Analytic Functions
Analytic Functions - Definition
Definition
A function \(f(z)\) is analytic at point \(z_0\) if:
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\(f(z)\) is differentiable at \(z_0\)
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\(f(z)\) is differentiable in some neighborhood of \(z_0\)
Cauchy-Riemann Equations
If \(f(z) = u(x,y) + iv(x,y)\) is analytic, then:
Key Point
Analytic functions are also called holomorphic or regular functions.
Properties of Analytic Functions
Important Properties
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Sum, difference, product of analytic functions is analytic
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Quotient is analytic (where denominator \(\neq 0\))
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Composition of analytic functions is analytic
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If \(f(z)\) is analytic and \(f'(z) \neq 0\), then \(f^{-1}(z)\) is analytic
Common Analytic Functions
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Polynomials: \(z^n, az + b\)
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Exponential: \(e^z\)
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Trigonometric: \(\sin z, \cos z\)
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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:
Consequences
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Path independence: \(\int_{C_1} f(z)dz = \int_{C_2} f(z)dz\) for same endpoints
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Deformation of contours: Can deform contour without changing integral value
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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:
Generalized Formula (Derivatives)
Key Applications
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Evaluate integrals of the form \(\oint_C \frac{f(z)}{z - z_0} dz\)
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Find derivatives of analytic functions
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Prove maximum modulus principle
Cauchy’s Formula - Example
Example
Evaluate \(\oint_C \frac{e^z}{z - i} dz\) where \(C: |z| = 2\)
Solution
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\(f(z) = e^z\) is analytic everywhere
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Singularity at \(z_0 = i\), and \(|i| = 1 < 2\), so \(i\) is inside \(C\)
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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 \(|z - z_0| < R\):
Radius of Convergence
\(R\) = distance from \(z_0\) to nearest singularity
Common Taylor Series
Laurent Series
Laurent Series
Laurent Series Expansion
If \(f(z)\) is analytic in an annulus \(r < |z - z_0| < R\), then:
Coefficients
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:
For \(0 < |z| < 1\):
Therefore:
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:
Types of Singularities
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Removable: \(\lim_{z \to z_0} f(z)\) exists and is finite
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Pole of order \(m\): \(\lim_{z \to z_0} (z-z_0)^m f(z)\) exists and \(\neq 0\)
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Essential: Neither removable nor pole
Residue Calculation Methods
Simple Pole (\(m = 1\))
If \(z_0\) is a simple pole:
Pole of Order \(m\)
Rational Functions
If \(f(z) = \frac{P(z)}{Q(z)}\) where \(P(z_0) \neq 0\) and \(Q(z_0) = 0\), \(Q'(z_0) \neq 0\):
Residue Theorem - Example
Example
Evaluate \(\oint_C \frac{z^2}{(z-1)(z-2)} dz\) where \(C: |z| = 3\)
Solution
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Singularities: \(z = 1\) (simple pole), \(z = 2\) (simple pole)
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Both are inside \(|z| = 3\)
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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: \(\oint_C f(z) dz = 2\pi i(-1 + 4) = 6\pi i\)
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:
Solution of Integrals - Examples
Example 1
\(\int_0^{2\pi} \frac{d\theta}{2 + \cos\theta}\)
Let \(z = e^{i\theta}\), then \(\cos\theta = \frac{z + z^{-1}}{2}\):
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.
GATE Quick Tips
Key Formulas to Remember
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Cauchy’s Formula: \(f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - z_0} dz\)
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Residue Theorem: \(\oint_C f(z) dz = 2\pi i \sum \text{Res}(f, z_k)\)
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Simple Pole Residue: \(\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)\)
Problem-Solving Strategy
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Identify singularities and their types
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Check which singularities are inside the contour
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Calculate residues at relevant singularities
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Apply residue theorem
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For real integrals, choose appropriate contour
Common Mistakes to Avoid
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Not checking if singularities are inside the contour
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Confusing order of poles
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Wrong substitution in trigonometric integrals