Ch1\ Function of Complex Variables: -

Complex umbers!
Z= (x, y)
x is read par of Z
y is imagining par of Z

 Z=x+iy 1

Note\ 𝒊 = √−𝟏 , 𝒊𝟐 = −𝟏

Ex: - 𝒙𝟐 + 𝟏 = 𝟎

Sol: - 𝒙𝟐 = −𝟏
+ +
𝒙=
−
√−𝟏 = −
𝒊
 (Imagining par & Z)
Agrand Diagram y

P(Z)
x = r cos 𝜃 2
r
y = r sin 𝜃
𝜃 X
𝜃 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑟𝑔𝑛𝑚𝑒𝑛𝑡 𝑜𝑓 𝑍

Ex: - 𝒁 = 𝟐 + 𝟑𝒊

yy z
2+3i

Pulwar coordinate are can obtain

𝑍 = 𝑟 [cos 𝜃 + 𝑖 sin 𝜃 ] 3

R= modulus of Z =mod Z is always positive

If r=0  Z=0
 |𝑍| = 𝑟 = √𝑋 2 + 𝑌 2 4

Equal (4) in equal (2) we

𝑦
sin 𝜃 =
√𝑥 2 + 𝑦 2
𝑥 5
cos 𝜃 =
√𝑥 2 +𝑦 2

H.w\ if z=2+3i find r and cos and sin?

Sol: - |𝑧| = 𝑟 = √𝑋2 + 𝑌2 = √4 + 9 = √13
𝑥 2
cos 𝜃 = =
√𝑥 2 +𝑦2 √13
𝑦 3
sin 𝜃 = =
√𝑥 2 +𝑦 2 √13

Complex conyuyte
𝑧 = 𝑥 + 𝑖𝑦
If 𝑧̅ = 𝑥 − 𝑖𝑦 6 y

Z=2+3i
 r = |𝒛| = |𝒛̅|
 𝒛̅ is mirror image of Z in read axis -x x
̅=2+3i
𝒁
Operation with Complex number: -
A. The sum [or defined] of Z1 and Z2 is defined
Z1+_ Z2 = (x1+_ iy1) +_ (x2 +_ iy2)
= (x1 +_ x2) + (y1 +_ y2) i

Note\ Complex number subject comulatine and Assoc inline law

of (sum).
a) Z1+Z2=Z2+Z1 (Com)
b) Z1+(Z2+Z3) =(Z1+Z2) +Z3 (asso)
Z1+z̅2=2x
Z1-z̅2 = 2iy
B. The product Z1 and Z2 if defined as
Z1. Z2=(X1+iy1) (X2+iy2)
= X1. X2+iX1.y2 +X2.y1i-y1. y2
=(X1.X2-y1.y2)+(X1.y1+X2.y2) i
Note\ Complex numbers sulsfi commulatine and Assoc inline
law of product.
a) Z1. Z2=Z2.Z1 (Com)
b) Z1(Z2. Z3) = (Z1. Z2) Z3
Z. z̅= (x+i. y). (x-i. y)
=𝑥 2 +iyx-iyx+𝑦 2
=𝑥 2 + 𝑦 2
 Z. z
̅=𝑥 2 + 𝑦 2 =𝑟 2
|𝒛𝟐 |=𝑟 2
|𝒛𝟐 |= Z. 𝑧1

|𝒛|=√𝒛. 𝒛 =z
H.w\ using the product prof that
Z1. Z2=r1.r2[cos(𝜃1 + 𝜃2) + 𝑖. sin(𝜃1 + 𝜃2)]?
Sol: -
Z1. Z2=r1.r2[cos(𝜃1 + 𝜃2) + 𝑖. sin(𝜃1 + 𝜃2)]
Z1. Z2=r1(cos 𝜃1 + i . sin 𝜃1). r2(cos 𝜃2 + i . sin 𝜃2)
=r1.r2(cos 𝜃1 + 𝑖 . sin 𝜃1). ( cos 𝜃2 + i . sin 𝜃2)
= r1. r2[cos 𝜃1. cos 𝜃2 + 𝑖 . cos 𝜃1 . sin 𝜃2 + 𝑖 . sin 𝜃1 . cos 𝜃2 − sin 𝜃1 . sin 𝜃2]
= r1. r2[cos 𝜃1. cos 𝜃2 − sin 𝜃1 . sin 𝜃2 + 𝑖 . (cos 𝜃1 . sin 𝜃2 + . sin 𝜃1 . cos 𝜃2)]
Then Z1. Z2=r1.r2[cos(𝜃1 + 𝜃2) + 𝑖. sin(𝜃1 + 𝜃2)]

C. The division of Z1 by Z2 is dif by

𝑥1+𝑖.𝑦1 𝑥2−𝑖.𝑦2
= .
𝑥2+𝑖.𝑦2 𝑥2−𝑖.𝑦2

(𝑥1+𝑖.𝑦1) . (𝑥2−𝑖.𝑦2)
= 𝑥22 +𝑦22

(𝑥1.𝑥2+𝑦1.𝑦2)+(𝑦1.𝑥2−𝑥1.𝑦2).𝑖
= 𝑥22 +𝑦22
z1 (𝑥1.𝑥2+𝑦1.𝑦2) (𝑦1.𝑥2−𝑥1.𝑦2)
= + .𝑖
z2 𝑥2 +𝑦2 𝑥2 +𝑦2
Interims of polor coord.
z1 𝑟1 (cos 𝜃1 + 𝑖. sin 𝜃1)
=
z2 𝑟2 (cos 𝜃2 + 𝑖. sin 𝜃2)

𝑟1 (cos 𝜃1 + 𝑖. sin 𝜃1) (cos 𝜃2 − sin 𝜃2)

= . .
𝑟2 (cos 𝜃2 + 𝑖. sin 𝜃2) (cos 𝜃2 − sin 𝜃2)

𝑟1 (cos 𝜃1 + 𝑖. sin 𝜃1). (cos 𝜃2 − sin 𝜃2)

= .
𝑟2 (cos 𝜃22 + sin 𝜃22 )

𝑟1
= . [cos 𝜃1. cos 𝜃2 − 𝑖 . cos 𝜃1 . sin 𝜃2 + 𝑖 . sin 𝜃1 . cos 𝜃2 + sin 𝜃1 . sin 𝜃2]
𝑟2
𝑟1
= . [(cos 𝜃1. cos 𝜃2 + sin 𝜃1 . sin 𝜃2) + 𝑖 . (sin 𝜃1 . cos 𝜃2 − . cos 𝜃1 . sin 𝜃2)]
𝑟2

H.w\ if Z1=-3+5i , Z2=2-3i find :-

z1
a. Z1+Z2 b. Z1-Z2 c. Z1. Z2 d. e. |𝑧1 | f. |𝑧1 . 𝑧2 |
z2

Sol: - a. Z1+Z2 = -3+5i + 2-3i = -1+2i

b. Z1-Z2 = -3+5i – 2+3i = -5+8i

c. Z1. Z2 =(-3+5i). (2-3i) = -6+9i+10i-15𝑖 2= 9+19i

z1 −3+5𝑖 (−3+5𝑖 ).(2+3𝑖) −6−9𝑖+10𝑖+15𝑖 2
d. = = =
z2 2−3𝑖 4+9 4+9
−21 𝑖
= +
13 13

e. |𝑧1 | = √𝑥 2 + 𝑦 2=√9 + 25 =√34

 f. |𝑧1 . 𝑧2 |= √92 + 192=√442

Exponential and circular function

𝑒 𝑖 𝜃 = cos 𝜃 + 𝑖 sin 𝜃 a

𝑒 −𝑖 𝜃 = cos 𝜃 − 𝑖 sin 𝜃 b

Z=r (cos 𝜃 + 𝑖 sin 𝜃)

Adding a and b we a blain

𝑒 𝑖 𝜃 + 𝑒 −𝑖 𝜃 = 2 cos 𝜃

𝑒 𝑖 𝜃 +𝑒 −𝑖 𝜃
 cos 𝜃 = 2
c

Subtracting a and b we a blain

𝑒 𝑖 𝜃 − 𝑒 −𝑖 𝜃 = 2 i sin 𝜃

𝑒 𝑖 𝜃 −𝑒 −𝑖 𝜃
 sin 𝜃 = 2𝑖
d

Equ. c And d difind of circular function

 𝜋
𝑖
Ex: - write Z=(4+3i)𝑒 3 in terms of u+iv where u and v are
read number?
𝜋
𝑖
Sol: - Z=(4+3i)𝑒 3

𝜋 𝜋
=(4+3i). ( cos 3 + 𝑖 sin 3 )
1 √3
=(4+3i). (2 + 2
𝑖)
3
2 3 √3
= 2 − √ 3𝑖 + 2 𝑖 − 2
3
√3 2 3
= (2 − ) + ( √3 + 2) 𝑖
2
3 4
4− √3 √3+3
= + 𝑖
2 2

H.w\ find
1) 𝑒 𝑛𝜋𝑖 = cos 𝑛𝜋 + 𝑖 sin 𝑛𝜋 = (−1)𝑛 + 0 = (−1)𝑛
2) 𝑒 2𝑛𝜋𝑖 = cos 2𝑛𝜋 + 𝑖 sin 2𝑛𝜋 = 1 + 0 = 1
(2𝑛+1)𝜋𝑖
(2𝑛+1) (2𝑛+1)
3) 𝑒 2 = cos 𝜋 + 𝑖 sin 𝜋 = 0 + 𝑖 (−1)𝑛 = 𝑖 (−1)𝑛
2 2
 Which represented by point p then the line of (p) is rotated
Angled of 𝛼 then the pointed is 𝜃 is then rotated angle of
(𝜃1 + 𝛼). y
𝜽(z2)
Z1=r𝑒 𝑖𝜃
P(z1)
𝑖(𝜃+𝑘 )
Z2=r𝑒
r
= r𝑒 𝑖𝜃 𝑒 𝑖𝑘 = Z1 𝑒 𝑖𝛼
𝜋 𝜋
𝑖 𝜽
When 𝛼 = 2 𝑡ℎ𝑒𝑡 Z2=Z1 𝑒 2 = i Z1 x
r
De moirres theorem: -
Let Z1= r1 (cos 𝜃1 + 𝑖 sin 𝜃1 )

Z2= r2 (cos 𝜃2 +𝑖 sin 𝜃2 )

Z1 Z2= r1 r2 [cos( 𝜃1 + 𝜃2 ) + 𝑖 sin(𝜃1 + 𝜃2 )]

Suilard if we had set of in complex number

Z1= r1 (cos 𝜃1 + 𝑖 sin 𝜃1 )

Z2= r2 (cos 𝜃2 +𝑖 sin 𝜃2 )

. .
. .
Zn= rn (cos 𝜃𝑛 +𝑖 sin 𝜃𝑛 )

Z1 Z2… Zn = r1 r2… rn [cos( 𝜃1 + 𝜃2 … 𝜃𝑛 ) + 𝑖 sin(𝜃1 + 𝜃2 … 𝜃𝑛 )]

 Special case!
When Z1 = Z2 = Z3 = Zn
r1 = r2 = r3 = rn
𝜃1 = 𝜃2 = 𝜃3 = 𝜃𝑛

Z1 Z2… Zn = r1 r2… rn [cos( 𝜃1 + 𝜃2 … 𝜃𝑛 ) + 𝑖 sin(𝜃1 + 𝜃2 … 𝜃𝑛 )]

z 𝑛 = cos( 𝜃𝑛 ) + 𝑖 sin( 𝜃𝑛 )

But Z=

z 𝑛 = (cos 𝜃 + i sin 𝜃)𝑛

equating eq a and b we have

cos( 𝑛𝜃) + 𝑖 sin( 𝑛𝜃) = (cos 𝜃 + i sin 𝜃)𝑛

Demores theorem
𝑎
n=even or odd or (read numbers)
𝑏
n≠ 0.4 Ex (0.3,0.9)
Ex: - find (cos 𝜃 + i sin 𝜃)−𝑚
Sol: - = cos( −𝑚𝜃) + 𝑖 sin( −𝑚𝜃)
= cos( 𝑚𝜃 ) + 𝑖 sin( −𝑚𝜃 )

= cos( 𝑚𝜃 ) − 𝑖 sin( 𝑚𝜃 ) or (cos 𝜃 + i sin 𝜃 )−𝑚

1 1 cos( 𝑚𝜃 )−𝑖 sin( 𝑚𝜃 )

=
(cos 𝜃+i sin 𝜃 )𝑚 = .
cos( 𝑚𝜃)+𝑖 sin( 𝑚𝜃 ) cos( 𝑚𝜃 )−𝑖 sin( 𝑚𝜃 )

= cos( 𝑚𝜃) − 𝑖 sin( 𝑚𝜃)

Application of demoires throm: -

Expansion of 𝑐𝑜𝑠 𝑛 𝜃, 𝑠𝑖𝑛𝑛 𝜃, cos(𝑛𝜃) , sin(𝑛𝜃)
Z=
If 1 1 a
= = cos 𝜃 − 𝑖 sin 𝜃
𝑧

1
Z+ = 2 cos 𝜃
𝑧 b
1
Z- 𝑧 = 2 i sin 𝜃
 Similarly, if we have
z 𝑛 = cos(𝑛𝜃 ) + 𝑖 sin (𝑛𝜃 )
c
1
= cos(𝑛𝜃 ) − 𝑖 sin (𝑛𝜃 )
𝑧𝑛

1
z 𝑛 + 𝑛 = 2cos(𝑛𝜃 )
𝑧 d
1
z 𝑛 − 𝑛 = 2 i sin(𝑛𝜃 )
𝑧

These results of eq. are usefut expanding powers of cos 𝜃 and

sin 𝜃 interms of multiple angles...
Ex: - a press 𝑐𝑜𝑠 6 𝜃 in temp multiple angles?
1
Sol: - Z+ 𝑧 = 2 cos 𝜃
1 1 6
[cos 𝜃 = (𝑧 + )]
2 𝑧
6
6
1 1
𝑐𝑜𝑠 𝜃= 6 (𝑧 + )
2 𝑧
6
1 6
1 4
1 2
1
𝑐𝑜𝑠 𝜃 = 6 [ (𝑧 + 6 ) + 6 (𝑧 + 4 ) + 15 (𝑧 + 2 ) + 20 ]
2 𝑧 𝑧 𝑧
1
𝑐𝑜𝑠 6 𝜃 = 6 [ 2 cos 6 𝜃 + 6.2 cos 4 𝜃 + 15.2 cos 2 𝜃 + 20 ]
2
1
𝑐𝑜𝑠 6 𝜃 = 5 [ cos 6 𝜃 + 6 cos 4 𝜃 + 15 + cos 2 𝜃 + 10 ]
2
Ex: - find to n complex rote of eq. z n−1 = 0 (n positivin teger)
Sol: - 𝑧𝑛 = 1
𝑧𝑛 = 𝑒2 𝑘 𝜋 𝑖

𝑒 2 𝑘 𝜋 𝑖 = cos 2𝜋𝑘 + 𝑖 sin 2𝜋𝑘 = 1 when k =0,1,2,3,4, .......

1
𝑧 = (1)𝑛
1
 = (cos 2𝜋𝑘 + 𝑖 sin 2𝜋𝑘 )𝑛
2𝜋𝑘 2𝜋𝑘
 = (cos 𝑛
+ 𝑖 sin
𝑛
) K≥𝑛−1
𝑤ℎ𝑒𝑛 (𝑛) = 3
k=0,1,2
Z1= cos 0 + 𝑖 sin 0 = 1
2𝜋 2𝜋
Z2= cos + 𝑖 sin = cos(120) + 𝑖 sin(120)
3 3

=cos(𝜋 − 60) + 𝑖 sin(𝜋 − 60)

1 √3
Z2= − + 𝑖
2 2
2𝜋2 2𝜋2
Z3= cos + 𝑖 sin
3 3
4𝜋 4𝜋
=cos + 𝑖 sin = cos(240) + 𝑖 sin(240)
3 3
=cos(𝜋 + 60) + 𝑖 sin(𝜋 + 60)
1 √3
Z3= + 𝑖
2 2
Ex: - find to rote of eq. 𝑧 4 − 1 = √𝑖 3

Sol: - 𝑧 4 − 1 = √𝑖 3
𝑖
𝑧 4 = 1 + √3

4
1 √3
𝑧 = 2( + 𝑖)
2 2
𝑧 4 = 2(cos 60 + 𝑖 sin 60)
1 1
𝑧 = 4 (cos 60 + 𝑖 sin 60)4
2

Evaluation of integer: -
Calculate complex
𝑏𝑡

𝑐 = ∫ 𝑒 9𝑥 cos 𝑏𝑥 𝑑𝑥
𝑎
When (a), (b) and (t) at constant
Let
𝑡

𝑠 = ∫ 𝑒 9𝑥 i sin(𝑏𝑥) 𝑑𝑥
𝑎
That forming to the complex’s numbers
𝑡

𝑐 + 𝑖𝑠 = ∫ 𝑒 𝑎𝑥 [cos (bx) + i sin(𝑏𝑥) 𝑑𝑥 ]

0

= ∫ 𝑒 𝑎𝑥 e𝑖𝑏𝑥 𝑑𝑥
0
𝑡

= ∫ 𝑒 𝑥(𝑎+𝑖𝑏) 𝑑𝑥
0
𝑡
1
= ∫(𝑎 + 𝑖𝑏). 𝑒 𝑥(𝑎+𝑖𝑏) 𝑑𝑥
𝑎 + 𝑖𝑏
0
1 𝑥(𝑎+𝑖𝑏) 𝑡
= |𝑒 |0
𝑎 + 𝑖𝑏

𝑒 𝑥(𝑎+𝑖𝑏)𝑡 − 𝑒 0
=
𝑎 + 𝑖𝑏

𝑒 𝑥(𝑎+𝑖𝑏)𝑡 − 1 𝑎 − 𝑖𝑏
= .
𝑎 + 𝑖𝑏 𝑎 − 𝑖𝑏

(𝑒 𝑥(𝑎+𝑖𝑏)𝑡 − 1)(𝑎 − 𝑖𝑏)

=
𝑎2 + 𝑏 2
 𝑎. 𝑒 𝑥(𝑎+𝑖𝑏)𝑡 − 𝑎 − 𝑖𝑏. 𝑒 𝑥(𝑎+𝑖𝑏)𝑡 + 𝑖𝑏
=
𝑎2 + 𝑏 2

𝒂. 𝒆𝒂𝒕 (𝐜𝐨𝐬 𝒃𝒕 + 𝒊 𝐬𝐢𝐧 𝒃𝒕 ) − 𝒂 − 𝒊𝒃. 𝒆𝒂𝒕 (𝐜𝐨𝐬 𝒃𝒕 + 𝒊 𝐬𝐢𝐧 𝒃𝒕 ) + 𝒊𝒃

=
𝒂𝟐 + 𝒃𝟐

𝒂. 𝒆𝒂𝒕 𝐜𝐨𝐬 𝒃𝒕 + 𝒂. 𝒆𝒂𝒕 𝒊 𝐬𝐢𝐧 𝒃𝒕 − 𝒂 − 𝒊𝒃. 𝒆𝒂𝒕 𝐜𝐨𝐬 𝒃𝒕 +. 𝒆𝒂𝒕 𝒊 𝐬𝐢𝐧 𝒃𝒕 + 𝒊𝒃

=
𝒂𝟐 + 𝒃𝟐

𝒆𝒂𝒕 (𝐚. 𝐜𝐨𝐬 𝒃𝒕 + 𝒊 𝐬𝐢𝐧 𝒃𝒕) − 𝒂 𝒆𝒂𝒕 (−𝒃 𝐜𝐨𝐬 𝒃𝒕 +. 𝒂 . 𝐬𝐢𝐧 𝒃𝒕) + 𝒃
= +𝒊.
𝒂𝟐 + 𝒃𝟐 𝒂𝟐 + 𝒃𝟐

𝒆𝒂𝒕 (𝐚. 𝐜𝐨𝐬 𝒃𝒕 + 𝒊 𝐬𝐢𝐧 𝒃𝒕) − 𝒂

𝑐 =
𝒂𝟐 + 𝒃𝟐

Function of complex variables: -

The general relationship is
W=f(z) where w=u+iv
Z=x+iy

Are both complex number as Z varies f(z) is said to be a function

of the complex variables(z).
Differentiatnal of function complex variables: -
We difined the derivative of function (f (z)) at a point (z,z0) by the
limit.
𝑑𝑤 𝑓(𝑧)−𝑓 (𝑧0 )
( ) = 𝑓̅ (𝑧0 ) = lim a
𝑑𝑧 𝑧,𝑧0 𝑧→𝑧0 𝑧−𝑧0

Where z-z0=∆𝑧
z=∆𝑧 +z0
z→z0
z-z0→ z0-z0
0z→0
𝑓(𝑧0 +∆𝑧)−𝑓 (𝑧0 )
 𝑓̅ (𝑧0 ) = lim ∆𝑧
b
𝑧0 →0

Note\ we know calculate the limet 𝑓̅ (𝑧0 ) in two distinct

methods firstly suppose (z) a (𝑧0) along alien throw (𝑧0) to real.
∆𝑧 = ∆𝑥 + 𝑖∆𝑦 y
∆𝑧 = ∆𝑥
𝑠𝑖𝑛𝑐𝑒 ∆𝑧 = ∆𝑥 + 𝑖∆𝑦 𝑡ℎ𝑒𝑛
𝑓𝑜𝑙𝑙𝑜𝑤𝑠 ∆𝑧 = ∆𝑥 𝑓𝑜𝑟 𝑡ℎ𝑖𝑠 z0 z
y0
𝑝𝑎𝑟𝑙𝑖𝑒 𝑏𝑎𝑡ℎ 𝑢𝑠𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 𝑤𝑒 ℎ𝑎𝑣𝑒
𝑒𝑞 (𝑧) 𝑝𝑎𝑟𝑙𝑖𝑐𝑙𝑢𝑟 𝑝𝑎𝑡ℎ
x0
x x
 𝑓̅ (𝑧0 ) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦)
𝑡ℎ𝑒𝑛 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑓𝑜𝑟𝑚 𝑒𝑞𝑢. (𝑏)
𝑢(𝑥0 + ∆𝑥, 𝑦0 ) + 𝑖𝑣 (𝑥0 + ∆𝑥, 𝑦0 ) − 𝑢(𝑥0 , 𝑦0 ) − 𝑖𝑣(𝑥0 , 𝑦0 )
𝑓̅ (𝑧0 ) = lim
∆𝑥→0 ∆𝑥

𝒖(𝒙𝟎 + ∆𝒙, 𝒚𝟎 ) − 𝒖(𝒙𝟎 , 𝒚𝟎 ) 𝒊𝒗 (𝒙𝟎 + ∆𝒙, 𝒚𝟎 ) − 𝒊𝒗(𝒙𝟎 , 𝒚𝟎 )

= 𝐥𝐢𝐦 + 𝐥𝐢𝐦
∆𝒙→𝟎 ∆𝒙 ∆𝒙→𝟎 ∆𝒙

𝑑𝑢 𝑑𝑣
𝑓̅ (𝑧0 ) = ( ) +𝑖 ( ) c
𝑑𝑥 𝑥=𝑥0 𝑑𝑥 𝑥=𝑥0

∆𝑧 = ∆𝑥 + 𝑖∆𝑦 y
∆𝑧 = 𝑖∆𝑦
y
z
Rnow we choose a second ∆𝑦

Path along a line throw(Z0) y z0

y0
Imagenry y-axis for this s y0
Path ∆𝑧 = 𝑖∆𝑦 , then eq(b) become x

𝑢(𝑥0 , 𝑦0 + ∆𝑦) + 𝑖𝑣 (𝑥0 , 𝑦0 + ∆𝑦) − 𝑢(𝑥0 , 𝑦0 ) − 𝑖𝑣(𝑥0 , 𝑦0 )

𝑓̅ (𝑧0 ) = lim
∆𝑦→0 𝑖∆𝑦

𝑢(𝑥0 , 𝑦0 + ∆𝑦) − 𝑢(𝑥0 , 𝑦0 ) 𝑣 (𝑥0 + ∆𝑥, 𝑦0 ) − 𝑣(𝑥0 , 𝑦0 )

𝑓̅ (𝑧0 ) = lim + 𝑖 lim
∆𝑦→0 𝑖∆𝑦 ∆𝑦→0 𝑖∆𝑦
 1 𝑑𝑢 𝑑𝑣
𝑓̅ (𝑧0 ) = ( ) + ( )
𝑖 𝑑𝑦 𝑦=𝑦 𝑑𝑦 𝑦=𝑦
0 0
d
𝑑𝑢 𝑑𝑣
𝑓̅ (𝑧0 ) = (−𝑖) ( ) + ( )
𝑑𝑦 𝑦=𝑦 𝑑𝑦 𝑦=𝑦
0 0

𝑓̅ (𝑧0 ) 𝑖𝑠 𝑡𝑜 𝑏𝑒 𝑢𝑛𝑖𝑞𝑢𝑒𝑙𝑦 𝑑𝑖𝑓𝑖𝑛𝑒𝑑 𝑎𝑡 𝑧0 𝑤ℎ𝑒𝑟𝑒 𝑝𝑎𝑡ℎ (3) 𝑎𝑛𝑑

𝑢 𝑚𝑢𝑠𝑡 𝑎𝑔𝑟𝑒𝑒 𝑞𝑞𝑛𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑟𝑒𝑎𝑙 𝑎𝑛𝑑 𝑖𝑚𝑎𝑔𝑒𝑛𝑟𝑦 𝑝𝑎𝑡𝑠 𝑒𝑞𝑢.
(𝑐 ), (4) , 𝑎𝑏𝑡𝑎𝑖𝑛

𝑑𝑢 𝑑𝑣
=
𝑑𝑥 𝑑𝑦
e
𝑑𝑢 𝑑𝑣
= −
𝑑𝑦 𝑑𝑥

𝑡ℎ𝑖𝑒𝑠 𝑎𝑞𝑒. 𝑎𝑟𝑒 𝑐𝑎𝑙𝑙𝑒𝑑 Cauchy -Rieman

𝑎𝑛𝑑 𝑡ℎ𝑒 𝑔𝑎𝑟𝑒 𝑛𝑒𝑐𝑐𝑒𝑠𝑠𝑒𝑟𝑦 𝑐𝑜𝑛𝑒𝑡𝑖𝑡𝑖𝑜𝑢𝑠 𝑓𝑜𝑟 𝑓(𝑧)𝑡𝑜 𝑏𝑒
𝑑𝑒𝑓𝑟𝑒𝑛𝑡𝑖𝑜𝑛 𝑎𝑏𝑙𝑒 𝑎𝑛𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑊 = 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖 𝑣(𝑥, 𝑦)
𝑤ℎ𝑖𝑐ℎ 𝑑𝑜𝑠𝑒 𝑛𝑜𝑡 𝑠𝑢𝑡𝑐𝑖𝑓𝑦 𝑡ℎ𝑒𝑠𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛
Ex: - is the following equation f(z)=𝑒 𝑧 differentiable function?
Sol: - f(z)=𝑒 𝑧 = 𝑒 𝑥+𝑖𝑦 = 𝑒 𝑥 [cos 𝑦 + 𝑖 sin 𝑦]
U (x, y) = 𝑒 𝑥 cos 𝑦
V (x, y) = 𝑒 𝑥 sin 𝑦
𝑑𝑢 𝑑𝑣
= 𝑒𝑥 cos 𝑦 . = 𝑒𝑥 sin 𝑦
𝑑𝑥 𝑑𝑥
𝑑𝑢 𝑥 sin 𝑦 . 𝑑𝑣
= −𝑒 = 𝑒𝑥 cos 𝑦
𝑑𝑦 𝑑𝑦

𝑑𝑢 𝑑𝑣
 𝑑𝑥
=
𝑑𝑦
𝑑𝑢 𝑑𝑣
 = −
𝑑𝑦 𝑑𝑥

 𝑒 𝑧 𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒

𝑑𝑢 𝑑𝑣
𝑓(𝑧) = 𝑖 +
𝑑𝑦 𝑑𝑦

= +𝑖 𝑒 𝑥 sin 𝑦 + 𝑒 𝑥 cos 𝑦

= 𝑒 𝑥 (cos 𝑦 + 𝑖 sin 𝑦)
H.w\ is the following equation f(z)=|𝑧| differentiable function?

H.w\ is the following equation f(z)=z differentiable function?

𝑥 𝑖𝑦
Sol: - 𝑓(𝑧) = −
𝑢 𝑣

𝑑𝑢 𝑑𝑣
=1 , =0
𝑑𝑥 𝑑𝑥

𝑑𝑢 𝑑𝑣
=0 , = −1
𝑑𝑦 𝑑𝑦

𝑑𝑢 𝑑𝑣 𝑑𝑢 𝑑𝑣
 𝑑𝑥
= −
𝑑𝑦
,
𝑑𝑦
=
𝑑𝑥