Department of Applied Mathematics and
Computational Sciences
II Sem B.E. Mechanical and Mechatronics
Complex variables and Transforms
Dr. V. Suresh kumar
Analytic function
Definition
A function 𝑓(𝑧) is said to be analytic at 𝑧 = 𝑧0 if 𝑓(𝑧) is differentiable at
𝑧 = 𝑧0 and it is differentiable at all the points in the neighborhood of 𝑧0 .
Cauchy-Riemann (CR) Equations
Let 𝑤 = 𝑓 𝑧 = 𝑢 𝑥, 𝑦 + 𝑖𝑣(𝑥, 𝑦). The CR equations given by
𝜕𝑢 𝜕𝑣
=
𝜕𝑥 𝜕𝑦
𝜕𝑢
𝜕𝑣
=−
𝜕𝑦
𝜕𝑥
𝑢𝑥 = 𝑣𝑦
𝑢y = −𝑣𝑥
THEOREM
Let f(z) = u(x, y) + iv(x, y) be defined and continuous in some
neighborhood of a point 𝑧 = 𝑥 + 𝑖𝑦 and differentiable at z itself.
Then at that point 𝑧, the first-order partial derivatives of 𝑢 and 𝑣 exist
and satisfy the Cauchy-Riemann equations.
If 𝑓(𝑧) is analytic in a domain 𝐷, then the first-order partial derivatives
of 𝑢 and 𝑣 exist and satisfies CR equations at all points of 𝐷.
Note
The above theorem is the necessary condition for the function to be analytic
Proof
path I
Let 𝑓 𝑧 = 𝑢 𝑥, 𝑦 + 𝑖𝑣(𝑥, 𝑦) be analytic at 𝑧
𝑦 + ∆𝑦
Therefore 𝑓 𝑧 is differentiable at 𝑧
𝑙𝑡
𝑓 ′ 𝑧 = ∆𝑧→0
𝑓 𝑧+∆𝑧 −𝑓(𝑧)
∆𝑧
𝑧 + ∆𝑧
∆𝑦
𝑦
path II
𝒛
∆𝑥
exists
𝑥
𝑥 + ∆𝑥
𝑢 𝑥 + ∆𝑥, 𝑦 + ∆𝑦 + 𝑖𝑣 𝑥 + ∆𝑥, 𝑦 + ∆𝑦 − 𝑢 𝑥, 𝑦 + 𝑖𝑣(𝑥, 𝑦)
= ∆𝑥 + 𝑖∆𝑦 → 0
∆𝑥 + 𝑖∆𝑦
= ∆𝑥 + 𝑖∆𝑦 → 0
Via path I
𝑢 𝑥 + ∆𝑥, 𝑦 + ∆𝑦 − 𝑢 𝑥, 𝑦 + 𝑖 𝑣 𝑥 + ∆𝑥, 𝑦 + ∆𝑦 − 𝑣(𝑥, 𝑦)
∆𝑥 + 𝑖∆𝑦
∆𝑥 = 0 & ∆𝑦 → 0
𝑓′(𝑧) = ∆𝑦 → 0
𝑢 𝑥, 𝑦 + ∆𝑦 − 𝑢 𝑥, 𝑦 + 𝑖 𝑣 𝑥, 𝑦 + ∆𝑦 − 𝑣(𝑥, 𝑦)
𝑖∆𝑦
1 𝑢 𝑥, 𝑦 + ∆𝑦 − 𝑢 𝑥, 𝑦
𝑣 𝑥, 𝑦 + ∆𝑦 − 𝑣(𝑥, 𝑦)
= ∆𝑦 → 0
+
𝑖
∆𝑦
∆𝑦
(1)
Proof Contd…
1
𝑢 𝑥, 𝑦 + ∆𝑦 − 𝑢 𝑥, 𝑦
𝑣 𝑥, 𝑦 + ∆𝑦 − 𝑣(𝑥, 𝑦)
= ∆𝑦 → 0
+ ∆𝑦 → 0
𝑖
∆𝑦
∆𝑦
=
1 𝜕𝑢 𝜕𝑣
+
𝑖 𝜕𝑦 𝜕𝑦
𝑓′
𝜕𝑢 𝜕𝑣
𝑧 = −𝑖
+
𝜕𝑦 𝜕𝑦
Via path II
(2)
∆𝑦 = 0 & ∆𝑥 → 0
(1) becomes
𝑓′(𝑧) = ∆𝑥 → 0
𝑢 𝑥 + ∆𝑥, 𝑦 − 𝑢 𝑥, 𝑦 + 𝑖 𝑣 𝑥 + ∆𝑥, 𝑦 − 𝑣(𝑥, 𝑦)
∆𝑥
𝑢 𝑥 + ∆𝑥, 𝑦 − 𝑢 𝑥, 𝑦
𝑣 𝑥 + ∆𝑥, 𝑦 − 𝑣(𝑥, 𝑦)
+ 𝑖 ∆𝑥 → 0
∆𝑥
∆𝑥
𝜕𝑢
𝜕𝑣
𝑓′(𝑧) =
+𝑖
(3)
𝜕𝑥
𝜕𝑥
= ∆𝑥 → 0
Proof Contd…
Since 𝑓 𝑧 is differentiable
(2) = (3)
𝜕𝑢 𝜕𝑣
𝜕𝑢
𝜕𝑣
−𝑖
+
=
+𝑖
𝜕𝑦 𝜕𝑦
𝜕𝑥
𝜕𝑥
Equating real and imaginary parts on both sides
𝜕𝑢 𝜕𝑣
=
𝜕𝑥 𝜕𝑦
𝜕𝑢
𝜕𝑣
=−
𝜕𝑦
𝜕𝑥
𝑢𝑥 = 𝑣𝑦
𝑢y = −𝑣𝑥
Example
Determine whether 𝑓 𝑧 = 𝑧 is analytic?
𝑢 + 𝑖𝑣 = 𝑥 − 𝑖𝑦
𝑢 = 𝑥,
𝑣 = −𝑦
𝑢𝑥 = 1
𝑣𝑥 = 0
𝑢𝑦 = 0
𝑣𝑦 = −1
CR equations not satisfied, 𝑓 𝑧 is not analytic
THEOREM
If two real-valued continuous functions 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦) of two real
variables 𝑥 and 𝑦 have continuous first order partial derivatives that satisfy the
C-R equations in some domain 𝐷 , then the complex function 𝑓(𝑧) =
𝑢(𝑥, 𝑦) + 𝑖 𝑣(𝑥, 𝑦) is analytic in 𝐷.
Note
The above theorem is the sufficient condition for the function to be analytic
Example
Determine whether 𝑓 𝑧 = 𝑒 𝑧 is analytic?
𝑢 + 𝑖𝑣 = 𝑒 𝑥+𝑖𝑦
= 𝑒 𝑥 (cos 𝑦 + 𝑖𝑠𝑖𝑛 𝑦)
𝑢 = 𝑒 𝑥 cos 𝑦 ,
𝑣 = 𝑒 𝑥 𝑠𝑖𝑛 𝑦
𝑢𝑥 = 𝑒 𝑥 cos 𝑦
𝑣𝑥 = 𝑒 𝑥 𝑠𝑖𝑛 𝑦
𝑢𝑦 = −𝑒 𝑥 sin 𝑦
𝑣𝑦 = 𝑒 𝑥 𝑐𝑜𝑠 𝑦
𝑢 and 𝑣 satisfies CR equations and 𝑢𝑥 , 𝑢𝑦 , 𝑣𝑥 , 𝑣𝑦 are continuous
Therefore 𝑓 𝑧 = 𝑒 𝑧 is analytic
CR equations in polar form
Let 𝑓 𝑧 = 𝑢 𝑟, 𝜃 + 𝑖𝑣(𝑟, 𝜃).
𝜕𝑢 1 𝜕𝑣
=
𝜕𝑟 𝑟 𝜕𝜃
𝜕𝑢
𝜕𝑣
= −𝑟
𝜕𝜃
𝜕𝑟
Example
Determine whether 𝑓 𝑧 = log 𝑧 is analytic?
𝑧 = 𝑟𝑒 𝑖𝜃
𝑓 𝑧 = log 𝑟𝑒 𝑖𝜃
= log 𝑟 + log 𝑒 𝑖𝜃
𝜕𝑢 1
=
𝜕𝑟 𝑟
𝜕𝑣
=0
𝜕𝑟
𝜕𝑢
=0
𝜕𝜃
𝜕𝑣
=1
𝜕𝜃
= log 𝑟 + 𝑖𝜃
𝑢 𝑟, 𝜃 = log 𝑟
𝑣 𝑟, 𝜃 = 𝜃
CR equations satisfied and partial derivatives
continuous except at 𝑟 = 0
log 𝑧 is analytic except at 𝑧 = 0
Problems for practice
Find where each of the following functions ceases to be analytic.
1.
𝑧
(𝑧 2 −1)
2.
𝑧 2 −4
(𝑧 2 +1)
3.
𝑧+𝑖
𝑧−𝑖 2
4. 𝑧 3
5. sin 𝑧
6. cosh z