Complex Numbers

Cartesian Form

Definition

A complex number $z$ is written:

$z = a + bi, \quad a, b \in \mathbb{'\{'}R{'\}'}$

where $a = \mathrm{Re}(z)$ is the real part and $b = \mathrm{Im}(z)$ is the imaginary part. The set of all complex numbers is $\mathbb{'\{'}C{'\}'}$.

Complex Arithmetic

Addition. $(a + bi) + (c + di) = (a + c) + (b + d)i$

Multiplication. $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$

Division. Multiply by the conjugate of the denominator:

$\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{c^2 + d^2} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}$

Complex Conjugate

The conjugate of $z = a + bi$ is $\bar{z} = a - bi$.

Properties:

• $z\bar{z} = a^2 + b^2 = |z|^2$
• $\overline{z_1 + z_2} = \bar{z}_1 + \bar{z}_2$
• $\overline{z_1 z_2} = \bar{z}_1 \bar{z}_2$
• $z + \bar{z} = 2\mathrm{Re}(z)$, $z - \bar{z} = 2i\,\mathrm{Im}(z)$

Modulus and Argument

The modulus (absolute value) of $z = a + bi$:

$|z| = \sqrt{a^2 + b^2}$

The argument $\arg(z) = \theta$ is the angle from the positive real axis, measured anticlockwise, where:

$\tan\theta = \frac{b}{a}$

The principal argument $\mathrm{Arg}(z)$ satisfies $-\pi \lt \theta \le \pi$. Use the quadrant of $(a, b)$ to determine the correct angle.

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Polar Form

Exponential Representation

Any non-zero complex number can be written in polar form:

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

where $r = |z|$ and $\theta = \arg(z)$.

Conversion

• Cartesian to polar: $r = \sqrt{a^2 + b^2}$, $\theta = \arctan\!\left(\dfrac{b}{a}\right)$ (adjusting for quadrant).
• Polar to Cartesian: $a = r\cos\theta$, $b = r\sin\theta$.

Example. Express $z = 1 + i\sqrt{3}$ in polar form.

$r = \sqrt{1 + 3} = 2, \qquad \theta = \arctan\!\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$

$z = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 2e^{i\pi/3}$

Multiplication and Division in Polar Form

If $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$:

$z_1 z_2 = r_1 r_2\, e^{i(\theta_1 + \theta_2)}$

$\frac{z_1}{z_2} = \frac{r_1}{r_2}\, e^{i(\theta_1 - \theta_2)}$

Multiplication: moduli multiply, arguments add. Division: moduli divide, arguments subtract.

Example. Compute $\dfrac{1 + i}{\sqrt{3} - i}$.

$1 + i = \sqrt{2}\,e^{i\pi/4}$, $\sqrt{3} - i = 2\,e^{-i\pi/6}$.

$\frac{1 + i}{\sqrt{3} - i} = \frac{\sqrt{2}}{2}\, e^{i(\pi/4 + \pi/6)} = \frac{\sqrt{2}}{2}\, e^{i5\pi/12}$

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De Moivre's Theorem

Statement

For any integer $n$:

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

In exponential notation: $\bigl(e^{i\theta}\bigr)^n = e^{in\theta}$.

Applications to Trigonometry

De Moivre's theorem allows us to express $\cos n\theta$ and $\sin n\theta$ as polynomials in $\cos\theta$ and $\sin\theta$.

Example. Express $\cos 3\theta$ and $\sin 3\theta$ in terms of $\cos\theta$ and $\sin\theta$.

$(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$

Expanding the left side using the binomial theorem:

$\cos^3\theta + 3i\cos^2\theta \sin\theta - 3\cos\theta \sin^2\theta - i\sin^3\theta$

Equating real and imaginary parts:

$\cos 3\theta = \cos^3\theta - 3\cos\theta \sin^2\theta = 4\cos^3\theta - 3\cos\theta$

$\sin 3\theta = 3\cos^2\theta \sin\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta$

Multiple Angle Formulas

Using $\cos^2\theta + \sin^2\theta = 1$ to eliminate $\sin\theta$:

$\cos 2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$

$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$

$\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$

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Roots of Complex Numbers

$n$-th Roots

The $n$ solutions to $z^n = w$ where $w = re^{i\theta}$ are:

$z_k = r^{1/n}\, \exp\!\left(\frac{i(\theta + 2k\pi)}{n}\right), \qquad k = 0, 1, 2, \ldots, n - 1$

These $n$ roots lie on a circle of radius $r^{1/n}$ in the complex plane, equally spaced at angles of $\dfrac{2\pi}{n}$.

Example. Find the cube roots of $-8$.

$-8 = 8e^{i\pi}$.

$z_k = 2\, \exp\!\left(\frac{i(\pi + 2k\pi)}{3}\right), \qquad k = 0, 1, 2$

• $z_0 = 2e^{i\pi/3} = 1 + i\sqrt{3}$
• $z_1 = 2e^{i\pi} = -2$
• $z_2 = 2e^{i5\pi/3} = 1 - i\sqrt{3}$

Roots of Unity

The $n$-th roots of unity are the solutions to $z^n = 1$:

$z_k = \exp\!\left(\frac{2k\pi i}{n}\right), \qquad k = 0, 1, \ldots, n - 1$

These lie on the unit circle, forming a regular $n$-gon.

Properties:

• The sum of all $n$-th roots of unity is zero: $\displaystyle\sum_{k=0}^{n-1} z_k = 0$.
• The product of all $n$-th roots of unity is $(-1)^{n+1}$.
• Primitive roots generate all other roots by repeated multiplication.

Example. The 4th roots of unity: $1, i, -1, -i$.

Example. The 5th roots of unity form a regular pentagon on the unit circle at angles $0, \dfrac{2\pi}{5}, \dfrac{4\pi}{5}, \dfrac{6\pi}{5}, \dfrac{8\pi}{5}$.

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Euler's Formula

Statement

$e^{i\theta} = \cos\theta + i\sin\theta$

This fundamental identity connects the exponential function with trigonometry.

Proof (via Maclaurin Series)

$e^{i\theta} = \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \cdots$

$= 1 + i\theta - \frac{\theta^2}{2!} - \frac{i\theta^3}{3!} + \frac{\theta^4}{4!} + \cdots$

$= \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right)$

$= \cos\theta + i\sin\theta$

Euler's Identity

Setting $\theta = \pi$:

$e^{i\pi} + 1 = 0$

This elegantly relates five fundamental constants: $e$, $i$, $\pi$, $1$, and $0$.

Connections Between Exponential and Trigonometric Functions

From Euler's formula:

$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$

$\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$

These are the starting point for many applications.

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Applications to Trigonometry

Linearising Products

Products of sines and cosines can be expressed as sums using Euler's formulas.

$\cos A \cos B = \frac{1}{2}\bigl[\cos(A + B) + \cos(A - B)\bigr]$

$\sin A \sin B = \frac{1}{2}\bigl[\cos(A - B) - \cos(A + B)\bigr]$

$\sin A \cos B = \frac{1}{2}\bigl[\sin(A + B) + \sin(A - B)\bigr]$

Sums of Sines and Cosines

$\sum_{k=1}^{n} \cos k\theta = \frac{\sin(n\theta/2)\cos\bigl((n+1)\theta/2\bigr)}{\sin(\theta/2)}$

$\sum_{k=1}^{n} \sin k\theta = \frac{\sin(n\theta/2)\sin\bigl((n+1)\theta/2\bigr)}{\sin(\theta/2)}$

Derivation. Using $\cos k\theta = \mathrm{Re}(e^{ik\theta})$ and summing the geometric series:

$\sum_{k=1}^{n} e^{ik\theta} = e^{i\theta}\cdot\frac{1 - e^{in\theta}}{1 - e^{i\theta}}$

Multiply numerator and denominator by $e^{-i\theta/2}$:

$= e^{i\theta/2}\cdot\frac{e^{in\theta/2}(e^{-in\theta/2} - e^{in\theta/2})}{e^{-i\theta/2} - e^{i\theta/2}} = e^{i\theta/2}\cdot\frac{e^{in\theta/2}(-2i\sin n\theta/2)}{-2i\sin\theta/2}$

$= \frac{e^{i(n+1)\theta/2} \sin(n\theta/2)}{\sin(\theta/2)}$

Taking real and imaginary parts gives the formulas above.

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Applications to Integration

Integrals of Powers of Sine and Cosine

Using $\cos\theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2}$ and $\sin\theta = \dfrac{e^{i\theta} - e^{-i\theta}}{2i}$, powers of trig functions can be expanded and integrated term by term.

Example. Evaluate $\displaystyle\int \cos^4\theta\,d\theta$.

$\cos^4\theta = \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^{\!4} = \frac{1}{16}\bigl(e^{4i\theta} + 4e^{2i\theta} + 6 + 4e^{-2i\theta} + e^{-4i\theta}\bigr)$

$= \frac{1}{16}(2\cos 4\theta + 8\cos 2\theta + 6) = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}$

$\int \cos^4\theta\,d\theta = \frac{1}{32}\sin 4\theta + \frac{1}{4}\sin 2\theta + \frac{3}{8}\theta + C$

Example. Evaluate $\displaystyle\int \sin^3\theta\,d\theta$.

$\sin^3\theta = \left(\frac{e^{i\theta} - e^{-i\theta}}{2i}\right)^{\!3} = \frac{-1}{8i}\bigl(e^{3i\theta} - 3e^{i\theta} + 3e^{-i\theta} - e^{-3i\theta}\bigr)$

$= \frac{1}{4}\bigl(3\sin\theta - \sin 3\theta\bigr)$

$\int \sin^3\theta\,d\theta = -\frac{3}{4}\cos\theta + \frac{1}{12}\cos 3\theta + C$

Integrals Involving Products

$\int \cos m\theta \cos n\theta\,d\theta = \frac{1}{2}\int \bigl[\cos(m+n)\theta + \cos(m-n)\theta\bigr]\,d\theta$

provided $m \ne \pm n$. When $m = n$:

$\int \cos^2 n\theta\,d\theta = \int \frac{1 + \cos 2n\theta}{2}\,d\theta = \frac{\theta}{2} + \frac{\sin 2n\theta}{4n} + C$

Orthogonality. For integers $m \ne n$:

$\int_0^{2\pi} \cos m\theta \cos n\theta\,d\theta = 0, \qquad \int_0^{2\pi} \sin m\theta \sin n\theta\,d\theta = 0$

$\int_0^{2\pi} \cos m\theta \sin n\theta\,d\theta = 0$

These orthogonality relations are fundamental to Fourier analysis.

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The Complex Plane (Argand Diagram)

Geometric Interpretation

The complex number $z = a + bi$ is represented by the point $(a, b)$ in the plane. The modulus $|z|$ is the distance from the origin, and $\arg(z)$ is the angle from the positive real axis.

Regions in the Complex Plane

Condition	Region
$	z
$	z
$	z - z_0
$\arg(z) = \alpha$	Half-line from origin at angle $\alpha$
$\alpha \lt \arg(z) \lt \beta$	Sector between angles $\alpha$ and $\beta$

Loci

$|z - a| = |z - b|$ represents the perpendicular bisector of the segment joining $a$ and $b$ in the complex plane.

$|z - a| = k|z - b|$ (with $k \ne 1$) represents a circle (Apollonius circle).

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Complex Differentiation and Integration

Differentiating Complex Functions

If $f(z) = u(x, y) + iv(x, y)$ is differentiable at $z_0$, it satisfies the Cauchy-Riemann equations:

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

A function satisfying these equations at every point of an open set is called analytic (or holomorphic) on that set.

Example. $f(z) = z^2 = (x + iy)^2 = (x^2 - y^2) + i(2xy)$.

$u = x^2 - y^2$, $v = 2xy$.

$\dfrac{\partial u}{\partial x} = 2x = \dfrac{\partial v}{\partial y}$ and $\dfrac{\partial u}{\partial y} = -2y = -\dfrac{\partial v}{\partial x}$. The Cauchy-Riemann equations are satisfied, so $z^2$ is analytic.

Integration in the Complex Plane

The integral of $f(z)$ along a contour $\gamma$ from $z_1$ to $z_2$ is:

$\int_\gamma f(z)\,dz$

For analytic functions, the integral is path-independent (by Cauchy's theorem): the integral over any closed contour of an analytic function is zero.

Applications to Real Integration

Certain difficult real integrals can be evaluated using complex analysis. A standard technique involves integrating $f(z)$ around a semicircular contour in the upper half-plane and applying the residue theorem.

Example. Evaluate $\displaystyle\int_{-\infty}^{\infty} \frac{1}{x^2 + 1}\,dx$.

The integrand has a simple pole at $z = i$ in the upper half-plane with residue:

$\mathrm{Res}\!\left(\frac{1}{z^2 + 1},\, i\right) = \lim_{z \to i} (z - i)\cdot\frac{1}{(z-i)(z+i)} = \frac{1}{2i}$

By the residue theorem, integrating over the semicircular contour and taking the radius to infinity:

$\int_{-\infty}^{\infty} \frac{1}{x^2 + 1}\,dx = 2\pi i \cdot \frac{1}{2i} = \pi$

This matches the known result: $[\arctan x]_{-\infty}^{\infty} = \dfrac{\pi}{2} - \left(-\dfrac{\pi}{2}\right) = \pi$.

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Polynomial Equations and the Fundamental Theorem

The Fundamental Theorem of Algebra

Every polynomial $p(z) = a_n z^n + a_{n-1}z^{n-1} + \cdots + a_0$ of degree $n \ge 1$ with complex coefficients has at least one root in $\mathbb{'\{'}C{'\}'}$. Equivalently, it factorises completely:

$p(z) = a_n(z - z_1)(z - z_2)\cdots(z - z_n)$

where $z_1, z_2, \ldots, z_n$ are the roots (not necessarily distinct).

Complex Roots of Real Polynomials

If a real polynomial has a complex root $z = a + bi$ (with $b \ne 0$), then its complex conjugate $\bar{z} = a - bi$ is also a root. Roots come in conjugate pairs.

Relationships Between Roots and Coefficients

For $p(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_1 z + a_0$ with roots $z_1, \ldots, z_n$:

• $\displaystyle\sum_{i=1}^{n} z_i = -a_{n-1}$
• $\displaystyle\sum_{i \lt j} z_i z_j = a_{n-2}$
• $\displaystyle\prod_{i=1}^{n} z_i = (-1)^n a_0$

Solving Cubic Equations Using Complex Numbers

Example. Solve $z^3 - 1 = 0$.

$z^3 = 1 = e^{2k\pi i}$ for $k = 0, 1, 2$.

$z_k = e^{2k\pi i/3}, \qquad k = 0, 1, 2$

$z_0 = 1$, $z_1 = e^{2\pi i/3} = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$, $z_2 = e^{4\pi i/3} = -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i$.

These are the cube roots of unity, forming an equilateral triangle on the unit circle.

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Complex Exponentials and Differential Equations

Solving ODEs with Complex Characteristic Roots

When the characteristic equation of a second order ODE has complex roots $\alpha \pm i\beta$, the general solution is:

$y = e^{\alpha x}(A\cos\beta x + B\sin\beta x)$

This is derived from the complex exponential solution $y = Ce^{(\alpha + i\beta)x}$:

$Ce^{\alpha x}e^{i\beta x} = Ce^{\alpha x}(\cos\beta x + i\sin\beta x)$

Taking real and imaginary parts gives the independent solutions $e^{\alpha x}\cos\beta x$ and $e^{\alpha x}\sin\beta x$.

Example. Solve $y'' + 4y' + 13y = 0$.

Characteristic equation: $\lambda^2 + 4\lambda + 13 = 0$.

$\lambda = \frac{-4 \pm \sqrt{16 - 52}}{2} = -2 \pm 3i$

$y = e^{-2x}(A\cos 3x + B\sin 3x)$

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Worked Problems

Problem: Loci in the Complex Plane

Sketch the region defined by $|z - 2 - 3i| \le 3$ and $\arg(z - 2 - 3i) \le \dfrac{\pi}{4}$.

$|z - (2 + 3i)| \le 3$ is the closed disk of radius $3$ centred at $2 + 3i$. Combined with the argument condition, this is the sector of the disk bounded by the positive real axis direction from $2 + 3i$ and the line at $\pi/4$ from that point.

Problem: De Moivre with Large Exponents

Find $\dfrac{(1 + i)^{10}}{(1 - i)^8}$.

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

$(1 + i)^{10} = (\sqrt{2})^{10}\, e^{10\pi i/4} = 32\, e^{5\pi i/2} = 32\, e^{i\pi/2} = 32i$

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

$\frac{(1 + i)^{10}}{(1 - i)^8} = \frac{32i}{16} = 2i$

warning

Common Pitfall

When computing arguments, always verify the quadrant. $\arctan(b/a)$ alone gives the correct angle only when $a \gt 0$. For other quadrants, add or subtract $\pi$ as needed. Using $\mathrm{atan2}(b, a)$ (if available) handles this automatically.

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Additional Worked Examples

Worked Example: Solving a Quartic Using Roots of Unity

Solve $z^4 + 4z^2 + 16 = 0$.

Solution

Let $w = z^2$. Then $w^2 + 4w + 16 = 0$.

$w = \frac{-4 \pm \sqrt{16 - 64}}{2} = \frac{-4 \pm \sqrt{-48}}{2} = -2 \pm 2i\sqrt{3}$

So $z^2 = -2 + 2i\sqrt{3}$ or $z^2 = -2 - 2i\sqrt{3}$.

For $z^2 = -2 + 2i\sqrt{3}$: Write in polar form.

$|w| = \sqrt{4 + 12} = 4, \qquad \arg(w) = \arctan\!\left(\frac{2\sqrt{3}}{-2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

$z = \pm 2\, e^{i\pi/3} = \pm\left(1 + i\sqrt{3}\right)$

For $z^2 = -2 - 2i\sqrt{3}$:

$|w| = 4, \qquad \arg(w) = -\frac{2\pi}{3}$

$z = \pm 2\, e^{-i\pi/3} = \pm\left(1 - i\sqrt{3}\right)$

The four roots are $1 + i\sqrt{3}$, $-1 - i\sqrt{3}$, $1 - i\sqrt{3}$, $-1 + i\sqrt{3}$.

Worked Example: De Moivre for Cosine Multiple Angle Formula

Express $\cos 5\theta$ as a polynomial in $\cos\theta$.

Solution

By De Moivre: $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta$.

Expanding the left side by the binomial theorem:

$(\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta$

Equating real parts:

$\cos 5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta$

Using $\sin^2\theta = 1 - \cos^2\theta$:

$\cos 5\theta = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2$

$= \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta$

$= 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$

Worked Example: Fifth Roots of a Complex Number

Find all solutions to $z^5 = 16\sqrt{2}(1 + i)$.

Solution

First express the right side in polar form:

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

$16\sqrt{2}(1 + i) = 16\sqrt{2} \cdot \sqrt{2}\, e^{i\pi/4} = 32\, e^{i\pi/4}$

The five roots are:

$z_k = 32^{1/5} \exp\!\left(\frac{i(\pi/4 + 2k\pi)}{5}\right) = 2 \exp\!\left(\frac{i(\pi + 8k\pi)}{20}\right), \quad k = 0, 1, 2, 3, 4$

• $z_0 = 2\, e^{i\pi/20}$
• $z_1 = 2\, e^{i9\pi/20}$
• $z_2 = 2\, e^{i17\pi/20}$
• $z_3 = 2\, e^{i25\pi/20} = 2\, e^{i5\pi/4}$
• $z_4 = 2\, e^{i33\pi/20}$

These lie on a circle of radius $2$ centred at the origin, equally spaced at angles of $\dfrac{2\pi}{5}$.

Worked Example: Integration Using Complex Exponentials

Evaluate $\displaystyle\int_0^{2\pi} \cos^6\theta\,d\theta$.

Solution

Using $\cos\theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2}$:

$\cos^6\theta = \frac{1}{64}(e^{i\theta} + e^{-i\theta})^6$

By the binomial theorem:

$(e^{i\theta} + e^{-i\theta})^6 = e^{6i\theta} + 6e^{4i\theta} + 15e^{2i\theta} + 20 + 15e^{-2i\theta} + 6e^{-4i\theta} + e^{-6i\theta}$

$= 2\cos 6\theta + 12\cos 4\theta + 30\cos 2\theta + 20$

Therefore:

$\cos^6\theta = \frac{1}{32}\cos 6\theta + \frac{3}{16}\cos 4\theta + \frac{15}{32}\cos 2\theta + \frac{5}{16}$

Integrating over $[0, 2\pi]$, the cosine terms vanish:

$\int_0^{2\pi} \cos^6\theta\,d\theta = \int_0^{2\pi} \frac{5}{16}\,d\theta = \frac{5}{16} \cdot 2\pi = \frac{5\pi}{8}$

Worked Example: Loci and Regions

Sketch and describe the region in the complex plane defined by $|z - 1 - i| \le 2$ and $|z + 1| \ge 1$.

Solution

$|z - (1 + i)| \le 2$ is the closed disk of radius $2$ centred at $(1, 1)$.

$|z - (-1)| \ge 1$ is the exterior of the open disk of radius $1$ centred at $(-1, 0)$ (including the boundary).

The region satisfying both conditions is the part of the disk $|z - 1 - i| \le 2$ that lies outside the disk $|z + 1| \lt 1$.

The distance between the two centres is $|(1 + i) - (-1)| = |2 + i| = \sqrt{5} \approx 2.24$. Since the radii are $2$ and $1$, and $\sqrt{5} \lt 2 + 1 = 3$, the disks overlap. The distance between centres $2.24$ is greater than the difference of radii $|2 - 1| = 1$, so neither disk is contained within the other. The region is a disk with a smaller circular cut-out near $(-1, 0)$.

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Common Pitfalls

1. Wrong quadrant for the argument. When $\mathrm{Re}(z) \lt 0$ and $\mathrm{Im}(z) \gt 0$, the point lies in the second quadrant and the argument is $\pi - \arctan\!\left(\dfrac{|\mathrm{Im}(z)|}{|\mathrm{Re}(z)|}\right)$, not $\arctan\!\left(\dfrac{\mathrm{Im}(z)}{\mathrm{Re}(z)}\right)$.

2. Forgetting all $n$ roots. An equation $z^n = w$ has exactly $n$ complex roots (counting multiplicity). Always list all $n$ roots by using $k = 0, 1, \ldots, n-1$.

3. De Moivre with negative $n$. $(\cos\theta + i\sin\theta)^{-1} = \cos(-\theta) + i\sin(-\theta) = \cos\theta - i\sin\theta$. This is equivalent to the conjugate, and the formula holds for all integers $n$.

4. Assuming $|z_1 + z_2| = |z_1| + |z_2|$. The triangle inequality states $|z_1 + z_2| \le |z_1| + |z_2|$, with equality only when $z_1$ and $z_2$ have the same argument (point in the same direction).

5. Conjugate confusion in division. To simplify $\dfrac{z_1}{z_2}$, multiply numerator and denominator by $\bar{z}_2$, not by $z_2$. Multiplying by $z_2$ does not make the denominator real.

6. Roots of unity sum misconception. The sum of all $n$-th roots of unity is zero, but the sum of a subset is generally not zero. For example, the sum of the primitive 4th roots of unity ($i$ and $-i$) is $0$, but the sum of the primitive 6th roots of unity is $-1$.

7. Incorrect use of Euler's formula for integration. When using $e^{i\theta}$ to integrate $\cos^n\theta$ or $\sin^n\theta$, the intermediate expressions involve complex exponentials. Always collect terms carefully and verify that the final integrand is real before integrating.

8. Confusing $z^n = 1$ with $z^n = -1$. The roots of $z^n = -1$ are obtained by writing $-1 = e^{i\pi}$ (not $e^{i0}$), giving $z_k = e^{i(\pi + 2k\pi)/n}$.

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Exam-Style Problems

1. Express $z = -\sqrt{3} + i$ in polar form and hence compute $z^{12}$.

2. Find the cube roots of $-27i$ and show that they form an equilateral triangle in the Argand diagram.

3. Use De Moivre's theorem to express $\sin 4\theta$ in terms of $\sin\theta$ and $\cos\theta$.

4. Solve $z^4 - 2z^3 + 5z^2 - 8z + 4 = 0$ given that $z = 1 + i$ is a root.

5. The complex number $z$ satisfies $|z - 2| = |z + 2i|$. Describe the locus of $z$ and find the complex number on this locus with the smallest modulus.

6. Evaluate $\displaystyle\int_0^{\pi} \sin^4\theta\,d\theta$ using complex exponentials.

7. Prove that $\cos^4\theta + \sin^4\theta = \dfrac{3}{4} + \dfrac{1}{4}\cos 4\theta$.

8. The 5th roots of unity are $1, \omega, \omega^2, \omega^3, \omega^4$. Show that $(1 - \omega)(1 - \omega^2)(1 - \omega^3)(1 - \omega^4) = 5$.

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Cross-References

• Trigonometric identities derived via De Moivre: see Trigonometry
• Differential equations with complex characteristic roots: see Differential Equations
• Sequences and series including power series for Euler's formula proof: see Sequences and Series
• Integration techniques for trigonometric powers: see Integration