Introduction to Complex Analysis

Complex analysis studies functions of a complex variable, and it is one of the most elegant subjects in mathematics. It underpins electrical engineering, fluid dynamics, signal processing and quantum physics. This introduction builds from complex arithmetic to analytic functions and contour integrals, with every step of each example shown.

Complex Numbers Refresher

A complex number has the form \( z = x + iy \), where \( x, y \in \mathbb{R} \) and \( i^2 = -1 \). We call \( x = \text{Re}(z) \) the real part and \( y = \text{Im}(z) \) the imaginary part.

Key operations:

• Conjugate: \( \bar{z} = x - iy \).
• Modulus: \( |z| = \sqrt{x^2 + y^2} \), the distance from the origin.
• Useful identity: \( z\bar{z} = x^2 + y^2 = |z|^2 \).

Worked Example 1: Complex Arithmetic

Find (a) \( (2 + 3i)(1 - i) \) and (b) \( \dfrac{3 + 4i}{1 - 2i} \).

(a) Multiply by expanding brackets:

$$ (2 + 3i)(1 - i) = 2 - 2i + 3i - 3i^2 $$

Since \( i^2 = -1 \), \( -3i^2 = 3 \):

$$ = 2 + 3 + (-2i + 3i) = 5 + i $$

(b) Divide by multiplying top and bottom by the conjugate of the denominator, \( 1 + 2i \):

$$ \frac{3 + 4i}{1 - 2i} \cdot \frac{1 + 2i}{1 + 2i} = \frac{(3 + 4i)(1 + 2i)}{1^2 + 2^2} $$

The numerator is

$$ 3 + 6i + 4i + 8i^2 = 3 + 10i - 8 = -5 + 10i $$

So

$$ \frac{-5 + 10i}{5} = -1 + 2i $$

Check: \( (-1 + 2i)(1 - 2i) = -1 + 2i + 2i - 4i^2 = -1 + 4i + 4 = 3 + 4i \). Correct.

Polar Form and Euler's Formula

Any non-zero complex number can be written using its modulus \( r = |z| \) and argument \( \theta \) (the angle from the positive real axis):

$$ z = r(\cos\theta + i\sin\theta) = re^{i\theta} $$

The second form uses Euler's formula, \( e^{i\theta} = \cos\theta + i\sin\theta \). Setting \( \theta = \pi \) gives the famous identity \( e^{i\pi} + 1 = 0 \).

Polar form makes multiplication easy: multiply the moduli and add the arguments.

De Moivre's Theorem

$$ (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta) $$

Worked Example 2: Powers Using De Moivre

Compute \( (1 + i)^8 \).

Step 1: Modulus.

$$ r = \sqrt{1^2 + 1^2} = \sqrt{2} $$

Step 2: Argument. The point \( (1, 1) \) lies in the first quadrant, so \( \theta = \tan^{-1}(1) = \frac{\pi}{4} \).

Step 3: Polar form.

$$ 1 + i = \sqrt{2}\, e^{i\pi/4} $$

Step 4: Raise to the 8th power.

$$ (1 + i)^8 = (\sqrt{2})^8 \, e^{i \cdot 8\pi/4} = 16 \, e^{2\pi i} $$

since \( (\sqrt{2})^8 = 2^4 = 16 \).

Step 5: Simplify. \( e^{2\pi i} = \cos 2\pi + i \sin 2\pi = 1 \), so

$$ (1 + i)^8 = 16 $$

Expanding with the binomial theorem would take far longer.

Analytic Functions and the Cauchy–Riemann Equations

Write a complex function as \( f(z) = u(x, y) + iv(x, y) \). The derivative

$$ f'(z) = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h} $$

must give the same value whichever direction \( h \) approaches 0 from in the complex plane. This is a much stronger requirement than real differentiability.

A function differentiable throughout an open region is called analytic (or holomorphic). If \( u \) and \( v \) have continuous partial derivatives, \( f \) is analytic exactly when the Cauchy–Riemann equations hold:

$$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} $$

Worked Example 3: Testing for Analyticity

(a) Is \( f(z) = z^2 \) analytic?

Expand with \( z = x + iy \):

$$ z^2 = x^2 + 2ixy + i^2 y^2 = (x^2 - y^2) + i(2xy) $$

So \( u = x^2 - y^2 \) and \( v = 2xy \). The partial derivatives are

$$ u_x = 2x, \quad v_y = 2x, \quad u_y = -2y, \quad v_x = 2y $$

Then \( u_x = v_y \) and \( u_y = -2y = -v_x \). Both equations hold everywhere, so \( z^2 \) is analytic on all of \( \mathbb{C} \), with \( f'(z) = u_x + iv_x = 2x + 2iy = 2z \).

(b) Is \( g(z) = \bar{z} \) analytic?

Here \( u = x \) and \( v = -y \), so \( u_x = 1 \) but \( v_y = -1 \). The first equation fails everywhere, so \( \bar{z} \) is not analytic anywhere, even though it is perfectly smooth as a map of the plane.

A First Contour Integral

Integrals in complex analysis are taken along curves called contours. Let \( C \) be the unit circle traversed anticlockwise, parametrised by \( z = e^{it} \) for \( 0 \leq t \leq 2\pi \). Then \( dz = ie^{it}\, dt \), and

$$ \oint_C \frac{1}{z}\, dz = \int_0^{2\pi} \frac{ie^{it}}{e^{it}}\, dt = \int_0^{2\pi} i \, dt = 2\pi i $$

By contrast, Cauchy's theorem states that if \( f \) is analytic on and inside a closed contour, then \( \oint_C f(z)\, dz = 0 \). The non-zero answer above arises because \( \frac{1}{z} \) is not analytic at \( z = 0 \), which lies inside the circle. This contrast leads directly to the residue theorem, one of the most powerful tools for evaluating real integrals.

Common Mistakes

• Forgetting \( i^2 = -1 \) when expanding products.
• Choosing the wrong quadrant for the argument. \( \tan^{-1}(y/x) \) alone cannot distinguish \( 1 + i \) from \( -1 - i \).
• Assuming smooth means analytic. \( \bar{z} \) and \( |z|^2 \) are smooth but fail Cauchy–Riemann.
• Ignoring orientation. Reversing a contour changes the sign of the integral.

Key Takeaways

• Divide complex numbers by multiplying by the conjugate of the denominator.
• Polar form \( re^{i\theta} \) and De Moivre's theorem make powers and roots simple.
• A function is analytic when it satisfies the Cauchy–Riemann equations (with continuous partials).
• \( \oint_C \frac{1}{z}\,dz = 2\pi i \) around the unit circle, while integrals of analytic functions around closed contours vanish.

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