Complex Variables Notes for GATE Electrical Engineering (EE)

Analytic Functions

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

  1. Sum, difference, product of analytic functions is analytic

  2. Quotient is analytic (where denominator \(\neq 0\))

  3. Composition of analytic functions is analytic

  4. 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

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

  1. Path independence: \(\int_{C_1} f(z)dz = \int_{C_2} f(z)dz\) for same endpoints

  2. Deformation of contours: Can deform contour without changing integral value

  3. 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

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

  1. \(f(z) = e^z\) is analytic everywhere

  2. Singularity at \(z_0 = i\), and \(|i| = 1 < 2\), so \(i\) is inside \(C\)

  3. 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

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

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

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

  1. Removable: \(\lim_{z \to z_0} f(z)\) exists and is finite

  2. Pole of order \(m\): \(\lim_{z \to z_0} (z-z_0)^m f(z)\) exists and \(\neq 0\)

  3. 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

  1. Singularities: \(z = 1\) (simple pole), \(z = 2\) (simple pole)

  2. Both are inside \(|z| = 3\)

  3. Residue at \(z = 1\):

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

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

Solution of Integrals

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

  1. Identify singularities and their types

  2. Check which singularities are inside the contour

  3. Calculate residues at relevant singularities

  4. Apply residue theorem

  5. 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