that
contains only points of S.
A point z2 is said to be boundary point of a set S whenever neighborhood of z2 contains
both interior and exterior points as well.
𝑶 𝑿
𝒀
.
𝒓𝟏
𝒓𝟐
𝒛𝟎
𝑶 𝑿
𝒀
𝒛𝟎
𝝐
𝑶 𝑿
𝒀
𝒛𝟎
𝝐
.
.
𝒛𝟏
𝒛𝟐
𝑺
.𝒛𝟎

 Open Set: A set is open if it contains none of the boundary points.
 Closed Set: A set is said to be closed set if it contains all of the boundary points.
 Connected Set:
An open set S is connected if each pair of points z0 and z2 in
it can be joined by a polygonal line, consisting of finite no.
of line segments joined end to end that lies entirely in S.
 Domain and Region:
A set S is said to be domain if set S is open and connected. Note that any neighborhood is
a Domain. A domain together with some, none or all of its boundary points is called
region.
 Bounded Region:
A region is said to be bounded, if it can be enclosed in a circle of finite radius.
 Compact region:
A region is said to be compact if it is closed and bounded.
 FORMULA TO FIND SQUARE ROOT OF COMPLEX NUMBER:
 Let z = x + iy be a complex number. Formula for finding square root of z is as below,
√z = √x + iy = ± [√
|z| + x
2
+ i(sign of y )√
|z| − x
2
]
METHOD – 2: SQUARE ROOT OF COMPLEX NUMBER
C 1 Find √−8 + 6i.
𝐀𝐧𝐬𝐰𝐞𝐫: ± (𝟏 + 𝟑𝐢)
H 2 Find the roots of the equation z2
+ 2iz + 2 − 4i = 0.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝟏 + 𝐢 , −𝟏 − 𝟑𝐢
𝑶
𝑿
𝒀
𝒛𝟏
𝒛𝟎
𝒛𝟐

H 3 Solve the Equation of z2
− (5 + i)z + 8 + i = 0.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝟑 + 𝟐𝐢 , 𝟐 − 𝐢
C 4 Find the roots of the equation z2
+ (2i − 3)z + 5 − i = 0.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝟐 − 𝟑𝐢 , 𝟏 + 𝐢
 DE-MOIVRE’S THEOREM:
 Statement: (cos θ + i sin θ) n
= cos nθ + i sin nθ ; n ∈ ℚ [i. e. (eiθ
)
n
= einθ
]
 Remarks
(1). (cos θ − i sin θ) n
= cos nθ − i sin nθ
(2). (sin θ ± i cos θ) n
≠ sin nθ ± i cos nθ
(3). (cos θ ± i sin α) n
≠ cos nθ ± i sin nα
(4). (sin θ ± i cos θ) n
= [cos (
π
2
− θ) ± i sin (
π
2
− θ)]
n
= cos n (
π
2
− θ) ± i sin n (
π
2
− θ)
 PROCEDURE TO FINDING OUT NTH ROOT OF A COMPLEX NUMBER:
 Let, z = r(cos θ + i sin θ) ; r > 0
 For, n ∈ ℕ
z
1
n = r
1
n[cos(θ + 2kπ) + i sin(θ + 2kπ)]
1
n
= r
1
n [cos (
θ + 2kπ
n
) + i sin (
θ + 2kπ
n
)]
= r
1
n e
i (
θ+2kπ
n
)
; k = 0,1,2, … , n − 1
Where, r
1
n is positive nth root of r.
 By putting k = 0,1,2, … , n − 1 , we have distinct roots of z
1
n.
 For k = n, n + 1, n + 2, … , we have repeated roots of z
1
n.

METHOD – 3: nth ROOT OF COMPLEX NUMBER
C 1 State De Moivre’s formula. Find & plot 5th roots of unity in complex plane.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐞
𝐢(
𝟐𝐤𝛑
𝟓
)
; 𝐤 = 𝟎, 𝟏, 𝟐, 𝟑, 𝟒
H 2 State De Moivre’s formula. Find & plot 6th roots of unity in complex plane.
𝐢(
𝐤𝛑
𝟑
)
; 𝐤 = 𝟎, 𝟏, 𝟐, 𝟑, 𝟒, 𝟓
H 3 Find the fourth root of -1.
𝐢(
𝟐𝐤+𝟏
𝟒
)𝛑
; 𝐤 = 𝟎, 𝟏, 𝟐, 𝟑
H 4 Find and plot the square root of 4i.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = ±√𝟐(𝟏 + 𝐢)
H 5 State De Moivre’s formula. Find the cube root of i.
𝐢(
𝟒𝐤+𝟏
𝟔
)
; 𝐤 = 𝟎, 𝟏, 𝟐
C 6 State De Moivre’s formula. Find and plot all root of √8i
3
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝟐 𝐞
𝐢(
𝟒𝐤+𝟏
𝟔
)𝛑
; 𝐤 = 𝟎, 𝟏, 𝟐
H 7 Find and plot all the roots of (1 + i)
1
3.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝟐
𝟏
𝟔 𝐞
𝐢(
𝛑
𝟏𝟐
+
𝟐𝐤𝛑
𝟑
)
; 𝐤 = 𝟎, 𝟏, 𝟐
C 8 Show that if c is any nth
root of Unity other than Unity itself , then 1 + c +
c2
+ ⋯ + cn−1
= 0 OR
Prove that the n roots of unity are in Geometric Progression. Also show
that their sum is zero.
T 9 Find the roots common to equation z4
+ 1 = 0 and z6
− i = 0.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 =
𝟏 − 𝐢
√𝟐
,
−𝟏 + 𝐢
√𝟐
H 10
Find the value of (
1
2
+
√3
2
i)
3
4
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐫 ⋅ 𝐞𝐢𝛉
= 𝐞𝐢
𝛑
𝟒 =
𝟏
√𝟐
+
𝐢
√𝟐

U N I T - 1 » C o m p l e x F u n c t i o n a n d C o n f o r m a l M a p p i n g [ 1 0 ]
C 11 Find real and imaginary part of (−1 − i)7
+ (−1 + i)7
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑𝐞(𝐳) = −𝟏𝟔; 𝐈𝐦(𝐳) = 𝟎
C 12
Simplify (
1 + sin θ + i cos θ
1 + sin θ − i cos θ
)
n
Answer: cos n (
π
2
− θ) + i sin n (
π
2
− θ)
 TRIGONOMETRIC ( CIRCULAR ) FUNCTIONS OF A COMPLEX NUMBER:
 By Euler’s Formula, eiz
= cos z + i sin z ⟹ e−iz
= cos z − i sin z
eiz
+ e−iz
= 2 cos z ⟹ cos z =
eiz
+ e−iz
2
eiz
− e−iz
= 2i sin z ⟹ sin z =
eiz
− e−iz
2i
Hyperbolic Function Of a Complex Number Relation between Circular and Hyperbolic Functions
cosh z =
ez
+ e−z
2
sin ix = i sinh x sinh ix = i sin x
sinh z =
ez
− e−z
2
cos ix = cosh x cosh ix = cos x
tanh z =
ez
− e−z
ez + e−z
tan ix = i tanh x tanh ix = i tan x
Hyperbolic Identities Inverse Hyperbolic Functions
cosh2
x − sinh2
x = 1 sinh−1
z = log (z + √z2 + 1)
sech2
x + tanh2
x = 1 cosh−1
z = log (z + √z2 − 1)
coth2
x − cosech2
x = 1 tanh−1
z =
1
2
log (
1 + z
1 − z
)
 Result: Show that
𝐬𝐢𝐧𝐡−𝟏
𝐳 = 𝐥𝐨𝐠(𝐳 + √𝐳𝟐 + 𝟏) , 𝐜𝐨𝐬𝐡−𝟏
𝐳 = 𝐥𝐨𝐠(𝐳 + √𝐳𝟐 − 𝟏) & 𝐭𝐚𝐧𝐡−𝟏
𝐳 =
𝟏
𝟐
𝐥𝐨𝐠 (
𝟏+𝐳
𝟏−𝐳
).
Proof :

Let w = sinh−1
z ⟹ z = sinh w =
ew−e−w
2
⟹ z =
e2w
− 1
2ew
⟹ e2w
− 2zew
− 1 = 0
⟹ ew
=
2z ± √4z2 + 4
2
= z + √z2 + 1
= log (z + √z2 + 1)
⟹ 𝐬𝐢𝐧𝐡−𝟏
𝐳 = 𝐥𝐨𝐠 (𝐳 + √𝐳𝟐 + 𝟏) … (𝐀)
Let w = cosh−1
z ⟹ z = cosh w =
ew+e−w
2
z =
e2w
+ 1
2ew
⟹ e2w
− 2zew
+ 1 = 0
⟹ ew
=
2z ± √4z2 − 4
2
= z + √z2 − 1
⟹ w = log (z + √z2 − 1)
⟹ 𝐜𝐨𝐬𝐡−𝟏
𝐳 = 𝐥𝐨𝐠 (𝐳 + √𝐳𝟐 − 𝟏) … (𝐁)
Let w = tanh−1
z ⟹ z = tanh w =
sinh w
cosh w
=
ew−e−w
ew+e−w
⟹ z =
ew
− e−w
ew + e−w
 Taking componendo and dividendo, we get
⟹
1 + z
1 − z
=
(ew
+ e−w) + (ew
− e−w)
(ew + e−w) − (ew − e−w)
=
2ew
2e−w
= e2w
⟹ 2w = log (
1 + z
1 − z
)
⟹ 𝐰 =
𝟏
𝟐
𝐥𝐨𝐠 (
𝟏 + 𝐳
𝟏 − 𝐳
) ⟹ 𝐭𝐚𝐧𝐡−𝟏
𝐳 =
𝟏
𝟐
𝐥𝐨𝐠 (
𝟏 + 𝐳
𝟏 − 𝐳
) … (𝐂)
Eqn. (A), (B) & (C) are required equations.

METHOD – 4: TRIGONOMETRIC FUNCTION OF COMPLEX NUMBER
H 1 Prove that sinh−1
x = log(x + √x2 + 1).
C 2 Prove that sin−1
z = −i ln(iz + √1 − z2)
C 3 Prove that tan−1
z =
i
2
log
i+z
i−z
.
T 4
Prove that sech−1
x = log [
1+√1−x2
x
].
C 5 Show that cos(iz
̅) = cos(iz)
̅̅̅̅̅̅̅̅̅ for all z.
H 6 Show that sin(iz)
̅̅̅̅̅̅̅̅̅ = sin(iz
̅) if and only if z = nπi (n ∈ Z).
C 7 Expand cosh(z1 + z2).
H 8 Expand sinh(z1 + z2).
 LOGARITHM OF A COMPLEX NUMBER:
 Polar representation of complex number, z = reiθ
⟹ z = rei(θ+2kπ)
⟹ log z = lnr + i(θ + 2kπ)
⟹ log z = ln(√x2 + y2) + i (2kπ + tan−1
(
y
x
)) ; k = 0, ±1, ±2, … is called “GENERAL
VALUE OF LOGARITHM”.
 If k = 0,
⟹ Log z = ln(√x2 + y2) + i tan−1
(
y
x
) is called “PRINCIPAL VALUE OF LOGARITHM”.
 Note:
In Complex analysis,
(1). Log is used for Complex Single-Valued Function.
(2). log is used for Complex Multi-Valued Function.
(3). ln is used for Real Valued Function.

METHOD – 5: LOGARITHM OF COMPLEX NUMBER
C 1 Define: log(x + iy). Determine log(1 − i).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐧 √𝟐 + 𝐢 (𝟐𝐤𝛑 −
𝛑
𝟒
)
H 2 Determine Log(1 + i)
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐧 √𝟐 + 𝐢 (
𝛑
𝟒
)
T 3
Find the principal value of z = [
e
2
(−1 − i√3)]
3πi
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑𝛑𝐢 (𝟏 −
𝟐𝛑𝐢
𝟑
)
C 4 Find all root s of the Equation log z =
iπ
2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐢
H 5 Prove that ii
= e−(4n+1)
π
2.
C 6 Find the value of (−i)i
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐞−(𝟒𝐧−𝟏)
𝛑
𝟐
 FUNCTION OF A COMPLEX VARIABLE:
 If corresponding to each value of a complex variable z = x + iy in a given region R,
there correspond one or more values of another complex variable w = u + iv then, w is
called a function of the complex variable z and is denoted by
w = f(z) = u + iv
 Where, u and v are the real and imaginary part of w respectively and u and v are
function of real variable x and y.
i. e. w = f(z) = u(x, y) + i v(x, y)
 LIMIT OF COMPLEX FUNCTION:
 A function f(z) is said to have a limit l, if for each +ve number ϵ, there is +ve number δ
such that

i. e. |f(z) − l| < ε whenever 0 < |z − z0| < δ
Symbolically, lim
z→z0
f(z) = l
 CONTINUITY OF COMPLEX FUNCTION:
 A complex valued function f(z) is said to be continuous at a point z = z0 if
(1). f(z0) exists (2). lim
z→z0
f(z) exist (3). lim
z→z0
f(z) = f(z0)
 Remark
f(z) = u(x, y) + i v(x, y) is continuous iff u(x, y) and v(x, y) are continuous.
If any one of these three conditions of continuity is not satisfied then f(z) is discontinuous
at z = z0.
 DIFFERENTIABILITY OF COMPLEX FUNCTION:
 Let w = f(z) be a continuous function and z0 be a fixed point then f(z) is said to be
differentiable at z0 if lim
z→z0
f(z)−f(z0)
z−z0
exists, then the derivative of f(z) at z0 is denoted by
f′(z0) and is defined as
f′(z0) = lim
z→z0
f(z) − f(z0)
z − z0
 Another form:
f′(z0) = lim
h→0
f(z + h) − f(z)
h
 Remark:
(1). The rules of differentiation are same as in calculus of real variables.
(2). If function is differentiable, then it is continuous.
𝑶 𝑿
𝒀
𝒛𝟎
𝜹
𝝐
𝒍 𝒖
𝒗
𝒛 − 𝒑𝒍𝒂𝒏𝒆 𝒘 − 𝒑𝒍𝒂𝒏𝒆

METHOD – 6: DIFFERENTIBILITY OF COMPLEX FUNCTION
C 1
Prove lim
z→1
iz
3
=
i
3
by definition.
C 2 Evaluate lim
z→i
z−i
z2+1
.
𝐀𝐧𝐬𝐰𝐞𝐫: −
𝐢
𝟐
T 3 Does lim
z→0
z
|z|
exist. ?
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐍𝐨𝐭 𝐞𝐱𝐢𝐬𝐭
H 4 Using the definition of limit, show that if f(z) = iz in the open disk |z| < 1,
then lim
z→1
f(z) = i.
C 5 Show that the limit of the function does not exist
f(z) = {
Im(z)
|z|
; z ≠ 0
0 ; z = 0
.
H 6
Discuss the continuity of f(z) = {
z
̅
z
, z ≠ 0
0 , z = 0
at origin.
Answer: Limit does not exist ; Discontinuous
C 7
Discuss the continuity of f(z) = {
z
̅2
z
, z ≠ 0
0 , z = 0
at z = 0.
Answer: Limit exists ; Continuous
H 8
Discuss continuity of f(z) = {
Re(z2)
|z|2
; z ≠ 0
0 ; z = 0
at z = 0.
Answer: Limit does not exist ; Discontinuous
H 9 Find out and give reason weather f(z) is continuous at z = 0,if
f(z) = {
Re(z2)
|z|
, z ≠ 0
0 , z = 0
.
Answer: Limit exists ; Continuous
C 10 Find the derivative of
z−i
z+i
at i.
𝟏
𝟐𝐢

C 11 Discuss the differentiability of f(z) = x2
+ iy2
.
Answer: differentiable except the line 𝐲 = 𝐱
H 12 Show that f(z) = z Im(z) is differentiable only at z = 0 and f′(0) = 0.
C 13 Show that f(z) = |z|2
is continuous at each point in the plane, but not
differentiable.
T 14 Show that f(z) = z
̅ is nowhere differentiable.
 ANALYTIC FUNCTION:
 A function f(z) is said to be analytic at point z0 = x0 + iy0 if the function is
differentiable at point z0 as well as it is differentiable everywhere in the neighborhood
of z0.
 Examples :
(1). f(z) =
1
z
is analytic at each non-zero point in the finite complex plane.
(2). f(z) = |z|2
is not analytic at any non-zero point because it is not differentiable at any
non-zero complex number.
(3). f(z) = z
̅ is nowhere analytic because it is nowhere differentiable.
 Remark:
(1). Analytic functions are also known as regular or holomorphic functions.
(2). A function f is analytic everywhere in domain D iff it is analytic at each point of
domain D.
(3). A function f is analytic everywhere in domain D then f is entire function in D.
 CAUCHY-RIEMANN EQUATIONS[C-R EQUATION]:
 If u(x, y) and v(x, y) are real single-valued functions of x ad y such that
∂u
∂x
,
∂u
∂y
,
∂v
∂x
and
∂v
∂y
are continuous in the region R,then

∂u
∂x
=
∂v
∂y
&
∂u
∂y
= −
∂v
∂x
are known as Cauchy-Riemann Equations.

 NECESSARY AND SUFFICIENT CONDITIONS FOR 𝐟(𝐳) TO BE ANALYTIC:
 The necessary and sufficient conditions for the function f(z) = u(x, y) + iv(x, y) to be
analytic in a region R are
(1)
∂u
∂x
,
∂u
∂y
,
∂v
∂x
and
∂v
∂y
are continuous functions of x and y in the region R.
(2)
∂u
∂x
=
∂v
∂y
&
∂u
∂y
= −
∂v
∂x
i.e. Cauchy-Riemann equations are satisfied.
 Remark:
(1). C.R. equations are necessary condition for differentiability but not sufficient.
(2). If f(z) = u(x, y) + iv(x, y) is an analytic function, then u(x, y) and v(x, y) are
conjugate functions.
(3). If a function is differentiable ⟹ function satisfies C.R. equation.If a function does
not satisfies C.R. equation ⟹ function is not differentiable.
(4). If function is differentiable at point (x0, y0) then derivative at z0 is given by
(5). f′(z0) = ux(x0, y0) + ivx(x0, y0). (Cartesian form)
(6). f′(z0) = e−iθ
(ur(r, θ) + ivr(r, θ)). (Polar form)
 CAUCHY-RIEMANN EQUATIONS IN POLAR FORM:
∂u
∂r
=
1
r
∂v
∂θ
and
∂v
∂r
= −
1
r
∂u
∂θ
METHOD – 7: ANALITICITY OF COMPLEX NUMBER
H 1 State necessary and sufficient Condition for function to be analytic and
prove that necessary condition. OR Write necessary condition for
differentiability of f(z).
H 2 Show that the function f(z) = xy + iy is continuous everywhere but is not
analytic.
C 3 Check Whether f(z) = z
̅ is analytic or not.
Answer: Nowhere analytic

H 4 Check Whether f(z) = 2x + ixy2
is analytic or not at any point.
C 5 State the necessary condition for f(z) to be analytic. For what values of z
is the function f(z) = 3x2
+ iy2
analytic ?
Answer: Except the line 𝐲 = 𝟑𝐱 function is nowhere analytic
H 6 Separate in real and imaginary part of f(z) = ez
. Also prove that it is
analytic everywhere.
Answer: 𝐑𝐞𝐟(𝐳) = 𝐞𝐱
𝐜𝐨𝐬 𝐲 ; 𝐈𝐦𝐟(𝐳) = 𝐞𝐱
𝐬𝐢𝐧 𝐲
H 7 Check Whether f(z) = ez
̅
is analytic or not at any point.
C 8 State Cauchy Riemann Theorem. Write CR equation in polar form and
verify it for f(z) =
z
z
̅
in polar form.
C 9
Is f(z) = √re
iθ
2 analytic? (r > 0, −π < θ < π)
Answer: Yes
H 10 f(z) = zn
= rn
einθ
for integer n.Verify C-R equation & find its derivative.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐀𝐧𝐚𝐥𝐲𝐭𝐢𝐜 𝐞𝐯𝐞𝐫𝐲𝐰𝐡𝐞𝐫𝐞 & 𝐟′(𝐳) = 𝐧 𝐳𝐧−𝟏
H 11 What is an analytic function? Show that f(z) = z3
is analytic everywhere.
Answer: Analytic everywhere
H 12 Check Whether f(z) = z
5
2 is analytic or not.
Answer: Analytic everywhere
H 13 Check Whether f(z) = z
3
2 is analytic or not. Also find derivative of it.
Answer: Analytic everywhere & 𝐟′(𝐳) =
𝟑
𝟐
⋅ 𝐳
𝟏
𝟐
H 14 Check Whether the function f(z) = sin z is analytic or not.
If analytic find it’s derivative.
Answer: Analytic everywhere & 𝐟′(𝐳) = 𝐜𝐨𝐬 𝐳
T 15 Examine the analyticity of sinh z.
Answer: Everywhere analytic

H 16
Show that for the function f(z) = {
z
̅2
z
; z ≠ 0
0 ; z = 0
is not differentiable at z = 0
even though Cauchy Reimann equation are satisfied at z = 0.
H 17 Show that the function f(z) = √|xy| satisfies the C-R equation at the
origin but f′(0) fails to exist.
C 18 If w is any function of x and y with differential co efficient of 1st and 2nd
order then prove that, 4
∂2w
∂z ∂z
̅
=
∂2w
∂x2
+
∂2w
∂y2
. Also prove that,
∂2w
∂z ∂z
̅
= 0; when
w is analytic function.
C 19 If f(z) is regular function of z, then prove that
(
∂2
∂x2
+
∂2
∂y2
) |f(z)|2
= 4|f′
(z)|2
C 20 If f(z) is analytic function, prove that (
∂2
∂x2
+
∂2
∂y2
) log|f′(z)| = 0.
C 21 If f(z) is analytic function, prove that
(
∂2
∂x2
+
∂2
∂y2
) |Re(f(z))|
2
= 2|f′(z)|2
.
 HARMONIC FUNCTIONS:
 A real valued function ϕ(x, y) is said to be harmonic function in domain D if
∂2
ϕ
∂x2
+
∂2
ϕ
∂y2
= 0. (Laplace Equation)
 All second order partial derivative ϕxx, ϕxy, ϕyx, ϕyy are continuous.
Theorem: If f(z) = u + iv is analytic in domain D then u & v are harmonic function in D.
 HARMONIC CONJUGATE:
 Let, u(x, y) and v(x, y) are harmonic function and they satisfy C.R. equations in certain
domain D then v(x, y) is harmonic conjugate of u(x, y).
 THEOREM:
 If f(z) = u + iv is analytic in D iff v(x, y) is harmonic conjugate of u(x, y).

 REMARK:
 If f(z) = u + iv is analytic function then v(x, y) is harmonic conjugate of u(x, y) but
u(x, y) is not harmonic conjugate of v(x, y). −u(x, y) is harmonic conjugate of v(x, y).
 MILNE-THOMSON’S METHOD:
 This method determines the analytic function f(z) when either u or v is given.
 We know that z = x + iy and z
̅ = x − iy
∴ x =
z + z
̅
2
& y =
z − z
̅
2i
 Now, f(z) = u(x, y) + iv(x, y) = u (
z+z
̅
2
,
z−z
̅
2i
) + iv (
z+z
̅
2
,
z−z
̅
2i
)
 Putting z
̅ = z, we get f(z) = u(z, 0) + iv(z, 0)
Which is same as f(z) = u(x, y) + iv(x, y) if we replace x by z and y by 0.
 Now, f(z) = u + iv
⟹ f′(z) =
∂u
∂x
+ i
∂v
∂x
=
∂u
∂x
− i
∂u
∂y
(By C. R. equations)
 Replacing x by z and y by 0,we get
f′(z) = ux(z, 0) − i uy(z, 0)
 Integrating both the sides, with respect to z, we get
f(z) = ∫ ux(z, 0) dz − ∫ i uy(z, 0) dz + c.
METHOD – 8: TO FIND HARMONIC FUNCTION
C 1 Define: Harmonic Function.
Show that u = x sin x cosh y − y cos x sinh y is harmonic.
H 2 Define: Harmonic Function.
Show that u =
x
x2+y2
is harmonic function for ℝ2
− (0,0).
C 3 Define: Harmonic Function.
Show that u(x, y) = x2
− y2
is harmonic. Find the corresponding analytic
function f(z) = u(x, y) + iv(x, y).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐳) = 𝐳𝟐
+ 𝐜

H 4 Define: Harmonic Function.
Show that is u(x, y) = x2
− y2
+ x harmonic.Find the corresponding
analytic function f(z) = u(x, y) + iv(x, y).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐳) = 𝐳𝟐
+ 𝐳 + 𝐜
C 5 Show that the function u = x3
− 3xy2
is harmonic and find the
corresponding analytic function.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐳) = 𝐳𝟑
+ 𝐜
H 6 Show that u(x, y) = 2x − x3
+ 3xy2
is harmonic in some domain and find
a harmonic conjugate v(x, y).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐯(𝐱, 𝐲) = 𝟐𝐲 − 𝟑𝐱𝟐
𝐲 + 𝐲𝟑
+ 𝐜
H 7 Determine a and b such that u = ax3
+ bxy is harmonic and find
Conjugate harmonic.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚 = 𝟎 ; 𝐛 ∈ ℂ
H 8 Define Harmonic Function. Show that the function u(x, y) = ex
cos y is
harmonic. Determine its harmonic conjugate v(x, y) and the analytic
function f(z) = u + iv.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐯(𝐱, 𝐲) = 𝐞𝐱
𝐬𝐢𝐧 𝐲 ; 𝐟(𝐳) = 𝐞𝐱
𝐜𝐨𝐬 𝐲 + 𝐢 𝐞𝐱
𝐬𝐢𝐧 𝐲
C 9 Determine the analytic function whose real part is
ex(x cos y − y sin y).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐳) = 𝐳 𝐞𝐳
+ 𝐜
H 10 Determine the analytic function whose real part is
e2x(x cos 2y − y sin 2y).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐳) = 𝐳 𝐞𝟐𝐳
+ 𝐜
T 11 Show that u(x, y) = ex2−y2
cos(2xy) is harmonic everywhere. Also find a
conjugate harmonic for u(x, y).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐯(𝐱, 𝐲) = 𝐞𝐱𝟐−𝐲𝟐
𝐬𝐢𝐧(𝟐𝐱𝐲)
C 12 Find the analytic function f(z) = u + iv, if
u − v = ex(cos y − sin y)
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐳) = 𝐞𝐳
+ 𝐜

H 13 Find the all analytic function f(z) = u + iv, if
u − v = (x − y)(x2
+ 4xy + y2).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐳) = −𝐢 𝐳𝟑
+ 𝐜
T 14 Find the analytic function f(z) if u − v =
ey−cos x sin x
cos hy−cosx
with f (
π
2
) =
3−i
2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐳) = 𝐜𝐨𝐭
𝐳
𝟐
+
𝟏
𝟏+𝐢
 DEFINITIONS:
 Conformal Mapping: Suppose two curves c1 and c2 intersect at point P in Z-plane and
the corresponding two curves c1′ and c2’ at P′ in the W-plane.
 If the angle of intersection of the curves at P is same as the angle of intersection of the
curve P′ in both magnitude and sense, then the transformation is said to be Conformal.
Note: If function is analytic, then mapping is conformal
 Fixed Point (Invariant Point): Fixed points of mapping w = f(z) are points that are
mapped onto themselves are “kept fixed” under the mapping.
 Bilinear linear transformations having α and β as fixed point is given by
w =
γz − αβ
z − (α + β) + γ
; for various values of γ.
 Critical Point: The point where f′(z) = 0 are called Critical Point. At critical point,
mapping is not conformal.
 Ordinary Point: The point where f′(z) ≠ 0 is called Ordinary Point.
METHOD – 9: FIXED POINT, CRITICAL POINT, ORDINARY POINT
C 1 Find Fixed point of bilinear trans.
(I) w =
z
2 − z
(II) w =
(2 + i)z − 2
z + i
(III) w =
z − 1
z + 1
(S − 16)
(𝐈)𝛂 = 𝟎, 𝛃 = 𝟏 (𝐈𝐈)𝛂 = 𝟏 + 𝐢, 𝛃 = 𝟏 − 𝐢 (𝐈𝐈𝐈)𝛂 = 𝐢, 𝛃 = −𝐢

H 2 Find fixed point of w =
z+1
z
and verify your result.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝛂 =
𝟏
𝟐
+
√𝟓
𝟐
, 𝛃 =
𝟏
𝟐
−
√𝟓
𝟐
C 3 Define Critical point & find critical point of the w = z + z2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = −
𝟏
𝟐
C 4 What does conformal mapping mean? At what points is the mapping by
w = z2
+
1
z2
not conformal?
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = ±𝟏, ±𝐢]
H 5 Check whether w =
1
z
is conformal mapping or not.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐘𝐞𝐬
METHOD – 10: ELEMENTARY TRANSFORMATION
C 1 Find and sketch the image of the region |z| > 1 under the transformation
w = 4z.
𝐀𝐧𝐬𝐰𝐞𝐫: |𝐰| > 𝟒
H 2 Determine & sketch the image of |z| = 1 under the transformation w =
z + i.
𝐀𝐧𝐬𝐰𝐞𝐫: [|𝐰 − 𝐢| = 𝟏]
H 3 Show that the region in the z- plane given by x > 0 , 0 < y < 2 has the
image −1 < u < 1 , v > 0 in the w-plane under the transformation w =
iz + 1.
H 4 Find the image of infinite strip 0 ≤ x ≤ 1 under the transformation w =
iz + 1. Sketch the region in ω − plane.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎 ≤ 𝐯 ≤ 𝟏
C 5 Determine and sketch the image of region 0 ≤ x ≤ 1 , 0 ≤ y ≤ π under
the transformation w = ez
.
𝐀𝐧𝐬𝐰𝐞𝐫: − 𝟏 ≤ 𝐮 ≤ 𝐞 ,𝟎 ≤ 𝐯 ≤ 𝐞

C 6 Find & sketch (plot) the image of the region x ≥ 1 under the
transformation w =
1
z
.
𝐀𝐧𝐬𝐰𝐞𝐫: |𝐰 −
𝟏
𝟐
| ≤
𝟏
𝟐
H 7 Find the image of infinite strip
1
4
≤ y ≤
1
2
under trans. w =
1
z
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑𝐞𝐠𝐢𝐨𝐧 𝐛𝐞𝐭𝐰𝐞𝐞𝐧 𝐭𝐡𝐞 𝐜𝐢𝐫𝐜𝐥𝐞𝐬 |𝐰 + 𝟐𝐢| ≤ 𝟐 &|𝐰 + 𝐢| ≥ 𝟏
T 8 Find image of critical |z| = 1 under transformation
w = f(z) =
z−i
1−iz
& find fixed points.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐯 = 𝟎
C 9 Find the image in the w −plane of the circle |z − 3| = 2 in the
z − plane under the inversion mapping w =
1
z
.
𝐀𝐧𝐬𝐰𝐞𝐫: |𝐰 −
𝟑
𝟓
| =
𝟒
𝟓√𝟐
T 10 Find the image of |z − 3i| = 3 under the mapping w =
1
z
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐯 =
𝟏
𝟔
H 11 Explain translation, rotation and magnification transformation. Find the
image of the |z − 1| = 1 under transformation w =
1
z
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮 =
𝟏
𝟐
C 12 Find the image of region bounded by 1 ≤ r ≤ 2 &
π
6
≤ θ ≤
π
3
in the z-
Plane under the transformation w = z2
. Show the region graphically.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏 ≤ 𝐫′
≤ 𝟒 &
𝛑
𝟑
≤ 𝛉′
≤
𝟐𝛑
𝟑
C 13 Determine the points where w = z +
1
z
is not conformal mapping. Also
find image of circle |z| = 2 under the transformation w = z +
1
z
.
𝐮𝟐
𝟐𝟓
+
𝐯𝟐
𝟗
=
𝟏
𝟒
T 14 Determine the region of w-plane into which the region bounded byx =
1, y = 1, x + y = 1 is mapped by the transformation w = z2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮𝟐
= 𝟏 − 𝟐𝐯

 BILINEAR TRANSFORMATION / LINEAR FRACTIONAL / MOBIUS TRANSFORMATION:
 A transformation of the form w =
az+b
cz+d
; Where a,b,c,d are complex constants and
ad − bc ≠ 0 is called a Bilinear Transformation.
 DETERMINATION OF BILINEAR TRANSFORMATION:
 If w1, w2, w3are the respective images of distinct points z1, z2, z3 then
(w − w1)(w2 − w3)
(w − w3)(w2 − w1)
=
(z − z1)(z2 − z3)
(z − z3)(z2 − z1)
METHOD – 11: BILINEAR TRANSFORMATION
C 1 Determine bilinear transformation which maps point 0, ∞, i into ∞, 1,0.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐰 =
𝐳 − 𝐢
𝐳
H 2 Define Mobius transformation. Determine the Mobius transformation that
maps. z1 = 0, z2 = 1, z3 = ∞ on to w1 = −1, w2 = −i, w3 = 1
respectively.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐰 = − (
𝟏 + 𝐢𝐳
𝟏 − 𝐢𝐳
)
H 3 Determine bilinear transformation which maps point 0, i, 1 into i, −1, ∞.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐰 = 𝐢 (
𝐢 − 𝐳
𝐢 + 𝐳
)
H 4 Determine the Linear Fractional Transformation that maps
z1 = 0, z2 = 1, z3 = ∞ onto w1 = −1, w2 = −i, w3 = 1 respectively.
𝐳 − 𝐢
𝐳 + 𝐢
C 5 Find the bilinear transformation which maps z = 1, i, −1 into the points
w = 0,1, ∞.
𝐢 − 𝐢𝐳
𝐳 + 𝟏

T 6 Find the bilinear transformation which maps z = 2,1,0 into the points
w = 1,0, i.
𝟐𝐳 − 𝟐
𝐳 − 𝐢𝐳 + 𝟐𝐢
T 7 Find bilinear transformation, which maps the points 1, −1, ∞ onto the
points 1 + i, 1 − i, 1.Also find fixed point.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐰 = 𝟏 +
𝐢
𝐳
H 8 Find the bilinear transformation that maps the points
z1 = 1, z2 = i, z3 = −1 on to w1 = −1, w2 = 0, w3 = 1 respectively.
Find image of |z| < 1 under this transformation.
𝐳 − 𝐢
𝐢𝐳 − 𝟏
C 9 Define a Linear Fractional Transformation. Find the bilinear
transformation that maps the points z1 = −1, z2 = 0, z3 = 1 on to
w1 = −i,w2 = 1, w3 = i respectively. Also find w for z = ∞.
𝐢 − 𝐳
𝐢 + 𝐳
; 𝐰𝐡𝐞𝐧 𝐳 → ∞, 𝐰 = −𝟏
T 10 Find bilinear transformation, which maps the point z = 1, i, −1 on to the
point w = i, 0, −i .Hence find the image of |z| < 1.
𝟏 + 𝐢𝐳
𝟏 − 𝐢𝐳
H 11 Find the Bilinear transformation which maps z = 1, i, −1 into w = 2, i, −2.
𝟐𝐢 − 𝟔𝐳
𝐢𝐳 − 𝟑
C 12 Find the Bilinear transformation which maps z = 1, i, −1 into w = i, 0,1.
(𝟏 + 𝐢) − (𝟏 − 𝐢)𝐳
𝟐
⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆

U N I T - 2 » C o m p l e x I n t e g r a l , S e q u e n c e a n d S e r i e s [ 2 7 ]
UNIT-2 » COMPLEX INTEGRAL, SEQUENCE AND SERIES
 INTRODUCTION:
Integrals of complex valued function of a complex variable are defined on curves in the
complex plane, rather than on interval of real line.
 Continuous arc:
The set of points (x, y) defined by x = f(t) , y = g(t), with parameter t in the interval
(a, b), define a continuous arc provided f and g are continuous functions.
 Smooth arc:
If f and g are differentiable on arc a ≤ t ≤ b and non-zero on open interval a < t < b is
called smooth arc.
 Simple Curve/Simple arc/Jordan arc:
A curve which does not intersect with itself. i.e. if z(t1) ≠ z(t2) when t1 ≠ t2.
 Simple Closed Curve:
A simple curve C except for the fact z(b) = z(a) ; where a & b are end points of interval.
 Contour:
A contour or piecewise smooth arc, is an arc consisting of a finite number of smooth arcs
join end to end.
If only initial and final values are same, a contour is called Simple closed contour.
 Simply connected:
A Simply connected domain D is simple closed path in D encloses only points of D.
Examples: A open disk, ellipse or any simple closed curve.
 Multiply connected:
A domain that is not simply connected is called multiply connected.
 Examples: An annulus is multiply connected.
Fig. A Fig. B Fig. C

 LINE INTEGRAL IN COMPLEX PLANE:
 A line integral of a complex function f(z) along the curve C is denoted by ∫ f(z)
c
dz.
 Note that, if C is closed path, then line integral of f(z) is denoted by ∮ f(z) dz
C
.
∮ f(z) dz
C
is known as Contour integral.
 Properties of Line Integral:
(1). Linearity
∫ [k1 f(z) + k2 g(z)] dz
C
= k1 ∫ f(z) dz
C
+ k2 ∫ g(z) dz
C
(2). Reversing the sense of integration
∫ f(z) dz
b
a
= − ∫ f(z) dz
a
b
(3). Partition of Path
∫ f(z) dz
C
= ∫ f(z) dz
C1
+ ∫ f(z) dz
C2
; where c = c1 ∪ c2
METHOD – 1: LINE INTEGRAL
C 1 Evaluate ∫ z2
dz
2+i
0
along the line joining the points (0,0) and (2,1).
𝟐
𝟑
+
𝟏𝟏
𝟑
𝐢
T 2 Evaluate ∫ (x2
− iy2)dz
c
, along the parabola y = 2x2
from (1,2) to (2,8).
𝟓𝟏𝟏
𝟑
−
𝟒𝟗
𝟓
𝐢
Simply connected Triply Connected
Doubly Connected

T 3 Evaluate ∫ Re(z)dz
c
where C is the shortest path from (1 + i) to (3 + 2i).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒 + 𝟐𝐢
H 4 Evaluate ∫ z
̅
c
dz from z = 1 − i to z = 3 + 2i along the straight line.
𝟏𝟏
𝟐
+ 𝟓𝐢
H 5 Evaluate ∫ z2
dz
c
where C is line joins point (0,0) to (4,2).
𝟏𝟔
𝟑
+
𝟖𝟖
𝟑
𝐢
C 6 Evaluate ∫ z
̅
4+2i
0
dz along the curve z = t2
+ it.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏𝟎 −
𝟖
𝟑
𝐢
H 7 Find the value of ∫ (z
̅)2
2+i
0
dz , along the real axis from 0 to 2 and then
vertically from 2 to 2 + i .
𝟏𝟒 + 𝟏𝟏𝐢
𝟑
T 8 Evaluate ∫ (z
̅)2
2+i
0
dz , along the shortest path joining the end points.
𝟏𝟎 − 𝟓𝐢
𝟑
C 9 Evaluate ∫ (x − y + ix2)dz
c
,Where c is along the imaginary axis from z =
0 to z = 1, z = 1 to z = 1 + i & z = 1 + i to z = 0.
𝟑𝐢 − 𝟏
𝟔
H 10 Evaluate ∫ {(x + y)dx + x2
ydy}
c
along y = x2
having (0,0),(3,9) end
points.
𝟓𝟏𝟑
𝟐
H 11 Evaluate ∫ z
̅
c
dz, where C is along the sides of triangle having vertices z =
0,1, i.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐢

C 12 Evaluate ∫ z2
dz
c
,Where c is the path joining the points 1 + i and 2 + 4i
along (a) the parabola x2
= y (b) the curve x = t, y = t2
.
𝟖𝟔
𝟑
− 𝟔𝐢
H 13 Evaluate ∫ z2
dz
c
,Where c is taken along triangle in z-plane having
vertices z = ±i , −1 taken in counter clockwise sense.
𝟐𝐢
𝟑
H 14 Evaluate ∫ |z|2
dz
c
When c is the boundary of the square with vertices
(0,0), (1,0), (1,1), (0,1).
𝐀𝐧𝐬𝐰𝐞𝐫: − 𝟏 + 𝐢
C 15 Evaluate ∫ f(z)dz
c
. Where f(z) is defined by f(z) = {1 ; y<0
4y ; y>0
.
c is the arc from z = −1 − i to z = 1 + i along the curve y = x3
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐 + 𝟑𝐢
C 16 Find the value of integral ∫ z
̅
c
dz where c is the right-hand half z =
2eiθ
; (−
π
2
≤ θ ≤
π
2
) of the circle |z| = 2, from
z = −2i to z = 2i.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒𝛑𝐢
 MAXIMUM MODULUS THEOREM:
 If f(z) is analytic inside and on a simple closed curve C, then maximum value of |f(z)|
occurs on C, unless f(z) must be constant.
METHOD – 2: MAXIMUM MODULUS THEOREM
C 1 Find an upper bound for the absolute value of the integral ∫ ez
c
dz, where
c is the line segment joining the points (0,0) and (1,2√2).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟑𝐞

C 2
Without using integration, show that |∮
ez
z+1
dz
C
| ≤
8πe4
3
; C: |z| = 4.
H 3 Find an upper bound for the absolute value of the integral∫
dz
z2+1
c
, where c
is the arc of a circle |z| = 2 that lies in the first quadrant.
𝛑
𝟑
T 4 Find an upper bound for the absolute value of the integral ∫ z2
dz,
c
where c
is the straight line segment from 0 to 1 + i.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐√𝟐
 CAUCHY’S INTEGRAL THEOREM (CAUCHY GOURSAT’S THEOREM):
 If f(z) is an analytic function in a simply connected domain D and f′(z) is continuous at
each point within and on a simple closed curve C in D, then
∮ 𝐟(𝐳)𝐝𝐳 = 𝟎
𝐂
 LIOUVILLE’S THEOREM:
 If f(z) is an analytic and bounded function for all 𝐳 in the entire complex plane, then
𝐟(𝐳) is constant.
METHOD – 3: CAUCHY INTEGRAL THEOREM
C 1 Evaluate ∮ ez2
dz
C
, where C is any closed contour.
Justify your answer.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟎
H 2 If C is any simple closed contour, in either direction, then show that
∫ exp(z3
)
c
dz = 0.

C 3 Evaluate ∮ (z2
+ 3)dz
C
, where C is any closed contour.
Justify your answer.
H 4 Evaluate ∮ (z2
− 2z − 3)dz
C
, where C is the circle |z| = 2.
C 5 Evaluate ∫
dz
z2
c
, c is along a unit circle.
H 6 Evaluate ∮
z
z−3
dz
C
, where C is the unit circle |z| = 1.
T 7 Evaluate ∮
ez
z+i
dz
C
, where C is the unit circle |z − 1| = 1.
C 8 Evaluate ∮
z+4
z2+2z+5
dz
C
, where C is the circle |z + 1| = 1.
 CAUCHY’S INTEGRAL FORMULA:
 If f(z) is an analytic within and on a simple closed curve C and z0 is any point interior to
C, then
∮
f(z)
z−z0
dz = 2πi f(z0)
C
;the integration being taken counterclockwise.
 In general, ∮
f(z)
(z−z0)n+1
dz
C
=
2πi
n!
fn(z0)
METHOD – 4: CAUCHY INTEGRAL FORMULA
C 1 Evaluate ∮
z2−4z+4
(z+i)
C
dz , where C is |z| = 2.
𝐀𝐧𝐬𝐰𝐞𝐫: (−𝟖 + 𝟔𝐢)𝛑

T 2 Evaluate ∮
sin 3z
z+
π
2
C
dz , where C is the circle is |z| = 5.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝛑𝐢
C 3 Evaluate ∮
cosz
(z−
π
4
)
3
C
dz where c is the circle |z −
π
2
| = 1, using Cauchy
Integral formula.
𝐀𝐧𝐬𝐰𝐞𝐫: √𝟐𝛑𝐢
C 4 Evaluate ∮
e2z
(z−ln 1.5)3
C
dz, where C is the square with vertices
(1,0), (0,1), (−1,0), (0, −1).
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟗𝛑𝐢
H 5 Evaluate ∮
5z+7
z2+2z−3
C
dz , where C is |z − 2| = 2.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟔𝛑𝐢
C 6 Evaluate ∮
dz
z2+1
C
, where C is |z + i| = 1, counterclockwise.
𝐀𝐧𝐬𝐰𝐞𝐫: − 𝛑
H 7 Evaluate ∮
2z+6
z2+4
C
dz , where C is |z − i| = 2.
𝐀𝐧𝐬𝐰𝐞𝐫: (𝟑 + 𝟐𝐢)𝛑
H 8 Using Cauchy Integral formula, Evaluate ∫
3z2+2
(z−2)(z2+4)
dz
C
, where C is
|z − 2| = 2.
𝟕𝛑𝐢
𝟐
C 9 Evaluate ∮
1
(z−1)2(z−3)
C
dz , where C is |z| = 2.
𝛑
𝟐
𝐢
T 10 Evaluate ∮
z−1
(z+1)2(z−2)
C
dz , where C is |z − i| = 2.
𝟐𝛑𝐢
𝟗
H 11 Evaluate ∫
z
z2+1
dz
c
, where c is the circle (i) |z +
1
2
| = 2 (ii)|z + i| = 1.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝛑𝐢, 𝛑𝐢

C 12 Evaluate ∫
1+z2
1−z2
c
dz, where c is unit circle centred at (1) z = −1 (2) z = i.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝛑𝐢, 𝟎
H 13 Evaluate ∫
z2+1
z2−1
c
dz, if c is unit circle of unit radius with centre (1) z = 1
(2) z = −1.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝛑𝐢, − 𝟐𝛑𝐢
H 14 State Cauchy-Integral theorem.
Evaluate ∮ (
3
z−i
−
6
(z−i)2
)
C
dz , Where C: |z| = 2.
H 15
Evaluate ∫
dz
z3(z + 4)
c:|z|=2
.
𝛑𝐢
𝟑𝟐
H 16 Evaluate ∮
ez
z(1−z)3
C
dz , where, C is (a)|z| =
1
2
(b)|z − 1| =
1
2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝛑𝐢, −𝛑𝐢𝐞
 SEQUENCES:
 A sequence is obtained by assigning to each positive integer n, a number zn, called a
term of the sequence, and is written z1, z2, z3, … OR {z1, z2, z3, … } OR {zn}.
 CONVERGENT SEQUENCE:
 A sequence is called convergent, if lim
n→∞
zn = c.
 In other words, for every ϵ > 0, we can find an N such that |zn − c| < 𝜖 for all n > N.
 A Divergent sequence is one that does not converge.
 Example:
(1). {
in
n
} = {i,
−1
2
,
−i
3
,
1
4
, … } is convergent with limit 0.
(2). {in} = {i, −1, −i, 1, … } is divergent.

 THEOREM:
 A sequence 𝑧1, 𝑧2, 𝑧3, … , 𝑧𝑛, … of complex numbers zn = xn + iyn (n = 1,2,3, … )
converges to c = a + ib if and only if the sequence of the real parts 𝑥1, 𝑥2, 𝑥3, ….
converges to a and the sequence of the imaginary parts 𝑦1, 𝑦2, 𝑦3, …. converges to b.
 Example:
 The sequence {zn = (1 −
1
𝑛2
) + i (2 +
4
𝑛
)} converges to 1 + 2i as {an = 1 −
1
𝑛2
}
converges to 1 and {bn = 2 −
4
𝑛
} converges to 2.
METHOD – 5: CONVERGENCE OF A SEQUENCE
C 1 Is sequence {i2𝑛} convergent or divergent?
Answer: Divergent
C 2 Is sequence {1 +
2in
n+1
} convergent or divergent?
Answer: Convergent
H 3 Is sequence {
1+3n2i
2n2−n
} convergent or divergent?
Answer: Convergent
H 4 Is sequence {
1+2𝑛2
𝑛2 −
n−1
n
i} convergent or divergent?
Answer: Convergent
C 5
Is sequence {e
nπi
3 } convergent or divergent?
Answer: Divergent
T 6 Is sequence {−(i4n)} convergent or divergent?
Answer: Convergent
C 7 Is sequence {sin(ni)} convergent or divergent?
Answer: Divergent

 SERIES:
 Given a sequence z1, z2, z3, … , we may form the sequence of the sums
s1 = z1 , s2 = z1 + z2, s3 = z1 + z2 + z3, …
and in general
sn = z1 + z2 + z3 + ⋯ + zn
 Here, sn is called the nth partial sum of the series.
 CONVERGENT SERIES:
 A series is called convergent, if lim
n→∞
sn = s.
 We may write, s = ∑ zm
∞
m=1 . Here, s is called the sum of series.
 THEOREM:
 A series of complex numbers with zm = xm + iym converges to s = u + iv if and only if
𝑥1 + 𝑥2 + 𝑥3 + ⋯ converges to u and the sequence of the imaginary parts 𝑦1 + 𝑦2 +
𝑦3 + ⋯ converges to v.
 THEOREM:
 If a series z1 + z2 + z3 + ⋯ converges, then lim
m→∞
zm = 0.
 If lim
m→∞
zm ≠ 0, then series z1 + z2 + z3 + ⋯ diverges.
 Note: If lim
m→∞
zm = 0, then series z1 + z2 + z3 + ⋯ may not converges.
 ABSOLUTE CONVERGENCE:
 A series z1 + z2 + z3 + ⋯ is called absolutely convergent, if the series of the absolute
values of the terms ∑ |zm|
∞
m=1 = |z1| + |z2| + |z3| + ⋯ is convergent.
 If a series is absolutely convergent, it is convergent.
 CONDITIONALLY CONVERGENCE:
 If z1 + z2 + z3 + ⋯ converges but |z1| + |z2| + |z3| + ⋯ diverges, then the series z1 +
z2 + z3 + ⋯ is called, more precisely, conditionally convergent.

 COMPARISON TEST:
 If a series z1 + z2 + z3 + ⋯ is given and we can find a convergent series b1 + b2 + b3 +
⋯ with nonnegative real terms such that |z1| ≤ b1, |z2| ≤ b2, |z3| ≤ b3, … then the
given series converges, even absolutely.
 GEOMETRIC SERIES:
 The geometric series
∑ qm
∞
m=0
= 1 + q + q2
+ ⋯
(1). Diverges, if |q| ≥ 1.
(2). Converges, if |q| < 1. Also ∑ qm
∞
m=0 =
1
1−q
.
 RATIO TEST:
 If a series z1 + z2 + z3 + ⋯ is given with zn ≠ 0 is such that lim
𝑛→∞
|
𝑧𝑛+1
𝑧𝑛
| = L.
(1). If L < 1,the series converges absolutely.
(2). If L > 1,the series diverges.
(3). If L = 1,test fails.
 ROOT TEST:
 If a series z1 + z2 + z3 + ⋯ is given with zn ≠ 0 is such that lim
𝑛→∞
√|zn| = L.
(1). If L < 1,the series converges absolutely.
(2). If L > 1,the series diverges.
(3). If L = 1,test fails.
METHOD – 6: CONVERGENCE OF SERIES
C 1 Check series ∑
(− i)n
ln 𝑛
∞
n=0 is convergent or divergent.
Answer: Divergent

H 2 Check series ∑
in
n2−𝑖
∞
Answer: Convergent
H 3 Check series ∑
(π+π i)2n+1
(2n+1) !
∞
Answer: Convergent
C 4 Check series ∑
(−1)n(1+i)2n
(2n) !
∞
Answer: Convergent
 POWER SERIES
 A series of the form
∑ an(z − z0)n
= a0 + a1(z − z0) + a2(z − z0)2
+ ⋯ + an(z − z0)n
+ ⋯
∞
n=0
in which the coefficients an ∈ ℂ ; n = 0,1, … and z0 is a fixed point in the complex z-
plane is called a power series in powers of (z − z0) or about z0 or centered at z0.
 CONVERGENCE OF A POWER SERIES IN A DISK
 The series converges everywhere in a circular disk |z − z0| < R and diverges
everywhere in the disk |z − z0| > R.
 Here, R is called the radius of convergence and the circle |z − z0| = R is called the circle
of convergence.
 RADIUS OF CONVERGENCE
 Let ∑ an(z − z0)n
∞
n=0 be a power series. Radius of convergence R for power series is
defined as below
 𝐑 = 𝐥𝐢𝐦
𝐧→∞
|
𝐚𝐧
𝐚𝐧+𝟏
| or 𝐑 = 𝐥𝐢𝐦
𝐧→∞
|𝐚𝐧|−
𝟏
𝐧
METHOD – 7: REDIUS OF CONVERGENCE
H 1
Find the radius of convergence of ∑(n + 2i)n
zn
∞
n=0
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑 = 𝟎

C 2
Find the radius of convergence of ∑ (
6n + 1
2n + 5
)
2
(z − 2i)n
∞
n=1
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑 = 𝟏
H 3
Find the radius of convergence of ∑(6 + 8i)n
zn
∞
n=1
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑 =
𝟏
𝟔 + 𝟖𝐢
T 4
Find the radius of convergence of ∑ zn
∞
n=0
.
H 5
Find the radius of convergence of ∑
2n
n!
zn
∞
n=0
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑 = ∞
C 6
Find the radius of convergence of ∑
zn
n (log n)3
∞
n=2
.
C 7
Find the radius of convergence of ∑ (1 +
1
n
)
n2
zn
∞
n=1
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑 =
𝟏
𝐞
 FUNCTION GIVEN BY POWER SERIES
 If any power series ∑ an(z − z0)n
∞
n=0 has a non – zero radius of convergence R, its sum is
a function of z, say f(z). We write,
f(z) = ∑ an(z − z0)n
∞
n=0
= a0 + a1(z − z0) + a2(z − z0)2
+ ⋯ |z − z0| < R.
 Here, f(z) is a called function given by power series.
 Note: A function cannot represented by two different power series at same centre.

 TAYLOR’S SERIES:
 Let f(z) be analytic everywhere inside a circle C with centre at z0 and radius R. then at
each point Z inside C,we have
f(z) = f(z0) + f′(z0)(z − z0) +
f′′(z0)
2!
(z − z0)2
+ ⋯ +
fn(z0)
n!
(z − z0)n
+ ⋯
 MACLAURIN’S SERIES:
 If we take z0 = 0, in Taylor’s series reduces to
f(z) = f(0) + f′(0)z +
f′′(0)
2!
z2
+ ⋯ +
fn
(0)
n!
zn
+ ⋯
 UNIFORM CONVERGENCE
 A series
∑ fm(z)
∞
m=0
= f0(z) + f1(z) + f2(z) + ⋯
With sum s(z) is called uniformly convergent in a region G if for every ϵ > 0. We can
find an N = N(ϵ) not depending on z, such that |s(z) − sn(z)| < ϵ, ∀ n > N(ϵ), z ∈ G.
 Uniformity of convergence is thus a property that always refers to an infinite set in the
z-plane, that is, a set consisting of infinitely many points.
 UNIFORM CONVERGENCE OF POWER SERIES
 A power series
∑ an(z − z0)n
∞
n=0
with a non – zero radius of convergence R is uniformly convergent in every circular disk
|z − z0| ≤ r of radius r < R.
METHOD – 8: TAYLOR’S SERIES AND MACLAURIN’S SERIES
C 1 Derive the Taylor’s series representation in
1
1 − z
= ∑
(z − i)n
(1 − i)n+1
∞
n=0
; where |z − i| < √2 .

H 2
Obtain the Taylor’s series f(z) = sin z in power of (z −
π
4
).
H 3
Determine Taylor’s series expansion of
2z2+9z+5
(z−3)(z+2)2
with center at z0 = 1.
C 4
Find Maclaurin series representation of f(z) = sin z in the region |z| < ∞.
C 5
Find the series expression for f(z) = tan−1
z at z = 0.
H 6
Expand f(z) =
z−1
z+1
as a Taylor’s series about the point z = 0.
⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆

U N I T - 3 » L a u r e n t ’ s S e r i e s a n d R e s i d u e s [ 4 2 ]
UNIT-3 » LAURENT’S SERIES AND RESIDUES
 LAURENT’S SERIES:
If f(z) is analytic within and on the ring ( annulus ) shaped region R bounded by two
concentric circles C1 and C2 of radii R1 and R2 (R2 < R1) resp. having center at the point
z = z0, then for all z in R, f(z) is uniquely represented by a convergent Laurent’s series
given by
f(z) = ∑ an(z − z0)n
∞
n=0
+ ∑ a−n(z − z0)−n
∞
n=1
Where, an =
1
2πi
∫
f(t)
(t − z0 )n+1
dt
C1
& a−n =
1
2πi
∫
f(t)
(t − z0 )−n+1
dt
C2
Here, ∑ a−n(z − z0)−n
∞
n=1
is known Pricipal Part of Laurent′
s series.
 GEOMETRIC SERIES:
1
1 − z
= ∑ zn
∞
n=0
(|z| < 1)
1
1 + z
= ∑(−1)n
zn
∞
n=0
(|z| < 1)
METHOD – 1: LAURENT’S SERIES
H 1 Show that when 0 < |z − 1| < 2,
z
(z − 1)(z − 3)
=
−1
2(z − 1)
− 3 ∑
(z − 1)n
2n+2
∞
n=0
C 2 Find the Laurent’s expansion in power of z that represent
f(z) =
1
z(z−1)
for domain (a) 0 < |z| < 1 (b) 0 < |z − 1| < 1.
H 3 Find the Laurent’s expansion of f(z) =
7z−2
(z+1)z(z−2)
in the region
1 < |z + 1| < 3.
C 4 Expand f(z) =
1
(z+1)(z−2)
in the region (i)|z| < 1 (ii)1 < |z| < 2
(iii) |z| > 2.

H 5
Expand f(z) =
1
(z−1)(z−2)
in the region (i)|z| < 1 (ii)1 < |z| < 2.
H 6 Expand f(z) = −
1
(z−1)(z−2)
in the region (a) |z| < 1 (b) 1 < |z| < 2 (c)
|z| > 2
H 7 Expand f(z) =
1
(z+2)(z+4)
for the region (a) |z| < 2 (b) 2 < |z| < 4 (c)
|z| > 4
C 8 Expand
1
z(z2−3z+2)
in a Laurent series about z = 0 for the regions
(a) 0 < |z| < 1 (b) |z| > 2
H 9 Find the Laurent’s series expansion of f(z) =
1
6−z−z2
in (i) the domain
|z| < 2 (ii) the domain 2 < |z| < 3 (iii) the domain |z| > 3.
H 10
Expand f(z) =
1
(z+1)(z+3)
in Laurent’s series in the interval 1 < |z| < 3
T 11 Write the two Laurent series expansion in powers of z that represent the
function f(z) =
1
z2(1−z)
in certain domains, also specify domains.
H 12
Find all Taylor’s and Laurent’s series of f(z) =
−2z+3
z2−3z+2
with centre at 0.
 DEFINITION:
 Singular point: A point z0 is a singular point if a function f(z) is not analytic at z0 but is
analytic at some points of each neighborhood of z0.
 Isolated point: A singular point z0 of f(z) is said to be isolated point if there is a
neighborhood of z0 which contains no singular points of f(z) except z0.
 In other words, f(z) is analytic in some deleted neighborhood, 0 < |z − z0| < ε.
e.g. f(z) =
z2+1
(z−1)(z−2)
has two isolated point z = 1 & z = 2.
 Poles: If principal part of Laurent’s series has finite number of terms,
i. e. f(z) = ∑ an(z − z0)n
∞
n=0
+
b1
z − z0
+
b2
(z − z0)2
+. … . . +
bn
(z − z0)n

then the singularity z = z0 is said to be pole of order n.
 If b1 ≠ 0 and b2 = b3 = ⋯ … = bn = 0,then
f(z) = ∑ an(z − z0)n
∞
n=0
+
b1
z − z0
the singularity z = z0 is said to be pole of order 1 or a simple pole.
 TYPES OF SINGULARITIES:
 Removable singularity: If in Laurent’s series expansion, If the principal part is zero,
i. e. f(z) = ∑ an(z − z0)n
∞
n=0 + 0
then the singularity z = z0 is said to be removable singularity. (i.e. f(z) is not defined at
z = z0 but lim
z→z0
f(z) exists.) e.g. f(z) =
sin z
z
is undefined at z = 0 but lim
z→0
sin z
z
= 1.
 So, z = 0 is a removable singularity.
 Essential singularity: If in the Laurent’s series expansion, the principal part contains
an infinite number of terms, then the singularity z = z0 is said to be an essential
singularity.
e.g. f(z) = sin
1
z
has an essential singularity at z = 0, As sin
1
z
=
1
z
−
1
3!z3
+
1
5!z5
+ ⋯
 RESIDUE OF A FUNCTION:
 If f(z) has a pole at the point z = z0 then the coefficient b1 of the term
1
z−z0
in the
Laurent’s series expansion of f(z) at z = z0 is called the residue of f(z) at z = z0.
 Residue of f(z) at z = z0 is denoted by Res
z=z0
f(z).
 TECHNIQUE TO FIND RESIDUE:
(1). If f(z) has a simple pole at z = z0 ,then Res(f(z0)) = lim
z→z0
(z − z0)f(z).
(2). If f(z) =
P(z)
Q(z)
has a simple pole at z = z0 ,then Res(f(z0)) =
P(z0)
Q′(z0)
.
(3). If f(z) has a pole of order n at z = z0 ,then
Res(f(z0)) =
1
(n−1)!
lim
z→z0
d(n−1)
dz(n−1)
[(z − z0)n
f(z)]

METHOD – 2: RESIDUES OF FUNCTION
C 1 Classify the poles of f(z) =
1
z2−z6
.
Answer:
𝐳𝟎 𝟎 ±𝟏 ±𝐢
𝐧 𝟐 𝟏 𝟏
Type of pole Double Simple Simple
C 2 Find the pole of order of the point z = 0 for the function f(z) =
sinz
z4 .
Answer: pole of order 𝐧 = 𝟑
C 3 Define residue at simple pole and find the sum of residues of the function
f(z) =
sin z
z cos z
at its poles inside the circle |z| = 2.
Answer:
𝐳𝟎 𝟎 𝛑
𝟐
⁄ i − 𝛑
𝟐
⁄
𝐧 𝟏 𝟏 𝟏
𝐑𝐞𝐬𝐢𝐝𝐮𝐞 𝟎 − 𝟐
𝛑
⁄ 𝟐
𝛑
⁄
C 4 Find the residue at z = 0 of f(z) =
1−ez
z3
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐧 = 𝟎 ; 𝐑𝐞𝐬𝐢𝐝𝐮𝐞 = −
𝟏
𝟐
H 5 Find the residue at z = 0 of f(z) = z cos
1
z
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐑𝐞𝐬𝐢𝐝𝐮𝐞 = −
𝟏
𝟐
H 6 Show that the singular point of the function f(z) =
1−cosh z
z3
is a pole.
Determine the order m of that pole and corresponding residue.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐦 = 𝟏 ; 𝐑𝐞𝐬𝐢𝐝𝐮𝐞 = −
𝟏
𝟐

C 7 Determine residue at poles (
2z+1
z2−z−2
).
Pole Residue
𝟐 5/3
−𝟏 1/3
H 8 Define residue at simple pole. Find the residues at each of its poles of
f(z) =
z2−2z
(z+1)2(z2+4)
in the finite z −plane.
𝐏𝐨𝐥𝐞 𝟐𝐢 −𝟐𝐢 −𝟏
𝐑𝐞𝐬𝐢𝐝𝐮𝐞 𝟕 − 𝐢
𝟐𝟓
⁄ 𝟕 + 𝐢
𝟐𝟓
⁄ −
𝟏𝟒
𝟐𝟓
 CAUCHY’S RESIDUE THEOREM:
 If f(z) is analytic in a closed curve C except at a finite number of singular points with in
C, then
∫ f(z)
C
dz = 2πi (sum of the residue at the singular points)
METHOD – 3: CAUCHY’S RESIDUE THEOREM
C 1 State Cauchy’s residue theorem. Evaluate ∫
5z−2
z(z−1)
C
dz , where C is the
circle |z| = 2.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟏𝟎𝛑𝐢
H 2 Evaluate ∫
2z2+3
z(z+1)(z+2)
C
dz ; C: |z| = 1.6 using Residue theorem.
𝐀𝐧𝐬𝐰𝐞𝐫: −𝟕𝛑𝐢
H 3 Evaluate ∮
cosπz2
(z−1)(z−2)
C
dz , where C is the circle |z| = 3.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒𝛑𝐢

C 4 Using residue theorem, Evaluate ∮
ez+z
z3−z
C
dz, Where C: |z| =
π
2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝛑𝐢 (𝐞 − 𝟐 +
𝟏
𝐞
)
H 5 Using residue theorem, Evaluate ∮
z2 sin z
4z2−1
C
dz, C: |z| = 2.
𝛑𝐢
𝟒
𝐬𝐢𝐧
𝟏
𝟐
C 6 Determine the poles of the function f(z) =
z2
(z−1)2(z+2)
and residue at each
pole.Hence evaluate ∫ f(z)
C
dz, where C: |z| = 3.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝛑𝐢
C 7 Find the value of the integral ∫
2z2+2
(z−1)(z2+9)
C
dz taken counterclockwise
around the circle C: |z − 2| = 2.
𝟒
𝟓
𝛑𝐢
H 8 Evaluate ∮
z3−z2+z−1
z3+4z
dz where c is the circle |z| = 3 and |z| = 1 using
Cauchy’s residue theorem.
𝐀𝐧𝐬𝐰𝐞𝐫: − 𝟐𝛑𝐢
H 9 Evaluate ∮
sin πz2+cosπz2
(z−1)2(z−2)
C
dz , where C is the circle |z| = 3.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒𝛑𝐢(𝛑 + 𝟏)
C 10 Evaluate ∫
dz
(z2+1)2
C
, where C: |z + i| = 1.
𝛑
𝟐
H 11 Evaluate ∮ e
3
zdz
C
, where C is |z| = 1.
C 12 Use residues to evaluate the integrals of the function
exp(−z)
z2
around the
circle |z| = 3 in the positive sense.
𝐀𝐧𝐬𝐰𝐞𝐫: − 𝟐𝛑𝐢

H 13 ∮
e2z
(z+1)3
C
dz where c4x2
+ 9y2
= 16 using residue theorem.
T 14 Evaluate ∮ tan z dz
C
, where C is the circle |z| = 2.
𝐀𝐧𝐬𝐰𝐞𝐫: − 𝟒𝛑𝐢
T 15 Evaluate ∮
dz
sinh 2z
dz
C
, Where C: |z| = 2.
𝐀𝐧𝐬𝐰𝐞𝐫: − 𝛑𝐢
 INTEGRATION AROUND THE UNIT CIRCLE:
 An integral of the type ∫ F
2π
0
(cos θ , sin θ) dθ, where F(cos θ , sin θ) is a rational function
of cos θ and sin θ can be evaluated by taking z = eiθ
.
Now, cos θ =
eiθ
+ e−iθ
2
=
1
2
(z +
1
z
) =
1
2
(
z2
+ 1
z
)
sin θ =
eiθ
− e−iθ
2i
=
1
2i
(z −
1
z
) =
1
2i
(
z2
− 1
z
)
 Here, z = eiθ
⟹ dz = ieiθ
dθ ⇒ dθ =
dz
i z
 Now, the given integral takes the form ∫ f(z)dz
c
, where f(z) is a rational function of z
and c is the unit circle |z| = 1. This complex integral can be evaluated using the residue
theorem.
 INTEGRATION AROUND A SMALL SEMI-CIRCLE(IMPROPER INTEGRALS OF RATIONAL
FUNCTIONS):
 Let, f(x) =
P(x)
Q(x)
; where P(x) and Q(x) are polynomials of degree m and n respectively.
 If Q(x) ≠ 0 ; for all real x and n ≥ m + 2, the Cauchy principal value of the integral is
P. V. ∫
P(x)
Q(x)
dx
∞
−∞
= 2πi ∑ Res
z=zj
P(z)
Q(z)
k
j=1
Where, 𝐳𝐣 are the poles of
𝐏(𝐳)
𝐐(𝐳)
that lie in the upper half plane.

 IMPROPER INTEGRALS INVOLVING TRIGONOMETRIC FUNCTIONS:
P. V. ∫
P(x)
Q(x)
cos αx dx
∞
−∞
= −2π ∑ Im [Res
z=zj
P(z)
Q(z)
]
k
j=1
P. V. ∫
P(x)
Q(x)
sin αx dx
∞
−∞
= 2π ∑ Re [Res
z=zj
P(z)
Q(z)
]
k
j=1
METHOD – 4: CONTOUR INTEGRATION BY USING RESIDUE THEOREM
C 1 Using the residue theorem, evaluate ∫
dθ
5−3 sin θ
2π
0
.
𝛑
𝟐
H 2 Using the residue theorem, evaluate ∫
4 dθ
5+4 sin θ
2π
0
.
𝟖𝛑
𝟑
H 3 Using the residue theorem, evaluate ∫
dθ
(2+cosθ)2
2π
0
.
𝟒𝛑
𝟑√𝟑
T 4 Using the residue theorem, evaluate ∫
dθ
(2+cosθ)
2π
0
.
2π
√3
H 5 Evaluate ∫
dθ
3−2 cosθ+sinθ
2π
0
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝛑
C 6 Evaluate ∫
dθ
17−8 cosθ
π
0
, by integrating around a unit circle.
𝛑
𝟏𝟓
T 7 Evaluate the Cauchy’s principle value of ∫
dx
(x2+1)(x2+9)
∞
−∞
.
𝛑
𝟏𝟐
C 8 Use residues to evaluate ∫
x2 dx
(x2+1)(x2+4)
∞
0
.
𝛑
𝟑𝟔

H 9 Evaluate ∫
3x+2
x(x−4)(x2+9)
∞
−∞
dx
𝟏𝟒𝛑
𝟕𝟓
H 10
Using contour Integration show that ∫
dx
1+x4
=
π
2√2
∞
0
.
T 11 Let a > b > 0. Prove that
∫
cos x dx
(x2 + a2)(x2 + b2)
∞
−∞
=
π
a2 − b2
(
e−b
b
−
e−a
a
) .
C 12 Using the theory of residue, evaluate ∫
cos x
x2+1
∞
−∞
dx .
𝟐𝛑
𝐞
H 13 Evaluate P.V. ∫
x cos x
x2+1
∞
−∞
dx .
T 14 Evaluate ∫
x sin x
x2+9
dx
∞
0
using residue.
𝛑
𝟐𝐞𝟑
T 15 Evaluate ∫
x cosx
x2+9
dx
∞
0
using residue.
⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆

U N I T - 4 » F i r s t O r d e r P a r t i a l D i f f e r e n t i a l E q u a t i o n [ 5 1 ]
UNIT-4 » FIRST ORDER PARTIAL DIFFERENTIAL EQUATION
 INTRODUCTION:
 A partial differential equation is a mathematical equation involving two or more
independent variables, unknown function and its partial derivative with respect to
independent variables.
 Partial differential equations are used to formulate the problems containing functions of
several variables, such as propagation of heat or sound, fluid flow, electrodynamics etc.
 DEFINITION: PARTIAL DIFFERENTIAL EQUATION:
 An equation which involves function of two or more variables and partial derivatives of that
function then it is called Partial Differential Equation.
e.g.
∂y
∂x
+
∂y
∂t
= 0.
 DEFINITION: ORDER OF DIFFERENTIAL EQUATION:
 The order of highest derivative which appears in differential equation is “Order of D.E”.
e.g. (
∂y
∂x
)
2
+
∂y
∂t
+ 5y = 0 has order 1.
 DEFINITION: DEGREE OF DIFFERENTIAL EQUATION:
 When a D.E. is in a polynomial form of derivatives, the highest power of highest order
derivative occurring in D.E. is called a “Degree Of D.E.”.
e.g. (
∂y
∂x
)
2
+
∂y
∂t
+ 5y = 0 has degree 2.
 NOTATION:
 Suppose z = f(x, y). For that , we shall use
∂z
∂x
= p ,
∂z
∂y
= q ,
∂2z
∂x2
= r,
∂2z
∂x ∂y
= s,
∂2z
∂y2
= t.
 FORMATION OF PARTIAL DIFFERENTIAL EQUATION:
 By Eliminating Arbitrary Constants
o Consider the function f(x, y, z, a, b) = 0. Where, a & b are independent arbitrary
constants.
o Step 1: f(x, y, z, a, b) = 0. ……(1)

o Step 2: fx(x, y, z, a, b) = 0. ……(2) and fy(x, y, z, a, b) = 0. ……(3)
o Step 3: Eliminate a & b from eq. (1), eq. (2) & eq. (3).
o We get partial differential equation of the form F(x, y, z, p, q) = 0
 By Eliminating Arbitrary Functions
 Type 1: Consider, the function (u, v) = 0 ; u and v are functions of x and y
o Step 1: Let, u = F(v).
o Step 2: Find ux & uy.
o Step 3: Eliminate the function F from ux & uy.
o Note: In such case, for elimination of function, substitution method is used.
 Type 2: Consider, the function z = f(x, y)
o Step 1: Find zx & zy.
o Step 2: Eliminate the function f from zx & zy.
o Note: In such case, for elimination of function, division of zx & zy is used.
METHOD – 1: EXAMPLE ON FORMATION OF PARTIAL DIFFERENTIAL EQUATION
H 1 Form the partial differential equation z = ax + by + ct .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐩𝐱 + 𝐪𝐲 + 𝐭
𝛛𝐳
𝝏𝒕
C 2 Form the partial differential equation z = (x − 2)2
+ (y − 3)2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒𝐳 = 𝐩𝟐
+ 𝐪𝟐
C 3 Form the partial differential equation for the equation
(x − a)(y − b) − z2
= x2
+ y2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟒(𝐱 + 𝐩𝐳)(𝐲 + 𝐪𝐳)−𝐳𝟐
= 𝐱𝟐
+ 𝐲𝟐
H 4 Eliminate the function f from the relation f(xy + z2
, x + y + z) = 0.
𝐩 + 𝟏
𝐪 + 𝟏
=
𝐲 + 𝟐𝐳𝐩
𝐱 + 𝟐𝐳𝐪

C 5 Form the partial differential equation of f(x + y + z, x2
+y2
+ z2) = 0.
𝐩 + 𝟏
𝐪 + 𝟏
=
𝐱 + 𝐳𝐩
𝐲 + 𝐳𝐪
T 6 From a partial differential equation by eliminating the arbitrary function
∅ from ∅(x + y + z, x2
+ y2
− z2) = 0.
𝐀𝐧𝐬𝐰𝐞𝐫: (𝐲 + 𝐳)𝐩 − (𝐱 + 𝐳)𝐪 = 𝐱 − 𝐲
H 7 Form the partial differential equation f(x2
− y2
, xyz) = 0.
𝐲𝐳 + 𝐱𝐲𝐩
𝐱𝐳 + 𝐱𝐲𝐪
= −
𝐱
𝐲
C 8 Form the partial differential equations by eliminating the arbitrary
function from f(x2
+ y2
, z − xy) = 0.
𝐱
𝐲
=
𝐩 − 𝐲
𝐪 − 𝐱
C 9 Form partial differential equation by eliminating the arbitrary function
from xyz = Ф(x + y + z).
𝐩 + 𝟏
𝐪 + 𝟏
=
𝐲𝐳 + 𝐱𝐲𝐩
𝐱𝐳 + 𝐱𝐲𝐪
H 10 Form the partial differential equation of z = f (
𝑥
𝑦
).
𝐩
𝐪
= −
𝐲
𝐱
H 11 Form the partial differential equation by eliminating the arbitrary
function from z = f(x2
− y2).
𝐩
𝐪
= −
𝐱
𝐲
T 12 Form the partial differential equation of z = xy + f(x2
+ y2).
𝐩 − 𝐲
𝐪 − 𝐱
=
𝐱
𝐲
C 13 Form partial differential equation of z = f(x + iy) + g(x − iy).
𝛛𝟐
𝐳
𝛛𝐱𝟐
+
𝛛𝟐
𝐳
𝛛𝐲𝟐
= 𝟎
 LAGRANGE’S DIFFERENTIAL EQUATION:
 A partial differential equation of the form Pp + Qq = R where P, Q and R are functions of
x, y, z, or constant is called lagrange linear equation of the first order.

 METHOD FOR OBTAINING GENERAL SOLUTION OF 𝐏𝐩 + 𝐐𝐪 = 𝐑:
 Step-1: From the A.E.
dx
P
=
dy
Q
=
dz
R
.
 Step-2: Solve this A.E. by the method of grouping or by the method of multiples or both to
get two independent solution u(x, y, z) = c1 and v(x, y, z) = c2.
 Step-3: The form F(u, v) = 0 or u = f(v) & v = f(u) is the general solution Pp + Qq = R .
 FOLLOWING TWO METHODS WILL BE USED TO SOLVE LANGRAGE’S LINEAR EQUATION
 Grouping Method
o This method is applicable only if the third variable z is absent in
dx
P
=
dy
Q
or it is
possible to eliminate z from
dx
P
=
dy
Q
.
o Similarly, if the variable x is absent in last two fractions or it is possible to eliminate
x from last two fractions
dy
Q
=
dz
R
, then we can apply grouping method.
 Multipliers Method
o In this method, we require two sets of multiplier l, m, n and l′
, m′
, n′
.
o By appropriate selection multiplier l, m, n (either constants or functions of x, y, z)
we may write
dx
P
=
dy
Q
=
dz
R
=
ldx + mdy + ndz
lP + mQ + nR
Such that, lP + mQ + nR = 0 .
o This implies ldx + mdy + ndz = 0
o Solving it we get u(x, y, z) = c1 … (1)
o Again we may find another set of multipliers l′
, m′
, n′
. So that, l′
P + m′
Q + n′
R = 0
o This gives, l′
dx + m′
dy + n′
dz = 0
o Solving it we get v(x, y, z) = c2 … (2)
o From (1) and (2), we get the general solution as F(u, v) = 0.

METHOD – 2: EXAMPLE ON LAGRANGE’S DIFFERENTIAL EQUATION
H 1 Solve x2
p + y2
q = z2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟 (
𝟏
𝐲
−
𝟏
𝐱
,
𝟏
𝐳
−
𝟏
𝐲
) = 𝟎
H 2 Solve y2
p − xyq = x(z − 2y) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱𝟐
+ 𝐲𝟐
, 𝐲𝐳 − 𝐲𝟐) = 𝟎
H 3 Solve xp + yq = 3z.
𝐱
𝐲
,
𝐲𝟑
𝐳
) = 𝟎
H 4 Find the general solution to the P.D.E. xp + yq = x − y .
𝐱
𝐲
, 𝐱 − 𝐲 − 𝐳) = 𝟎
C 5 Solve (z − y)p + (x − z)q = y − x .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱 + 𝐲 + 𝐳, 𝐱𝟐
+𝐲𝟐
+ 𝐳𝟐) = 𝟎
H 6 Solve x(y − z)p + y(z − x)q = z(x − y) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟(𝐱 + 𝐲 + 𝐳, 𝐱𝐲𝐳) = 𝟎
C 7 Solve (x2
− y2
− z2)p + 2xyq = 2xz .
𝐲
𝐳
,
𝐱𝟐
+ 𝐲𝟐
+ 𝐳𝟐
𝐳
) = 𝟎
T 8 Solve (y + z)p + (x + z)q = x + y .
𝐱 − 𝐲
𝐲 − 𝐳
, (𝐱 + 𝐲 + 𝐳)(𝐱 − 𝐲)𝟐
) = 𝟎
T 9 Solve x2(y − z)p + y2(z − x)q = z2(x − y) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐟 (𝐱𝐲𝐳,
𝟏
𝐱
+
𝟏
𝐲
+
𝟏
𝐳
) = 𝟎
C 10 Solve (x2
− yz)p + (y2
− zx)q = z2
− xy .
𝐱 − 𝐲
𝐲 − 𝐳
,
𝐲 − 𝐳
𝐳 − 𝐱
) = 𝟎
 NON LINEAR PARTIAL DIFFERENTIAL EQUATION OF FIRST ORDER:
 A partial differential equation in which p & q occur in more than one order is known as Non
Linear Partial Differential Equation.

 Type 1: Equation Of the form f(p, q) = 0.
o Step 1: Substitute p = a & q = b.
o Step 2: Convert b = g(a).
o Step 3: Complete Solution : z = ax + by + c ⟹ z = ax + g(a)y + c
 Type 2: Equation Of the form f(x, p) = g(y, q).
o Step 1: f(x, p) = g(y, q) = a
o Step 2: Solving equations for p & q. Assume p = F(x) & q = G(y).
o Step 3: Complete Solution : z = ∫ F(x) dx + ∫ G(y) dy + b.
 Type 3: Equation Of the form z = px + qy + f(p, q) (Clairaut’s form. ) W-15
o Step 1: Substitute p = a & q = b.
o Step 2: Complete Solution : z = ax + by + f(a, b).
 Type 4: Equation Of the form f(z, p, q) = 0.
o Step 1: Assume q = ap
o Step 2: Solve the Equation in dz = p dx + q dy
METHOD – 3: EXAMPLE ON NON-LINEAR PDE
C 1 Solve p2
+ q2
= 1 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐚𝐱 ± (√𝟏 − 𝐚𝟐) 𝐲 + 𝐜
H 2 Solve √p + √q = 1 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐚𝐱 + (𝟏 − √𝐚)
𝟐
𝐲 + 𝐜
H 3 Find the complete integral of q = pq + p2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐚𝐱 +
𝐚𝟐
𝟏 − 𝐚
𝐲 + 𝐜 ; 𝐚 ≠ 𝟏
C 4 Solve p2
+q2
= npq .
𝐧𝐚 ± 𝐚√𝐧𝟐 − 𝟒
𝟐
𝐲 + 𝐜
T 5 Find the complete integral of p2 = q + x .
𝟐
𝟑
(𝐱 + 𝐚)
𝟑
𝟐 + 𝐚𝐲 + 𝐛

C 6 Solve p2
+ q2
= x + y .
𝟐
𝟑
(𝐚 + 𝐱)
𝟑
𝟐 +
𝟐
𝟑
(𝐲 − 𝐚)
𝟑
𝟐 + 𝐛
T 7 Solve p2
− q2
= x − y .
𝟐
𝟑
(𝐚 + 𝐱)
𝟑
𝟐 +
𝟐
𝟑
(𝐚 + 𝐲)
𝟑
𝟐 + 𝐛
H 8 Solve p − x2
= q + y2
.
𝐱𝟑
𝟑
+ 𝐚𝐲 −
𝐲𝟑
𝟑
+ 𝐛
C 9 Solve z = px + qy + p2
q2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐚𝐱 + 𝐛𝐲 + 𝐚𝟐
𝐛𝟐
H 10 Solve qz = p(1 + q) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠(𝐚𝐳 − 𝟏) = 𝐱 + 𝐚𝐲 + 𝐛
C 11 Solve pq = 4z .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚𝐳 = (𝐱 + 𝐚𝐲 + 𝐛)𝟐
 CHARPIT’S METHOD:
 Consider, f(x, y, z, p, q) = 0.
o Step 1: Find value of p & q by using the relation
dx
∂f
∂p
=
dy
∂f
∂q
=
dz
p
∂f
∂p
+ q
∂f
∂q
=
dp
− (
∂f
∂x
+ p
∂f
∂z
)
=
dq
− (
∂f
∂y
+ q
∂f
∂z
)
( lagrange − Charpit eqn
)
o Step 2: Find value of p & q.
o Step 3: Complete Solution : z = ∫ p dx + ∫ q dy + c.
METHOD – 4: EXAMPLE ON NON-LINEAR PDE
C 1 Using Charpit’s method solve z = px + qy + p2
+ q2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐚𝐱 + 𝐛𝐲 + 𝐚𝟐
+ 𝐛𝟐
C 2 Solve by charpit’s method px + qy = pq .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝟐𝐚𝐳 = (𝐚𝐱 + 𝐲)𝟐
+ 𝐛

H 3 Solve by charpit’s method (p2
+ q2)y = qz .
𝐀𝐧𝐬𝐰𝐞𝐫: √𝐳𝟐 − 𝐚𝟐𝐲𝟐 = 𝐚𝐱 + 𝐛
T 4 Solve py = 2 yx + log q .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐚𝐳 = 𝐚𝐱𝟐
+ 𝐚𝟐
𝐱 + 𝐞𝐚 𝐲
+ 𝐚𝐛
C 5 Solve z2
= p q x y .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐛 𝐱𝐚
𝐲
𝟏
𝐚
C 6 Solve x2
p2
+ y2
q2
= z2
.
𝐀𝐧𝐬𝐰𝐞𝐫: √𝐚𝟐 + 𝟏 𝐥𝐨𝐠 𝒛 = 𝐚 𝐥𝐨𝐠 𝐱 + 𝐥𝐨𝐠 𝐲 + 𝐛
C 7 Solve (p2
+ q2)y = q z.
𝐀𝐧𝐬𝐰𝐞𝐫: √𝐳𝟐 − 𝐚𝟐𝐲𝟐 = 𝐚 𝐱 + 𝐛
H 8 Solve by charpit’s method p = (z + qy)2
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳𝐲 = 𝐚𝐱 + 𝐳√𝐚√𝐲 + 𝐛
H 9 Solve by charpit’s method z = p2
x − q2
y .
𝐀𝐧𝐬𝐰𝐞𝐫: √𝟏 + 𝐚 √𝐳 = √𝐚 √𝐱 + √𝐲 + 𝐛
T 10 Solve by charpit’s method p x y + p q + q y = y z.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐥𝐨𝐠(𝐳 − 𝐚𝐱) = 𝐲 − 𝐚 𝐥𝐨𝐠(𝐚 + 𝐲) + 𝐛
⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆ ⋆

U N I T - 5 » H i g h e r o r d e r P a r t i a l D i f f e r e n t i a l E q u a t i o n [ 5 9 ]
UNIT-5 » HIGHER ORDER PARTIAL DIFFERENTIAL EQUATION
 LINEAR PDE WITH CONSTANT CO-EFFICIENT:
 The nth order linear partial differential equation with constant co-efficient is
a0
∂n
z
∂xn
+ a1
∂n
z
∂xn−1 ∂y
+ ⋯ an
∂n
z
∂yn
= F(x, y) … … … (A)
Where, a0, a1, … , an are constants.
 NOTATIONS:
 Replacing
∂
∂x
= D and
∂
∂y
= D′
in Eq. (A) , it can be written in operator form as below,
a0Dn
z + a1Dn−1
D′
z + ⋯ + anD′n
z = F(x, y) 𝐎𝐑 [f(D, D′)]z = F(x, y)
 AUXILIARY EQUATION:
 The auxiliary equation for nth order PDE a0Dn
z + a1Dn−1
D′
z + ⋯ + anD′n
z = F(x, y)
is derived by replacing D by m , D’ by 1 and F(x, y) by 0.
 COMPLEMENTARY FUNCTION (C.F.--𝐳𝐜):
 A general solution of [f(D, D′)]z = 0 is called complementary function of [f(D, D′)]z = F(x, y).
 PARTICULAR INTEGRAL (P.I.--𝐳𝐩):
 A particular integral of [f(D, D′)]z = F(x, y) is P. I. =
1
f(D,D′)
F(x, y).
 GENERAL SOLUTION OF PDE:
 G. S. = C. F. +P. I = zc + zp
 METHOD FOR FINDING C.F. OF PARTIAL DIFFERENTIAL EQUATION:
 Consider, a0Dn
z + a1Dn−1
D′
z + ⋯ + anD′n
z = F(x, y)
 The Auxiliary equation is a0mn
z + a1mn−1
z + ⋯ + anz = 0.
 Let m1, m2, … be the roots of auxiliary equation.
Case Nature of the “n” roots General Solutions
1. m1 ≠ m2 ≠ m3 ≠ m4 ≠ ⋯ z = f1(y + m1x) + f2(y + m2x) + f3(y + m3x) + ⋯

2.
m1 = m2 = m
m3 ≠ m4 ≠ ⋯
z = f1(y + mx) + x f2(y + mx) + f3(y + m3x) + ⋯
3.
m1 = m2 = m3 = m
m4 ≠ m5, …
z = f1(y + mx) + x f2(y + mx)
+x2
f3(y + mx) + f4(y + m4x) + ⋯
METHOD – 1: EXAMPLE ON SOLUTION OF HIGHER ORDERED PDE
C 1
Solve
∂2
z
∂x2
−
∂2
z
∂x ∂y
− 6
∂2
z
∂y2
= 0 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏(𝐲 − 𝟐𝐱) + 𝐟𝟐(𝐲 + 𝟑𝐱)
C 2
Solve
∂2
u
∂x2
+
∂2
u
∂y2
= 0 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮 = 𝐟𝟏(𝐲 + 𝐢𝐱) + 𝐟𝟐(𝐲 − 𝐢𝐱)
H 3 Solve 2 r + 5 s + 2 t = 0.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏 (𝐲 −
𝟏
𝟐
𝐱) + 𝐟𝟐(𝐲 − 𝟐𝐱)
C 4
Solve
∂3
z
∂x3
+ 3
∂3
z
∂x2 ∂y
− 4
∂3
z
∂y3
= 0 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏(𝐲 − 𝟐𝐱) + 𝐱 𝐟𝟐(𝐲 − 𝟐𝐱) + 𝐟𝟑(𝐲 + 𝐱)
H 5
Solve
∂3
z
∂x3
− 3
∂3
z
∂x2 ∂y
+ 2
∂3
z
∂y3
= 0 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏(𝐲 + 𝐱) + 𝐟𝟐[𝐲 + (𝟏 + √𝟑)𝐱] + 𝐟𝟑[𝐲 + (𝟏 − √𝟑)𝐱]
 METHOD FOR FINDING PARTICULAR INTEGRAL:
 For partial differential equation the value of Particular integral can be find by following
methods.
(1) General Method
(2) Short-cut Method
 GENERAL METHOD
o Consider the partial differential equation f(D, D′)z = F(x, y)
o Particular integral P. I. =
1
f(D,D′)
F(x, y)

o Suppose, f(x, y) is factorized into n linear factors.
P. I. =
1
f(D, D′)
F(x, y) =
1
(D − m1D′)(D − m2D′) … (D − mnD′)
F(x, y)
Which can be evaluated by
1
D − mD′
F(x, y) = ∫ F(x, c − mx)dx
o Where, c is replaced by y + mx after integration.
 SHORTCUT METHOD
 Case-1 F(x, y) = eax+by
P. I. =
1
f(D, D′)
eax+by
=
1
f(a, b)
eax+by
, if f(a, b) ≠ 0
 If f(a, b) = 0 then m =
a
b
is a root of auxiliary equation repeated r times.
f(D, D′) = (D −
a
b
D′
)
r
g(D, D′
)
P. I. =
1
(D −
a
b
D′)
r
g(D, D′)
eax+by
=
xr
r!
1
g(a, b)
eax+by
, g(a, b) ≠ 0
 Case-2 F(x, y) = sin(ax + by)
P. I. =
1
f(D2, DD′, D′2
)
sin(ax + by) =
1
f(−a2, −ab, −b2)
sin(ax + by)
Where, f(−a2
, −ab, −b2
) ≠ 0
 If f(−a2
, −ab, −b2) = 0, then use general method for finding P.I.
 Case-3 F(x, y) = cos(ax + by)
P. I. =
1
f(D2, DD′, D′2
)
cos(ax + by) =
1
f(−a2, −ab, −b2)
cos(ax + by)
Where, f(−a2
, −ab, −b2
) ≠ 0
 If f(−a2
, −ab, −b2) = 0, then use general method for finding P.I.
 Case-4 F(x, y) = xm
yn
P. I. =
1
f(D, D′)
xm
yn
= [f(D, D′
)]−1
xm
yn

 Expand [f(D, D′
)]−1
by using binomial expansion according to the following rules:
o If n < m, expand in power of
D′
D
.
o If m < n, expand in power of
D
D′
.
 Case-5 f(x, y) = eax+by
V(x, y)
P. I. =
1
f(D, D′)
eax+by
V(x, y) = eax+by
1
f(D + a, D′ + b)
V(x, y)
METHOD – 2: EXAMPLE ON SOLUTION OF HIGHER ORDERED PDE
C 1
Solve
∂z
∂x
+
∂z
∂y
= 2 ex+y
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏(𝐲 − 𝐱) +
𝟑
𝟐
𝐞𝐱+𝐲
C 2
Solve
∂2
z
∂x2
− 2
∂2
z
∂x ∂y
+
∂2
z
∂y2
= ex+4 y
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏(𝐲 + 𝐱) + 𝐱 𝐟𝟐(𝐲 + 𝐱) +
𝟏
𝟗
𝐞𝐱+𝟒 𝐲
H 3
Solve
∂3
z
∂x3
− 2
∂3
z
∂x2 ∂y
= 2e2x
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝛟𝟏(𝐲) + 𝐱𝛟𝟐(𝐲) + 𝛟𝟑(𝐲 + 𝟐𝐱) +
𝟏
𝟒
𝐞𝟐𝐱
H 4 Find complete solution
∂3z
∂x3
− 3
∂3z
∂2x ∂y
+ 4
∂3z
∂y3
= ex+2y
𝐀𝐧𝐬𝐰𝐞𝐫: 𝛟𝟏(𝐲 − 𝐱) + 𝛟𝟐(𝐲 + 𝟐𝐱) + 𝐱𝛟𝟑(𝐲 + 𝟐𝐱) +
𝐞𝐱+𝟐𝐲
11
T 5
Solve
∂2
z
∂x2
− 4
∂2
z
∂ x ∂y
+ 4
∂2
z
∂y2
= e2x+3y
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝛟𝟏(𝐲 + 𝟐𝐱) + 𝐱𝛟𝟐(𝐲 + 𝟐𝐱) +
𝐞𝟐𝐱+𝟑𝐲
16
C 6 Solve (D + D′)z = cos x .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟(𝐲 − 𝐱) + 𝐬𝐢𝐧 𝐱

C 7
Solve
∂2
z
∂x2
− 4
∂2
z
∂y2
= cos 2x cos 3y .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏(𝐲 − 𝟐𝐱) + 𝐟𝟐(𝐲 + 𝟐𝐱) +
𝟏
𝟑𝟐
𝐜𝐨𝐬 𝟐𝐱 𝐜𝐨𝐬 𝟑𝐲
H 8
Solve
∂2
z
∂x2
−
∂2
z
∂ x ∂y
= sin x sin 2y .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏(𝐲) + 𝐟𝟐(𝐲 + 𝟐𝐱) −
𝟏
𝟔
[𝐜𝐨𝐬(𝐱 − 𝟐𝐲) + 𝟑 𝐜𝐨𝐬(𝐱 + 𝟐𝐲)]
H 9 Solve (D2
− 3DD′
+ 2D′2
)z = cos(x + 2y) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝛟𝟏(𝐲 − 𝟐𝐱) + 𝛟𝟐(𝐲 − 𝐱) −
𝟏
𝟑
𝐜𝐨𝐬(𝐱 + 𝟐𝐲)
C 10
Solve 2
∂2
z
∂x2
− 5
∂2
z
∂x ∂y
+ 2
∂2
z
∂y2
= sin(2x + y) .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏(𝟐𝐲 + 𝐱) + 𝐟𝟐(𝐲 + 𝟐𝐱) −
𝟏
𝟑
𝐱 𝐜𝐨𝐬(𝐲 + 𝟐𝐱)
C 11
Solve
∂2
z
∂x2
+ 3
∂2
z
∂x ∂y
+ 2
∂2
z
∂y2
= x + y .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝛟𝟏(𝐲 − 𝟐𝐱) + 𝛟𝟐(𝐲 − 𝐱) −
𝐱𝟑
𝟑
+
𝐱𝟐
𝐲
𝟐
H 12
Solve
∂2
z
∂x2
− 4
∂2
z
∂y2
= 3x − 4y .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏(𝐲 + 𝟐𝐱) + 𝐟𝟐(𝐲 − 𝟐𝐱) +
𝐱𝟑
𝟐
− 𝟐 𝐱𝟐
𝐲
C 13 Solve (D3
+ D2
D′
− DD′2
− D′3
)z = ex
cos 2𝑦
𝐳 = 𝐟𝟏(𝐲 − 𝐱) + 𝐱𝐟𝟐(𝐲 − 𝐱) + 𝐟𝟑(𝐲 + 𝐱) +
𝐞𝐱
𝟐𝟓
(𝐜𝐨𝐬 𝟐𝒚 + 𝟐 𝐬𝐢𝐧 𝟐𝒚)
H 14
Solve
∂2
z
∂x2
− 3
∂2
z
∂x ∂y
+ 2
∂2
z
∂y2
= (1 + 2x) ex+2y
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏(𝐲 + 𝟐𝐱) + 𝐟𝟐(𝐲 + 𝐱) +
(𝟔𝐱 + 𝟏𝟏)
𝟗
𝐞𝐱+𝟐𝐲
T 15
Solve
∂2
z
∂x2
+
∂2
z
∂x ∂y
− 6
∂2
z
∂y2
= y sin x .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳 = 𝐟𝟏(𝐲 − 𝟑𝐱) + 𝐟𝟐(𝐲 + 𝟐𝐱) − 𝐜𝐨𝐬 𝐱 − 𝐲 𝐬𝐢𝐧 𝐱

 METHOD OF SEPARATION OF VARIABLES:
 Step 1: Let u(x, y) = X(x) ∙ Y(y)
 Step 2: Find
∂u
∂x
,
∂u
∂y
,
∂2u
∂x2
,
∂2u
∂x ∂y
,
∂2u
∂y2
as requirement and substitute in given Partial Differential
Eqn.
 Step 3: Convert it into Separable Variable equation and equate with constant say k
individually.
 Step 4: Solve each Ordinary Differential Equation.
 Step 5: Put value of X(x) & Y(y) in equation u(x, y) = X(x) ∙ Y(y).
METHOD – 3: EXAMPLE ON SEPARATION OF VARIABLES
H 1 Solve the equation by method of separation of variables
∂u
∂x
= 4
∂u
∂y
,
where u(0, y) = 8e−3y
.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮(𝐱, 𝐲) = 𝟖 𝐞−𝟏𝟐𝐱−𝟑𝐲
C 2 Solve
∂u
∂x
= 2
∂u
∂t
+ u subject to the condition u(x, 0) = 6e−3x
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮(𝐱, 𝐭) = 𝟔 𝐞−𝟑𝐱−𝟐𝐭
H 3 Solve the equation ux = 2ut + u given u(x, 0) = 4e−4x
by the method of
separation of variable.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮(𝐱, 𝐭) = 𝟒𝐞−𝟒𝐱−
𝟓
𝟐
𝐭
C 4 Using method of separation of variables solve
∂u
∂x
+
∂u
∂y
= 2(x + y)u.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮(𝐱, 𝐲) = 𝐜𝟏𝐜𝟐𝐞𝟐𝐱𝟐+𝟐𝐲𝟐+𝐤𝐱−𝐤𝐲
H 5 Solve x
∂u
∂x
− 2y
∂u
∂y
= 0 using method of separation variables.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮(𝐱, 𝐲) = 𝐜𝟏 𝐜𝟐 𝐱𝐤
𝐲
𝐤
𝟐
H 6 Using the method of separation variables, solve the partial differential
equation uxx = 16uy .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮(𝐱, 𝐲) = (𝐜𝟏𝐞√𝐤𝐱
+ 𝐜𝟐𝐞−√𝐤𝐱
) 𝐜𝟑 𝐞
𝐤𝐲
𝟏𝟔
H 7 Using method of separation of variables solve
∂2u
∂x2 =
∂u
∂y
+ 2u.
) 𝐜𝟑 𝐞(𝐤−𝟐)𝐲

C 8 Solve two dimensional Laplace’s equation
∂2u
∂x2
+
∂2u
∂y2
= 0,using the method
separation of variables.
) (𝐜𝟑 𝐜𝐨𝐬 √𝐤𝐲 + 𝐜𝟒 𝐬𝐢𝐧 √𝐤𝐲)
H 9 Using the method of separation of variables, solve the partial differential
equation
∂2u
∂x2
= 16
∂2u
∂y2
.
) (𝐜𝟑𝐞
√𝐤
𝟒
𝐲
+ 𝐜𝟒𝐞−
√𝐤
𝟒
𝐲
)
H 10 Solve the method of separation of variables
∂2u
∂x2
− 4
∂u
∂x
+
∂u
∂y
= 0 .
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐮(𝐱, 𝐲) = (𝐜𝟏𝐞(𝟐+√𝟒+𝐤)𝐱
+ 𝐜𝟐𝐞(𝟐−√𝟒+𝐤)𝐱
) 𝐜𝟑𝐞−𝐤𝐲
C 11 Solve
∂2z
∂x2
− 2
∂z
∂x
+
∂z
∂y
= 0 by the method of separation variable.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐳(𝐱, 𝐲) = (𝐜𝟏𝐞(𝟏+√𝟏+𝐤)𝐱
+ 𝐜𝟐𝐞(𝟏−√𝟏+𝐤)𝐱
) 𝐜𝟑𝐞−𝐤𝐲
 CLASSIFICATION OF SECOND ORDER PARTIAL DIFFERENTIAL EQUATION:
 The general form of a non-homogeneous second order P.D.E.
A(x, y)
∂2
z
∂x2
+ B(x, y)
∂2
z
∂x ∂y
+ C(x, y)
∂2
z
∂y2
+ f (x, y, z,
∂z
∂x
,
∂z
∂y
) = F(x, y) … … (1)
 Equation (1) is said to be
Elliptic, If B2
− 4AC < 0 Parabolic, If B2
− 4AC = 0 Hyperbolic, If B2
− 4AC > 0
METHOD – 4: EXAMPLE ON CLASSIFICATION OF 2ND ORDER PDE
C 1 Classify the 2nd order P.D.E. t
∂2u
∂t2
+ 3
∂2u
∂x ∂t
+ x
∂2u
∂x2
+ 17
∂u
∂x
= 0.
𝐀𝐧𝐬𝐰𝐞𝐫: Hyperbolic, if 𝐱𝐭 >
𝟗
𝟒
; Parabolic, if 𝐱𝐭 =
𝟗
𝟒
; Elliptical, if 𝐱𝐭 <
𝟗
𝟒
H 2 Classify the 2nd order P.D.E. 4
∂2u
∂t2
− 9
∂2u
∂t ∂x
+ 5
∂2u
∂x2
= 0.
𝐀𝐧𝐬𝐰𝐞𝐫: 𝐇𝐲𝐩𝐞𝐫𝐛𝐨𝐥𝐢𝐜

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