Complex analysis deals with complex-valued functions of complex variables. Some key concepts covered in the document include: - Complex functions can be expressed in either Cartesian (z = x + iy) or polar (z = reiθ) form. - For a complex function f(z) to be differentiable, it must satisfy the Cauchy-Riemann conditions. - Analytic functions are differentiable at every point in their domain. For example, ez is analytic everywhere while z̅ is analytic nowhere. - Branch cuts arise for multi-valued functions like √z, with the line between θ = 2π and θ = 4π representing one branch cut.

1. NPTEL – Physics – Mathematical Physics - 1
Lecture 40
Summary
Complex Analysis
𝑧 = 𝑥 + 𝑖𝑦
= 𝑟𝑒𝑖𝜃
= 𝑟 (𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃)
𝑥, 𝑦 are real
polar form
Cauchy Riemann Condition (CR condition)
Let a complex function be 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦). The derivative 𝑓(𝑧) to exist at
a particular point z demands that,
𝑢𝑦 = −𝑣𝑥
} CR condition
Polar representation of the CR condition
𝑢𝑥 = 𝑣𝑦
𝑟𝑢𝑟 = 𝑣𝜃
𝑢𝜃 = −𝑟𝑣𝑟
}
Analytic functions
A function 𝑓(𝑧) is said to be analytic in a region if (a) it is defined, (b) and is
differentiable at every point in the region.
Example. 𝑓(𝑧) = 𝑒𝑧 is analytic in the entire finite z plane, whereas 𝑓(𝑧) = 𝑧̅ is
analytic nowhere.
𝑓(𝑧) = 𝑥 – 𝑖𝑦 𝑢 = 𝑥, 𝑣 = −𝑦
CR conditions are not satisfied.
Thus CR conditions are equivalent to analyticity
Branch cut
Let us consider a function 𝑓(𝑧) = √𝑧, 𝑧 = 𝑟𝑒𝑖𝜃 ,
𝑖𝜃
𝑓(𝑧) = √𝑟𝑒 ⁄2
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2. NPTEL – Physics – Mathematical Physics - 1
The value of the function at A is 𝑓(𝑧) = √𝑟𝑒𝑖 𝜃1⁄2. After a complete revolution,
𝑓(𝑧) = √𝑟𝑒
𝑖(𝜃1+2𝜋⁄2
= −√𝑟𝑒𝑖 𝜃1⁄2
So it does not return to the same value. So the function is multivalued and √𝑟𝑒
𝑖𝜃1⁄2 is a
value on one of the branches and between 2𝜋 ≤ 𝜃 ≤ 4𝜋, the function assumes another
branch. The line OB is called the branch cut.
Complex integration Cauchy – Goursat Theorem
If 𝑓(𝑧) is analytic in a region R and on it’s boundary 𝐶, then ∮ 𝑓(𝑧)𝑑𝑧 = 0 Also known as
cauchy’s integral theorem.
Cauchy’s Integral formula
𝑓(𝑎) =
1 𝑓(𝑧)
∮ 𝑑𝑧
2𝜋𝑖 𝑧 𝑧−𝑎
𝑓(𝑧) is analytic inside and on the boundary c of a simply connected region R, excepting
at a point a.
Residue theorem
If 𝑓(𝑧) is analytic inside and on a simple closed curve c except for a pole of order m at 𝑧
= 𝑎, inside c, prove that
1 1
2𝜋𝑖
∮ 𝑓(𝑧)𝑑𝑧 = 𝐿𝑖𝑚
𝑧→𝑎 (𝑚 − 1)! 𝑑𝑧𝑚−1
𝑑𝑚−1
[(𝑧 − 𝑎)𝑚𝑓(𝑧)]
𝑐
If there are two different poles of order 𝑚1and 𝑚2
1 1
2𝜋𝑖
∮ 𝑓(𝑧)𝑑𝑧 = 𝐿𝑖𝑚
𝑧→𝑎1 (𝑚1−1)! 𝑑𝑧𝑚1−1
𝑐
𝑑𝑚1−1
[(𝑧 − 𝑎1)𝑚1 𝑓(𝑧)] +
𝐿𝑖𝑚
𝑧→𝑎2 (𝑚2−1)! 𝑑𝑧𝑚2−1
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1 𝑑𝑚2−1
[(𝑧 − 𝑎 )𝑚2 𝑓(𝑧)]
2
Example
Integrate,
∞ 𝑑𝑥
𝐼 =
∫
0 1+𝑥3
The integrand I is not even so we cannot extend it up to -∞.

3. NPTEL – Physics – Mathematical Physics - 1
Let us evaluate the integral
𝑧3 = −1
𝐽 = ∮
𝑑𝑧
1+𝑧3 over the Contour below
𝑧 = 3
√−1 are the
poles
3 𝑖𝜋
= √𝑒𝑖𝜋 = 𝑒 ⁄6
0 0
𝑧 = 𝑟𝑒𝑖𝜃
1 + 𝑧3 = 1 + 𝑟3𝑒3𝑖𝜃
𝑑𝑧 = 𝑑𝑟𝑒𝑖𝜃
∞ 𝑑𝑥
𝐽 =
∫
1+𝑥3
2𝜋⁄
+ ∫
3
𝑑𝑧
1+𝑧3 ∞ 1+𝑧3
0 𝑑𝑧
+ ∫
0
So,
0 𝑒 𝑑𝑟
𝐽 = 𝐼
+ ∫
𝑖
𝜃
∞ 1+𝑟3𝑒3𝑖𝜃
𝜃 = 2𝜋⁄
3
= 𝐼 + ( − √
1
2
3
2 ∞
⁄ ) ∫
𝑖 0 𝑑𝑟
1+𝑟3
𝑒 ⁄3 = 𝑐𝑜𝑠 + 𝑖𝑠𝑖𝑛
𝑖2𝜋 2𝜋
3
2𝜋
3
𝐽 = (1 − 𝑒 ⁄3) 𝐼
2𝜋𝑖
= − √
1 𝑖
3
2 2
𝑒2𝜋𝑖 = 𝑐𝑜𝑠2𝜋 + 𝑖𝑠𝑖𝑛2𝜋
= 1
Now 𝐽 has a simple pole at 𝑧 = 𝑒
𝑖
𝜋
⁄3
𝐽 = 2𝜋𝑖
3𝑒
2𝜋
𝑖
⁄3
𝐼 =
2𝜋
3√3
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