Complex Variables Notes for GATE Electrical Engineering (EE)

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 \(z_0\)

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

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

• Polynomials: \(z^n, az + b\)

• Exponential: \(e^z\)

• Trigonometric: \(\sin z, \cos z\)

• Logarithmic: \(\log z\) (with branch cuts)

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

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

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

• Evaluate integrals of the form \(\oint_C \frac{f(z)}{z - z_0} dz\)

• Find derivatives of analytic functions

• Prove maximum modulus principle

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

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:

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

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

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

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

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

Using partial fractions:

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

Therefore:

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:

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

If \(z_0\) is a simple pole:

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

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

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: \(\oint_C f(z) dz = 2\pi i(-1 + 4) = 6\pi i\)

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

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

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

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

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

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

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

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