limits, derivatives

1. Complex Analysis
( 15UMTC61)
-K.Anitha M.Sc., M.Phil.,
-A.Sujatha M.Sc., M.Phil.,
Department of Mathematics ( SF )
V.V.Vanniaperumal College for Women,
Virudhunagar.

2. Limits
• Let f be a function defined at all points z in some neighbourhood of 𝑧0
except possibly for the point 𝑧0itself.
• The limit of f(z) as z approaches 𝑧0is a number 𝑤0,
• ie, lim
𝑧→𝑧0
𝑓 𝑧 = 𝑤0

3. Uniqueness of limit
• When limit of a function f ( z ) exists at a point 𝑧0 , it must be unique.
Proof:
Suppose f ( z ) has two different limits 𝑤0 and 𝑤1
Then for a positive number 𝜀 , there are + ve number’s 𝛿0 and 𝛿1 such that
𝑓 𝑧 − 𝑤0 < 𝜀 whenever 0 < 𝑧 − 𝑧0 < 𝛿0
And 𝑓 𝑧 − 𝑤1 < 𝜀 whenever 0 < 𝑧 − 𝑧0 < 𝛿1

4. Let 𝛿 = min (𝛿0 , 𝛿1) . Then for
0 < 𝑧 − 𝑧0 < 𝛿0
• 𝑓 𝑧 − 𝑤0 − 𝑓 𝑧 − 𝑤1 ≤ 𝑓 𝑧 − 𝑤0 + 𝑓 𝑧 − 𝑤1
< 2𝜀
ie, 𝑤1 − 𝑤0 < 2𝜀
But 𝑤1 − 𝑤0 is a constant and 𝜀 can be chosen arbitrarily small.
If we choose 𝜀 = 0 , then 𝑤1 − 𝑤0= 0
ie , 𝑤1 = 𝑤0
Hence the limit of a function is unique.

5. Continuity
• The function f ( z ) is continuous at a point 𝑧0 , if
i) lim
𝑧→𝑧0
𝑓(𝑧) exists
ii) f(𝑧0) exists
iii) lim
𝑧→𝑧0
𝑓(𝑧) = f(𝑧0)
A function of a complex variable is said to be continuous in a region R if it is
continuous at each point in R.

6. Properties
i) Sum of two continuous functions is a continuous function .
ii) Product of the continuous functions is a continuous function .
iii) Quotient of two continuous functions is continuous where the denominator
not equal to zero.
iv) Polynomial is continuous in the entire plane.

7. Derivatives
• The derivative of f at 𝑧0, denoted by 𝑓/(𝑧0) = lim
𝑧→𝑧0
𝑓 𝑧 −𝑓(𝑧0)
𝑧−𝑧0
provided this
limit exists.
• The function f is said to be differentiable at 𝑧0 when its derivative at 𝑧0
exists.

8. Analytic Functions
• A single valued function w = f ( z ) in a domain D is said to be analytic at a
point z = a in a D if there exists a neighbourhood 𝑧 − 𝑎 < 𝛿 at all points
of which the function is differentiable.
Singular Point
The point at which f ( z ) is not differentiable are called singular points of the
function.

9. Ex. Given u = 𝑦3
- 3𝑥2
y find f(z)such that f(z)
is analytic
• Solution
• dv =
𝜕𝑣
𝜕𝑥
𝑑𝑥 +
𝜕𝑣
𝜕𝑦
dy
=
−𝜕𝑢
𝜕𝑦
𝑑𝑥 +
𝜕𝑢
𝜕𝑥
dy ( C R equations)
v = (
−𝜕𝑢
𝜕𝑦
𝑑𝑥 +
𝜕𝑢
𝜕𝑥
dy )
𝜕𝑢
𝜕𝑥
= -6xy ;
𝜕𝑢
𝜕𝑦
= 3𝑦2
-3𝑥2
v = (−3𝑦2+3𝑥2)𝑑𝑥 − 6𝑥𝑦 𝑑𝑦

10. = 3𝑥2
dx – (3𝑦2
dx + 6xy dy )
= 𝑑 ( 𝑥3) - d ( 3x𝑦2 )
v = 𝑥3
- 3x𝑦2
+ c
Hence f ( z ) = u + iv = 𝑦3- 3𝑥2y + i (𝑥3 - 3x𝑦2)
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