Complex Variables Notes for GATE Electrical Engineering (EE)

Analytic Functions - Definition

Definition

A function \(f(z)\) is analytic at point \(z_0\) if:

\(f(z)\) is differentiable at \(z_0\)
\(f(z)\) is differentiable in some neighborhood of \(z_0\)

Cauchy-Riemann Equations

If \(f(z) = u(x,y) + iv(x,y)\) is analytic, then:

\[\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\]

Key Point

Analytic functions are also called holomorphic or regular functions.

Properties of Analytic Functions

Important Properties

Sum, difference, product of analytic functions is analytic
Quotient is analytic (where denominator \(\neq 0\))
Composition of analytic functions is analytic
If \(f(z)\) is analytic and \(f'(z) \neq 0\), then \(f^{-1}(z)\) is analytic

Common Analytic Functions

Polynomials: \(z^n, az + b\)
Exponential: \(e^z\)
Trigonometric: \(\sin z, \cos z\)
Logarithmic: \(\log z\) (with branch cuts)

Cauchy’s Integral Theorem

Statement

If \(f(z)\) is analytic in a simply connected domain \(D\) and \(C\) is any closed contour in \(D\), then:

\[\oint_C f(z) dz = 0\]

Consequences

Path independence: \(\int_{C_1} f(z)dz = \int_{C_2} f(z)dz\) for same endpoints
Deformation of contours: Can deform contour without changing integral value
Existence of primitive: \(F'(z) = f(z)\) exists

GATE Tip

Use this theorem to evaluate complex integrals by choosing convenient paths!

Cauchy’s Integral Formula

Basic Formula

If \(f(z)\) is analytic inside and on a simple closed contour \(C\), and \(z_0\) is inside \(C\), then:

\[f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - z_0} dz\]

Generalized Formula (Derivatives)

\[f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} dz\]

Key Applications

Evaluate integrals of the form \(\oint_C \frac{f(z)}{z - z_0} dz\)
Find derivatives of analytic functions
Prove maximum modulus principle

Cauchy’s Formula - Example

Example

Evaluate \(\oint_C \frac{e^z}{z - i} dz\) where \(C: |z| = 2\)

Solution

\(f(z) = e^z\) is analytic everywhere
Singularity at \(z_0 = i\), and \(|i| = 1 < 2\), so \(i\) is inside \(C\)
Using Cauchy’s formula:

\[\oint_C \frac{e^z}{z - i} dz = 2\pi i \cdot e^i = 2\pi i \cdot (\cos 1 + i\sin 1)\]

GATE Strategy

Always check: Is the singularity inside the contour?

Taylor Series

Taylor Series Expansion

If \(f(z)\) is analytic at \(z_0\), then for \(|z - z_0| < R\):

\[f(z) = \sum_{n=0}^{\infty} a_n(z - z_0)^n\]

where \(a_n = \frac{f^{(n)}(z_0)}{n!}\)

Radius of Convergence

\(R\) = distance from \(z_0\) to nearest singularity

Common Taylor Series

\[\begin{aligned} e^z &= \sum_{n=0}^{\infty} \frac{z^n}{n!} = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots \\ \sin z &= \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!} = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots \\ \cos z &= \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots \end{aligned}\]

Laurent Series

Laurent Series Expansion

If \(f(z)\) is analytic in an annulus \(r < |z - z_0| < R\), then:

\[f(z) = \sum_{n=-\infty}^{\infty} a_n(z - z_0)^n\]

\[= \underbrace{\sum_{n=-\infty}^{-1} a_n(z - z_0)^n}_{\text{Principal part}} + \underbrace{\sum_{n=0}^{\infty} a_n(z - z_0)^n}_{\text{Regular part}}\]

Coefficients

\[a_n = \frac{1}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} dz\]

where \(C\) is any contour in the annulus around \(z_0\).

Important

The coefficient \(a_{-1}\) is called the residue of \(f(z)\) at \(z_0\).

Laurent Series - Example

Example

Find Laurent series of \(f(z) = \frac{1}{z(z-1)}\) around \(z = 0\)

Solution

Using partial fractions:

\[\frac{1}{z(z-1)} = \frac{1}{z} \cdot \frac{1}{z-1} = \frac{1}{z} \cdot \frac{-1}{1-z}\]

For \(0 < |z| < 1\):

\[\frac{1}{1-z} = 1 + z + z^2 + z^3 + \cdots\]

Therefore:

\[f(z) = \frac{-1}{z}(1 + z + z^2 + z^3 + \cdots) = -\frac{1}{z} - 1 - z - z^2 - \cdots\]

Note

Residue at \(z = 0\) is \(a_{-1} = -1\)

Residue Theorem

Statement

If \(f(z)\) is analytic inside and on a simple closed contour \(C\) except for isolated singularities \(z_1, z_2, \ldots, z_n\) inside \(C\), then:

\[\oint_C f(z) dz = 2\pi i \sum_{k=1}^{n} \text{Res}(f, z_k)\]

Types of Singularities

Removable: \(\lim_{z \to z_0} f(z)\) exists and is finite
Pole of order \(m\): \(\lim_{z \to z_0} (z-z_0)^m f(z)\) exists and \(\neq 0\)
Essential: Neither removable nor pole

Residue Calculation Methods

Simple Pole (\(m = 1\))

If \(z_0\) is a simple pole:

\[\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)\]

Pole of Order \(m\)

\[\text{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}}[(z - z_0)^m f(z)]\]

Rational Functions

If \(f(z) = \frac{P(z)}{Q(z)}\) where \(P(z_0) \neq 0\) and \(Q(z_0) = 0\), \(Q'(z_0) \neq 0\):

\[\text{Res}(f, z_0) = \frac{P(z_0)}{Q'(z_0)}\]

Residue Theorem - Example

Example

Evaluate \(\oint_C \frac{z^2}{(z-1)(z-2)} dz\) where \(C: |z| = 3\)

Solution

Singularities: \(z = 1\) (simple pole), \(z = 2\) (simple pole)
Both are inside \(|z| = 3\)
Residue at \(z = 1\):

\[\text{Res}(f, 1) = \lim_{z \to 1} (z-1) \frac{z^2}{(z-1)(z-2)} = \frac{1^2}{1-2} = -1\]
Residue at \(z = 2\):

\[\text{Res}(f, 2) = \lim_{z \to 2} (z-2) \frac{z^2}{(z-1)(z-2)} = \frac{2^2}{2-1} = 4\]
Result: \(\oint_C f(z) dz = 2\pi i(-1 + 4) = 6\pi i\)

Real Integrals using Residue Theorem

Type 1: \(\int_0^{2\pi} F(\cos\theta, \sin\theta) d\theta\)

Substitute: \(z = e^{i\theta}\), \(\cos\theta = \frac{z + z^{-1}}{2}\), \(\sin\theta = \frac{z - z^{-1}}{2i}\), \(d\theta = \frac{dz}{iz}\)

Transform to: \(\oint_{|z|=1} F\left(\frac{z + z^{-1}}{2}, \frac{z - z^{-1}}{2i}\right) \frac{dz}{iz}\)

Type 2: \(\int_{-\infty}^{\infty} f(x) dx\)

Use semicircular contour in upper half-plane. If \(f(z) \to 0\) as \(|z| \to \infty\) in upper half-plane:

\[\int_{-\infty}^{\infty} f(x) dx = 2\pi i \sum \text{Res}(f, z_k)\]

where \(z_k\) are poles in upper half-plane.

Solution of Integrals - Examples

Example 1

\(\int_0^{2\pi} \frac{d\theta}{2 + \cos\theta}\)

Let \(z = e^{i\theta}\), then \(\cos\theta = \frac{z + z^{-1}}{2}\):

\[\oint_{|z|=1} \frac{1}{2 + \frac{z + z^{-1}}{2}} \frac{dz}{iz} = \oint_{|z|=1} \frac{2dz}{iz(4 + z + z^{-1})} = \oint_{|z|=1} \frac{2dz}{i(z^2 + 4z + 1)}\]

Poles at \(z = -2 \pm \sqrt{3}\). Only \(z = -2 + \sqrt{3}\) is inside \(|z| = 1\).

Example 2

\(\int_{-\infty}^{\infty} \frac{dx}{x^2 + 1}\)

Poles at \(z = \pm i\). Only \(z = i\) is in upper half-plane.

\[\text{Res}(f, i) = \lim_{z \to i} (z-i) \frac{1}{z^2 + 1} = \frac{1}{2i}\]

\[\int_{-\infty}^{\infty} \frac{dx}{x^2 + 1} = 2\pi i \cdot \frac{1}{2i} = \pi\]

GATE Quick Tips

Key Formulas to Remember

Cauchy’s Formula: \(f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - z_0} dz\)
Residue Theorem: \(\oint_C f(z) dz = 2\pi i \sum \text{Res}(f, z_k)\)
Simple Pole Residue: \(\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)\)

Problem-Solving Strategy

Identify singularities and their types
Check which singularities are inside the contour
Calculate residues at relevant singularities
Apply residue theorem
For real integrals, choose appropriate contour

Common Mistakes to Avoid

Not checking if singularities are inside the contour
Confusing order of poles
Wrong substitution in trigonometric integrals