Complex Analysis: CO4 & CO5 - LNMIIT Study Guide

CO4

The Complex Plane

Interpret the geometry of the complex plane and understand the analyticity of complex valued functions

1a. What is a Complex Number?

A complex number extends the one-dimensional real number line into a two-dimensional plane. Instead of just moving left and right, we can now move in any direction!

Complex Number Representation

A complex number \(z\) can be written in multiple forms:

Rectangular: \(z = x + iy\) where \(x = \text{Re}(z)\), \(y = \text{Im}(z)\)
Polar: \(z = r(\cos\theta + i\sin\theta)\) where \(r = |z| = \sqrt{x^2 + y^2}\), \(\theta = \arg(z)\)
Euler's Form: \(z = re^{i\theta}\)

Interactive Complex Plane Explorer

Click anywhere on the canvas below to see all representations of a complex number:

Click on the canvas to explore complex numbers!

Euler's Formula: The Most Beautiful Equation

Euler's Formula

\[e^{i\theta} = \cos\theta + i\sin\theta\]

Why is this true? Consider the Taylor series expansions:

\[e^{i\theta} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \frac{(i\theta)^4}{4!} + \cdots\]
Since \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), we can separate real and imaginary parts:
\[= \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)\]
\[= \cos\theta + i\sin\theta\]

Rotating Vector Animation

Watch how \(e^{i\theta}\) rotates around the unit circle, with real and imaginary projections

De Moivre's Theorem

\[(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)\] Or equivalently: \[(e^{i\theta})^n = e^{in\theta}\]

Geometric Meaning: Raising a complex number to the \(n\)-th power multiplies its angle by \(n\) and raises its magnitude to the \(n\)-th power.

Interactive De Moivre Visualizer

Power n: 2

Adjust the slider to see how \(z^n\) rotates as \(n\) changes

Example: Computing \((1 + i)^{10}\)

Step 1: Convert to polar form
\(1 + i = \sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) = \sqrt{2}e^{i\pi/4}\)

Step 2: Apply De Moivre's theorem
\((1 + i)^{10} = (\sqrt{2})^{10}e^{i \cdot 10\pi/4} = 32e^{i5\pi/2}\)

Step 3: Simplify the angle
\(5\pi/2 = 2\pi + \pi/2\), so \(e^{i5\pi/2} = e^{i\pi/2} = i\)

Result: \((1 + i)^{10} = 32i\)

nth Roots of Unity

The \(n\)-th roots of unity are the solutions to \(z^n = 1\). They are equally spaced around the unit circle, separated by angles of \(2\pi/n\).

Formula for nth Roots of Unity

The \(n\)-th roots of unity are: \[z_k = e^{2\pi ik/n} = \cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}\] for \(k = 0, 1, 2, \ldots, n-1\)

Interactive nth Roots Visualizer

Number of roots (n): 5

Adjust the slider to see how the roots are distributed on the unit circle

1b. Functions of Complex Variables

A complex function \(f(z) = u(x,y) + iv(x,y)\) maps one complex plane to another. Unlike real functions that map a line to a line, complex functions map a plane to a plane!

Complex Function Representation

If \(z = x + iy\), then \(f(z) = u(x,y) + iv(x,y)\) where:

\(u(x,y)\) is the real part of \(f\)
\(v(x,y)\) is the imaginary part of \(f\)

Limit and Continuity in 2D

In the complex plane, a point can be approached from infinitely many directions! For \(\lim_{z \to z_0} f(z)\) to exist, the limit must be the same along ALL paths.

⚠️ Key Difference from Real Analysis

In real analysis, we only approach from left or right. In complex analysis, we can approach from any direction in the plane. This makes complex differentiability much more restrictive and powerful!

Example: \(f(z) = z^2\)

Let \(z = x + iy\), then:
\(f(z) = (x + iy)^2 = x^2 - y^2 + 2ixy\)

So \(u(x,y) = x^2 - y^2\) and \(v(x,y) = 2xy\)

This function maps circles to ellipses and rotates the plane!

1c. Differentiability & Analyticity: The Big Idea

Complex differentiability is MUCH stronger than real differentiability. If a complex function is differentiable at a point, it's automatically infinitely differentiable there!

The Cauchy-Riemann Equations

A function \(f(z) = u(x,y) + iv(x,y)\) is differentiable at \(z_0 = x_0 + iy_0\) if and only if the partial derivatives of \(u\) and \(v\) exist and satisfy: \[\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \text{and} \quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\]

Geometric Meaning: The Cauchy-Riemann equations ensure that the function preserves angles (conformal mapping) and has the same derivative regardless of the direction of approach.

Why C-R Equations Matter

The Cauchy-Riemann equations tell us that:

The function must preserve angles (conformal)
The function must preserve orientation
The Jacobian matrix has a special structure (it's a scaled rotation)
The function is infinitely differentiable if it's differentiable once!

Interactive C-R Equation Checker

Select Function:

Select a function to check C-R equations

Worked Example: Is \(f(z) = z^2\) analytic?

Step 1: Write in terms of \(x\) and \(y\)
\(f(z) = (x + iy)^2 = x^2 - y^2 + 2ixy\)
So \(u(x,y) = x^2 - y^2\) and \(v(x,y) = 2xy\)

Step 2: Compute partial derivatives
\(\frac{\partial u}{\partial x} = 2x\), \(\frac{\partial u}{\partial y} = -2y\)
\(\frac{\partial v}{\partial x} = 2y\), \(\frac{\partial v}{\partial y} = 2x\)

Step 3: Check C-R equations
\(\frac{\partial u}{\partial x} = 2x = \frac{\partial v}{\partial y}\) ✓
\(\frac{\partial u}{\partial y} = -2y = -\frac{\partial v}{\partial x}\) ✓

Conclusion: Both C-R equations are satisfied everywhere, so \(f(z) = z^2\) is entire (analytic everywhere in \(\mathbb{C}\))

Worked Example: Is \(f(z) = \bar{z}\) (conjugate) analytic?

Step 1: Write in terms of \(x\) and \(y\)
\(f(z) = \overline{x + iy} = x - iy\)
So \(u(x,y) = x\) and \(v(x,y) = -y\)

Step 2: Compute partial derivatives
\(\frac{\partial u}{\partial x} = 1\), \(\frac{\partial u}{\partial y} = 0\)
\(\frac{\partial v}{\partial x} = 0\), \(\frac{\partial v}{\partial y} = -1\)

Step 3: Check C-R equations
\(\frac{\partial u}{\partial x} = 1 \neq -1 = \frac{\partial v}{\partial y}\) ✗

Conclusion: The C-R equations are NOT satisfied, so \(f(z) = \bar{z}\) is nowhere analytic!

Definitions

Analytic at a point: \(f\) is differentiable in some neighborhood of that point
Entire function: \(f\) is analytic everywhere in \(\mathbb{C}\)
Holomorphic: Another word for analytic

1d. Harmonic Functions

If \(f = u + iv\) is analytic, then both \(u\) and \(v\) satisfy Laplace's equation. Such functions are called harmonic.

Harmonic Functions

A function \(u(x,y)\) is harmonic if it satisfies Laplace's equation: \[\nabla^2 u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\]

Key Fact: If \(f = u + iv\) is analytic, then both \(u\) and \(v\) are harmonic.

Harmonic Conjugate

Given a harmonic function \(u(x,y)\), we can find another harmonic function \(v(x,y)\) such that \(f = u + iv\) is analytic. We call \(v\) the harmonic conjugate of \(u\).

Worked Example: Find the harmonic conjugate of \(u(x,y) = x^2 - y^2\)

Step 1: Use the first C-R equation
\(\frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} = 2x\)

Step 2: Integrate with respect to \(y\)
\(v(x,y) = \int 2x \, dy = 2xy + g(x)\)
where \(g(x)\) is an arbitrary function of \(x\) alone

Step 3: Use the second C-R equation
\(\frac{\partial v}{\partial x} = 2y + g'(x) = -\frac{\partial u}{\partial y} = -(-2y) = 2y\)

Step 4: Solve for \(g(x)\)
\(2y + g'(x) = 2y \implies g'(x) = 0 \implies g(x) = C\) (constant)

Result: \(v(x,y) = 2xy + C\)
So \(f(z) = (x^2 - y^2) + i(2xy) = z^2\) (choosing \(C = 0\))

⚠️ Common Mistake

Don't forget the arbitrary function when integrating! The harmonic conjugate is unique only up to an additive constant.

1e. Elementary Complex Functions

1. The Complex Exponential: \(e^z\)

Complex Exponential

\[e^z = e^{x+iy} = e^x(\cos y + i\sin y) = e^x e^{iy}\]

Domain: All of \(\mathbb{C}\)
Range: \(\mathbb{C} \setminus \{0\}\) (all complex numbers except 0)
Analytic: Entire function
Periodic: \(e^{z + 2\pi i} = e^z\) (period \(2\pi i\))
Never zero: \(e^z \neq 0\) for any \(z\)

Example: Compute \(e^{1+i\pi}\)

\(e^{1+i\pi} = e^1 \cdot e^{i\pi} = e \cdot (\cos\pi + i\sin\pi) = e \cdot (-1 + 0i) = -e\)

2. Complex Trigonometric Functions

Complex Sine and Cosine

\[\sin z = \frac{e^{iz} - e^{-iz}}{2i}, \quad \cos z = \frac{e^{iz} + e^{-iz}}{2}\]

Key Difference from Real Trig: Complex sine and cosine are UNBOUNDED!

Domain: All of \(\mathbb{C}\)
Range: All of \(\mathbb{C}\)
Analytic: Entire functions
Example: \(\sin(iy) = i\sinh(y)\) can be arbitrarily large!

Example: Show that \(|\sin z|\) can be greater than 1

Let \(z = i\). Then:
\(\sin(i) = \frac{e^{i \cdot i} - e^{-i \cdot i}}{2i} = \frac{e^{-1} - e^{1}}{2i} = \frac{1/e - e}{2i}\)

\(|\sin(i)| = \frac{|e - 1/e|}{2} = \frac{e - 1/e}{2} \approx 1.175 > 1\)

3. Complex Logarithm: \(\log z\)

Complex Logarithm

\[\log z = \ln|z| + i\arg(z)\]

Multi-valued: Since \(\arg(z)\) is defined up to multiples of \(2\pi\), the logarithm is multi-valued!

Principal value: \(\text{Log } z = \ln|z| + i\text{Arg}(z)\) where \(-\pi < \text{Arg}(z) \leq \pi\)
Branch cut: Usually along the negative real axis
Analytic: In \(\mathbb{C} \setminus (-\infty, 0]\) (complex plane minus negative real axis)

Example: Find all values of \(\log(-1)\)

Step 1: Write in polar form
\(-1 = e^{i(\pi + 2\pi k)}\) for any integer \(k\)

Step 2: Apply logarithm
\(\log(-1) = i(\pi + 2\pi k) = i\pi(1 + 2k)\) for \(k \in \mathbb{Z}\)

Principal value: \(\text{Log}(-1) = i\pi\) (taking \(k = 0\))

⚠️ Branch Cuts

Multi-valued functions like \(\log z\) and \(\sqrt{z}\) require us to choose a "branch" to make them single-valued. The branch cut is where we make a discontinuity to avoid going around the origin multiple times.

CO5

Contour Integrals & Residue Theory

Explain fundamental concepts of contour integrals and evaluate improper real integrals using residue theory

2a. Path Integrals — The Big Picture

A contour integral is an integral along a path in the complex plane. Unlike real integrals that go from left to right, contour integrals can follow any curve!

Contour Integral Definition

Let \(C\) be a smooth curve parameterized by \(z(t) = x(t) + iy(t)\) for \(a \leq t \leq b\). Then: \[\int_C f(z) \, dz = \int_a^b f(z(t)) z'(t) \, dt\]

Steps to compute:

Parameterize the contour \(C\): \(z = z(t)\)
Compute \(dz = z'(t) dt\)
Substitute into the integral: \(\int_a^b f(z(t)) z'(t) \, dt\)
Evaluate the resulting integral

Interactive Contour Visualizer

Watch as the contour is traced in the complex plane

ML Inequality

ML Inequality (Estimation Lemma)

If \(|f(z)| \leq M\) for all \(z\) on contour \(C\), and \(L\) is the length of \(C\), then: \[\left|\int_C f(z) \, dz\right| \leq M \cdot L\]

Why it's useful: Helps us bound integrals without computing them exactly, especially useful for showing that certain integrals vanish as the contour grows.

Worked Example: Compute \(\int_C z \, dz\) along two different paths

Path 1: Straight line from \(0\) to \(1+i\)
Parameterize: \(z(t) = t(1+i)\) for \(0 \leq t \leq 1\)
\(z'(t) = 1+i\)
\(\int_C z \, dz = \int_0^1 t(1+i)(1+i) \, dt = (1+i)^2 \int_0^1 t \, dt = 2i \cdot \frac{1}{2} = i\)

Path 2: From \(0\) to \(1\), then from \(1\) to \(1+i\)
First segment: \(z(t) = t\) for \(0 \leq t \leq 1\), \(z'(t) = 1\)
\(\int_0^1 t \cdot 1 \, dt = \frac{1}{2}\)
Second segment: \(z(t) = 1 + it\) for \(0 \leq t \leq 1\), \(z'(t) = i\)
\(\int_0^1 (1+it) \cdot i \, dt = i\int_0^1 (1+it) \, dt = i\left[t + \frac{it^2}{2}\right]_0^1 = i(1 + \frac{i}{2}) = i - \frac{1}{2}\)
Total: \(\frac{1}{2} + i - \frac{1}{2} = i\)

Conclusion: Both paths give the same result! This is because \(f(z) = z\) is analytic, so the integral is path-independent.

2b. Cauchy-Goursat Theorem

Cauchy-Goursat Theorem

If \(f\) is analytic in and on a simple closed contour \(C\), then: \[\oint_C f(z) \, dz = 0\]

Geometric Intuition: Analytic functions have "no curl" — they're like conservative vector fields. Going around a closed loop brings you back to where you started with no net change.

When Does It Fail?

The theorem fails when there's a singularity (point where \(f\) is not analytic) inside the contour. This is actually the key to the Residue Theorem!

Example: \(\oint_{|z|=1} \frac{1}{z} \, dz\)

The function \(f(z) = \frac{1}{z}\) has a singularity at \(z = 0\), which is inside the contour.

Parameterize: \(z = e^{i\theta}\) for \(0 \leq \theta \leq 2\pi\)
\(dz = ie^{i\theta} d\theta\)

\(\oint_{|z|=1} \frac{1}{z} \, dz = \int_0^{2\pi} \frac{1}{e^{i\theta}} \cdot ie^{i\theta} \, d\theta = \int_0^{2\pi} i \, d\theta = 2\pi i \neq 0\)

2c. Cauchy Integral Formula

Cauchy Integral Formula

If \(f\) is analytic inside and on a simple closed contour \(C\), and \(z_0\) is inside \(C\), then: \[f(z_0) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - z_0} \, dz\]

Why This Is Incredible: You can find the value of \(f\) anywhere inside just from knowing its values on the boundary! This has no analogue in real analysis.

Generalized Cauchy Integral Formula

The \(n\)-th derivative of \(f\) at \(z_0\) is: \[f^{(n)}(z_0) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}} \, dz\]

Worked Example: Compute \(\oint_{|z|=2} \frac{e^z}{z-1} \, dz\)

Step 1: Identify the form
This matches \(\oint_C \frac{f(z)}{z - z_0} \, dz\) with \(f(z) = e^z\) and \(z_0 = 1\)

Step 2: Check if \(z_0\) is inside \(C\)
\(|1| = 1 < 2\), so \(z_0=1\) is inside \(|z|=2\) ✓

Step 3: Apply Cauchy Integral Formula
\(\oint_{|z|=2} \frac{e^z}{z-1} \, dz = 2\pi i \cdot f(1) = 2\pi i \cdot e^1 = 2\pi i e\)

2d. Power Series

Power Series

A power series centered at \(z_0\) has the form: \[f(z) = \sum_{n=0}^{\infty} a_n(z - z_0)^n\] It converges inside a circle of radius \(R\) (the radius of convergence).

Key Theorem: Analyticity ⟺ Power Series

A function is analytic at \(z_0\) if and only if it can be represented by a power series in some neighborhood of \(z_0\). This is much stronger than in real analysis!

Taylor Series for Common Functions

\[e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!} = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\] \[\sin z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!} = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots\] \[\cos z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots\] \[\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n = 1 + z + z^2 + z^3 + \cdots \quad (|z| < 1)\]

Interactive Taylor Series Visualizer

Number of terms: 5

Watch how the Taylor series for \(e^z\) converges as we add more terms

2e. Laurent Series: The Key to Residues

Laurent series extend Taylor series by allowing negative powers. This is crucial for understanding singularities and computing residues!

Laurent Series

In an annulus \(r < |z - z_0| < R\), a function can be represented as: \[f(z)=\sum_{n=-\infty}^{\infty} a_n(z - z_0)^n=\underbrace{\sum_{n=-\infty}^{-1} a_n(z - z_0)^n}_{\text{Principal Part}} + \underbrace{\sum_{n=0}^{\infty} a_n(z - z_0)^n}_{\text{Analytic Part}}\]

The coefficient \(a_{-1}\) is called the residue — it's the key to everything!

Three Methods to Find Laurent Series

Method 1: Direct Substitution Using Known Series

Example: Find Laurent series of \(f(z) = \frac{e^z}{z^2}\) about \(z = 0\)

Step 1: Use Taylor series for \(e^z\)
\(e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\)

Step 2: Divide by \(z^2\)
\(\frac{e^z}{z^2} = \frac{1}{z^2} + \frac{1}{z} + \frac{1}{2!} + \frac{z}{3!} + \frac{z^2}{4!} + \cdots\)

Result: Principal part: \(\frac{1}{z^2} + \frac{1}{z}\), Residue: \(a_{-1} = 1\)

Method 2: Partial Fractions

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

Step 1: Partial fraction decomposition
\(\frac{1}{z(z-1)} = \frac{A}{z} + \frac{B}{z-1}\)
Solving: \(A = -1\), \(B = 1\)

Step 2: Expand each term
\(\frac{1}{z(z-1)} = -\frac{1}{z} + \frac{1}{z-1} = -\frac{1}{z} - \frac{1}{1-z}\)

Step 3: Use geometric series for \(|z| < 1\)
\(-\frac{1}{z} - (1 + z + z^2 + z^3 + \cdots) = -\frac{1}{z} - 1 - z - z^2 - \cdots\)

Result: Residue: \(a_{-1} = -1\)

Method 3: Long Division

Example: Find Laurent series of \(f(z) = \frac{z}{z^2-1}\) about \(z = 0\)

Step 1: Rewrite
\(\frac{z}{z^2-1} = \frac{z}{-(1-z^2)} = -z \cdot \frac{1}{1-z^2}\)

Step 2: Use geometric series
\(-z(1 + z^2 + z^4 + z^6 + \cdots) = -z - z^3 - z^5 - z^7 - \cdots\)

Result: No principal part, so residue = 0

2f. Classification of Singularities

Understanding singularities is crucial for applying the Residue Theorem. There are three types:

Types of Isolated Singularities

1. Removable Singularity

All negative power coefficients are zero: \(a_{-n} = 0\) for all \(n \geq 1\)

Example: \(f(z) = \frac{\sin z}{z}\) at \(z = 0\)

\(\frac{\sin z}{z} = \frac{1}{z}\left(z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots\right) = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots\)

No negative powers! Can define \(f(0) = 1\) to make it analytic.

2. Pole of Order m

Finitely many negative terms, highest is \((z-z_0)^{-m}\)

Examples:

\(\frac{1}{z^2}\) has a pole of order 2 at \(z = 0\)
\(\frac{1}{(z-1)^3}\) has a pole of order 3 at \(z = 1\)
\(\frac{z+1}{z(z-2)}\) has simple poles (order 1) at \(z = 0\) and \(z = 2\)

3. Essential Singularity

Infinitely many negative terms

Example: \(f(z) = e^{1/z}\) at \(z = 0\)

\(e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2!z^2} + \frac{1}{3!z^3} + \cdots\)

Infinitely many negative powers!

Visual Memory Aid

Removable

No negative powers

... + a₀ + a₁z + a₂z² + ...

Pole

Finite negative powers

a₋ₘ/zᵐ + ... + a₋₁/z + a₀ + ...

Essential

Infinite negative powers

... + a₋₃/z³ + a₋₂/z² + a₋₁/z + ...

⚠️ How to Identify Singularity Type

Find the Laurent series expansion
Look at the principal part (negative powers)
Count: 0 negative terms → removable, finite → pole, infinite → essential

2g. The Residue Theorem: The Grand Finale

The Residue Theorem is one of the most powerful tools in complex analysis. It connects contour integrals to the residues at singularities.

Cauchy Residue Theorem

If \(f\) is analytic inside and on a simple closed contour \(C\) except for isolated singularities at \(z_1, z_2, \ldots, z_n\) inside \(C\), then: \[\oint_C f(z) \, dz = 2\pi i \sum_{k=1}^{n} \text{Res}(f, z_k)\]

In words: The integral around a closed contour equals \(2\pi i\) times the sum of all residues inside!

Quick Formulas for Computing Residues

Residue Formulas

1. Simple Pole (order 1): \[\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)\] 2. Pole of Order m: \[\text{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}}\left[(z - z_0)^m f(z)\right]\] 3. From Laurent Series: \[\text{Res}(f, z_0) = a_{-1} \text{ (coefficient of } \frac{1}{z-z_0}\text{)}\]

Worked Example 1: \(\oint_{|z|=3} \frac{1}{(z-1)(z-2)} \, dz\)

Step 1: Identify singularities
Simple poles at \(z = 1\) and \(z = 2\), both inside \(|z| = 3\)

Step 2: Compute residue at \(z = 1\)
\(\text{Res}(f, 1) = \lim_{z \to 1} (z-1) \cdot \frac{1}{(z-1)(z-2)} = \lim_{z \to 1} \frac{1}{z-2} = \frac{1}{1-2} = -1\)

Step 3: Compute residue at \(z = 2\)
\(\text{Res}(f, 2) = \lim_{z \to 2} (z-2) \cdot \frac{1}{(z-1)(z-2)} = \lim_{z \to 2} \frac{1}{z-1} = \frac{1}{2-1} = 1\)

Step 4: Apply Residue Theorem
\(\oint_{|z|=3} \frac{1}{(z-1)(z-2)} \, dz = 2\pi i(-1 + 1) = 0\)

Worked Example 2: \(\oint_{|z|=1} \frac{e^z}{z^3} \, dz\)

Step 1: Identify singularity
Pole of order 3 at \(z = 0\)

Step 2: Find Laurent series
\(\frac{e^z}{z^3} = \frac{1}{z^3}\left(1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots\right) = \frac{1}{z^3} + \frac{1}{z^2} + \frac{1}{2!z} + \frac{1}{3!} + \cdots\)

Step 3: Extract residue
\(\text{Res}(f, 0) = a_{-1} = \frac{1}{2!} = \frac{1}{2}\)

Step 4: Apply Residue Theorem
\(\oint_{|z|=1} \frac{e^z}{z^3} \, dz = 2\pi i \cdot \frac{1}{2} = \pi i\)

Interactive Residue Finder

Select Function:

Select a function to find its residue

2h. Evaluating Real Improper Integrals: The Payoff!

This is WHY engineers and physicists learn complex analysis! We can evaluate difficult real integrals by cleverly using contour integration in the complex plane.

Type 1: Rational Functions

Method for \(\int_{-\infty}^{\infty} \frac{P(x)}{Q(x)} dx\)

Requirements:

Degree of \(Q\) is at least 2 more than degree of \(P\)
\(Q(x) \neq 0\) for real \(x\)

Method:

Close the contour with a semicircle in the upper half-plane
Find all poles in the upper half-plane
Compute residues at those poles
Apply: \(\int_{-\infty}^{\infty} \frac{P(x)}{Q(x)} dx = 2\pi i \sum \text{Residues}\)

Worked Example: \(\int_{-\infty}^{\infty} \frac{1}{1+x^2} dx = \pi\)

Step 1: Extend to complex function
\(f(z) = \frac{1}{1+z^2} = \frac{1}{(z-i)(z+i)}\)

Step 2: Identify poles
Poles at \(z = i\) and \(z = -i\). Only \(z = i\) is in the upper half-plane.

Step 3: Compute residue at \(z = i\)
\(\text{Res}(f, i) = \lim_{z \to i} (z-i) \cdot \frac{1}{(z-i)(z+i)} = \frac{1}{i+i} = \frac{1}{2i}\)

Step 4: Apply Residue Theorem
\(\int_{-\infty}^{\infty} \frac{1}{1+x^2} dx = 2\pi i \cdot \frac{1}{2i} = \pi\)

Verification: This matches \(\arctan(x)\big|_{-\infty}^{\infty} = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi\) ✓

Type 2: Integrals with Trigonometric Functions

Method for \(\int_{-\infty}^{\infty} f(x) \cos(ax) dx\) or \(\int_{-\infty}^{\infty} f(x) \sin(ax) dx\)

Method:

Consider \(\int_{-\infty}^{\infty} f(x) e^{iax} dx\)
Close contour in upper half-plane (if \(a > 0\))
Use Jordan's Lemma to show semicircular part vanishes
Extract real or imaginary part for cosine or sine

Worked Example: \(\int_{-\infty}^{\infty} \frac{\cos x}{1+x^2} dx = \frac{\pi}{e}\)

Step 1: Consider complex version
\(\int_{-\infty}^{\infty} \frac{e^{ix}}{1+x^2} dx\), then take real part

Step 2: Extend to \(f(z) = \frac{e^{iz}}{1+z^2}\)
Pole at \(z = i\) in upper half-plane

Step 3: Compute residue
\(\text{Res}(f, i) = \lim_{z \to i} (z-i) \cdot \frac{e^{iz}}{(z-i)(z+i)} = \frac{e^{i \cdot i}}{2i} = \frac{e^{-1}}{2i}\)

Step 4: Apply Residue Theorem
\(\int_{-\infty}^{\infty} \frac{e^{ix}}{1+x^2} dx = 2\pi i \cdot \frac{e^{-1}}{2i} = \pi e^{-1}\)

Step 5: Take real part
\(\int_{-\infty}^{\infty} \frac{\cos x}{1+x^2} dx = \text{Re}\left(\pi e^{-1}\right) = \frac{\pi}{e}\)

⚠️ Jordan's Lemma

When using \(e^{iaz}\) with \(a > 0\), close the contour in the upper half-plane. Jordan's Lemma guarantees that the integral over the semicircular arc vanishes as the radius goes to infinity.

Section 4

Exam Preparation

Key theorems, common question types, and mistakes to avoid for Quiz-2

Key Theorems to Memorize (Flashcards)

Click each card to reveal the theorem statement or application

Common Mistakes to Avoid

⚠️ Mistake 1: Forgetting Multi-valuedness

\(\log z\) and \(\sqrt{z}\) are multi-valued! Always specify which branch you're using.

⚠️ Mistake 2: Wrong Order in C-R Equations

It's \(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\) and \(\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\). Note the negative sign!

⚠️ Mistake 3: Forgetting to Check Which Poles Are Inside

Only include residues of poles INSIDE the contour. Check \(|z_k| < R\) for \(|z|=R\).

⚠️ Mistake 4: Confusing Pole Order with Residue

A pole of order 3 doesn't mean residue = 3! You must compute \(a_{-1}\) from Laurent series.

⚠️ Mistake 5: Wrong Contour Direction

Residue Theorem assumes counterclockwise orientation. If clockwise, add a negative sign!

Quick Reference Formula Sheet

        Euler's Formula: \(e^{i\theta} = \cos\theta + i\sin\theta\)

        De Moivre: \((\cos\theta + i\sin\theta)^n = \cos(n\theta) +
        i\sin(n\theta)\)

        C-R Equations: \(\frac{\partial u}{\partial x} = \frac{\partial
        v}{\partial y}\), \(\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}\)

        Cauchy-Goursat: \(\oint_C f(z) dz = 0\) if \(f\) analytic in and on
        \(C\)

        Cauchy Integral: \(f(z_0) = \frac{1}{2\pi i}\oint_C
        \frac{f(z)}{z-z_0} dz\)

        Residue Theorem: \(\oint_C f(z) dz = 2\pi i \sum \text{Res}(f,
        z_k)\)

        Simple Pole Residue: \(\text{Res}(f, z_0) = \lim_{z \to z_0}
        (z-z_0)f(z)\)

        ML Inequality: \(|\int_C f(z) dz| \leq M \cdot L\)
      

CO4 & CO5 Study Guide

The Complex Plane

1a. What is a Complex Number?

Complex Number Representation

Interactive Complex Plane Explorer

Euler's Formula: The Most Beautiful Equation

Euler's Formula

Rotating Vector Animation

De Moivre's Theorem

De Moivre's Theorem

Interactive De Moivre Visualizer

Example: Computing \((1 + i)^{10}\)

nth Roots of Unity

Formula for nth Roots of Unity

Interactive nth Roots Visualizer

1b. Functions of Complex Variables

Complex Function Representation

Limit and Continuity in 2D

⚠️ Key Difference from Real Analysis

Example: \(f(z) = z^2\)

1c. Differentiability & Analyticity: The Big Idea

The Cauchy-Riemann Equations

Why C-R Equations Matter

Interactive C-R Equation Checker

Worked Example: Is \(f(z) = z^2\) analytic?

Worked Example: Is \(f(z) = \bar{z}\) (conjugate) analytic?

Definitions

1d. Harmonic Functions

Harmonic Functions

Harmonic Conjugate

Worked Example: Find the harmonic conjugate of \(u(x,y) = x^2 - y^2\)

⚠️ Common Mistake

1e. Elementary Complex Functions

1. The Complex Exponential: \(e^z\)

Complex Exponential

Example: Compute \(e^{1+i\pi}\)

2. Complex Trigonometric Functions

Complex Sine and Cosine

Example: Show that \(|\sin z|\) can be greater than 1

3. Complex Logarithm: \(\log z\)

Complex Logarithm

Example: Find all values of \(\log(-1)\)

⚠️ Branch Cuts

Contour Integrals & Residue Theory

2a. Path Integrals — The Big Picture

Contour Integral Definition

Interactive Contour Visualizer

ML Inequality

ML Inequality (Estimation Lemma)

Worked Example: Compute \(\int_C z \, dz\) along two different paths

2b. Cauchy-Goursat Theorem

Cauchy-Goursat Theorem

When Does It Fail?

Example: \(\oint_{|z|=1} \frac{1}{z} \, dz\)

2c. Cauchy Integral Formula

Cauchy Integral Formula

Generalized Cauchy Integral Formula

Worked Example: Compute \(\oint_{|z|=2} \frac{e^z}{z-1} \, dz\)

2d. Power Series

Power Series

Key Theorem: Analyticity ⟺ Power Series

Taylor Series for Common Functions

Interactive Taylor Series Visualizer

2e. Laurent Series: The Key to Residues

Laurent Series

Three Methods to Find Laurent Series

Method 1: Direct Substitution Using Known Series

Method 2: Partial Fractions

Method 3: Long Division

2f. Classification of Singularities

Types of Isolated Singularities

1. Removable Singularity

2. Pole of Order m

3. Essential Singularity

Visual Memory Aid

Removable

Pole

Essential

⚠️ How to Identify Singularity Type

2g. The Residue Theorem: The Grand Finale