Complex Numbers

Cartesian Form

Cartesian form complex numbers ($z$) are separated to real ($a$) and imaginary parts ($bi$):

\begin`\{aligned}` z = a + bi, \quad a,b \in \mathbb{'\{'}R{'\}'} \end`\{aligned}`

Polar Form

Complex numbers ($z$) in polar form are defined with argument ($\theta$) away from $0$ in a unit circle, scaled by the magnitude ($r$):

\begin`\{aligned}` z = r(\cos \theta + i \sin \theta) \end`\{aligned}`

Euler's Form

Complex numbers ($z$) in Euler's form are defined by extending polar form, using Euler's number:

\begin`\{aligned}` z = re^{i\theta} \end`\{aligned}`

Conversions

Cartesian to Polar Form

As complex numbers is form by two coordinates (real and imaginary), the magnitude can by obtained by the Pythagorean identity:

\begin`\{aligned}` r = |z| = \sqrt{a^2+b^2} \end`\{aligned}`

Similarly, the argument ($\theta$) can be determined by the two coordinates by the definition of $\tan(\theta) = \frac{b}{a}$:

\begin`\{aligned}` \theta = \arg(z) = \arctan \frac{b}{a} \end`\{aligned}`

Since $\tan \theta$ is undefined at $a = 0$, case piecewise definition can define $\theta$ at $a = 0$:

\begin`\{aligned}` \theta = \arg(z) = \begin`\{cases}` \arctan \frac{b}{a} & a \neq 0\\ \frac{\pi}{2} & a = 0, b \gt{} 0\\ -\frac{\pi}{2} & a = 0, b \lt{} 0\\ \end`\{cases}` \end`\{aligned}`

Polar to Cartesian

Since polar coordinates is already defined by the argument, conversion to cartesian is simply evaluating the definition of cartesian coordinate ($\mathrm{cis }\theta$):

\begin`\{aligned}` z = r(\cos \theta + i \sin \theta) \end`\{aligned}`

Euler to Polar

The equivalence of Euler and polar form can be shown by their Taylor series expansion:

\begin`\{aligned}` re^{i\theta} = r\left(1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \cdots \right)

Separating real and imaginary parts:

\begin`\{aligned}` re^{i\theta} &= r\left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots \right) + ir\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots \right)\\ &= r\cos\theta + ir\sin\theta \end`\{aligned}`

This gives Euler's formula:

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

A special case at $\theta = \pi$:

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

This identity links five fundamental constants: $e$, $i$, $\pi$, $1$, and $0$.

Operations on Complex Numbers

Addition and Subtraction

Add/subtract the real and imaginary parts separately:

\begin`\{aligned}` (a + bi) + (c + di) &= (a + c) + (b + d)i\\ (a + bi) - (c + di) &= (a - c) + (b - d)i \end`\{aligned}`

Multiplication

Use the distributive property and $i^2 = -1$:

\begin`\{aligned}` (a + bi)(c + di) &= ac + adi + bci + bdi^2\\ &= (ac - bd) + (ad + bc)i \end`\{aligned}`

In polar form, multiplication is simpler — multiply the magnitudes and add the arguments:

\begin`\{aligned}` r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} &= r_1 r_2 \cdot e^{i(\theta_1 + \theta_2)} \end`\{aligned}`

Division

Multiply numerator and denominator by the conjugate of the denominator:

\begin`\{aligned}` \frac{a + bi}{c + di} &= \frac{a + bi}{c + di} \cdot \frac{c - di}{c - di}\\ &= \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} \end`\{aligned}`

In polar form, divide the magnitudes and subtract the arguments:

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

Complex Conjugate

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

Key properties:

\begin`\{aligned}` z \cdot \bar{z} &= a^2 + b^2 = |z|^2\\ z + \bar{z} &= 2a \quad (\mathrm{real})\\ z - \bar{z} &= 2bi \quad (\mathrm{imaginary}) \end`\{aligned}`

The Argand Diagram

An Argand diagram represents complex numbers in a 2D plane:

• The horizontal axis represents the real part ($\mathrm{Re}(z)$)
• The vertical axis represents the imaginary part ($\mathrm{Im}(z)$)

A complex number $z = a + bi$ is plotted at the point $(a, b)$.

Modulus

The modulus (magnitude) is the distance from the origin to the point:

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

Argument

The argument is the angle measured anticlockwise from the positive real axis:

$$\arg(z) = \arctan\left(\frac{b}{a}\right)$$

Principal argument: $\arg(z) \in (-\pi, \pi]$

tip

When finding $\arg(z)$, always draw a quick Argand diagram to check which quadrant the point is in. A common mistake is to use the calculator value directly without adjusting for the correct quadrant.

Geometric Interpretation of Operations

• Addition: vector addition (parallelogram rule)
• Conjugation ($\bar{z}$): reflection in the real axis
• Negation ($-z$): reflection in the origin
• Multiplication by $i$: rotation $90^\circ$ anticlockwise

Complex Plane (Argand Diagram)

Drag points to see how addition, conjugation, and multiplication affect the position on the Argand diagram.

De Moivre's Theorem

De Moivre's theorem states that for any integer $n$:

\begin`\{aligned}` \left(r(\cos\theta + i\sin\theta)\right)^n = r^n\left(\cos(n\theta) + i\sin(n\theta)\right) \end`\{aligned}`

Or in Euler's form:

$$\left(re^{i\theta}\right)^n = r^n e^{in\theta}$$

Applications of De Moivre's Theorem

Raising to a power:

$$(1 + i)^6 = \left(\sqrt{2}\, e^{i\pi/4}\right)^6 = (\sqrt{2})^6 \cdot e^{i \cdot 6\pi/4} = 8 \cdot e^{i \cdot 3\pi/2} = 8\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = -8i$$

Trigonometric identities: Expanding $(\cos\theta + i\sin\theta)^n$ using the binomial theorem and equating real and imaginary parts gives expressions for $\cos(n\theta)$ and $\sin(n\theta)$.

Roots of Complex Numbers

$n$th Roots

Every non-zero complex number has exactly $n$ distinct $n$th roots. To find the $n$th roots of $z = re^{i\theta}$:

\begin`\{aligned}` z^{1/n} &= r^{1/n} \cdot e^{i(\theta + 2k\pi)/n}, \quad k = 0, 1, 2, \ldots, n-1 \end`\{aligned}`

Roots of Unity

The $n$th roots of unity ($z^n = 1$) lie on the unit circle at equal angular intervals of $\frac{2\pi}{n}$:

$$e^{2k\pi i / n}, \quad k = 0, 1, 2, \ldots, n-1$$

Worked Example: Cube Roots of Unity

Find all solutions to $z^3 = 1$.

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

• $k = 0$: $z_0 = e^0 = 1$
• $k = 1$: $z_1 = e^{2\pi i/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$
• $k = 2$: $z_2 = e^{4\pi i/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$

Note that $z_0 + z_1 + z_2 = 0$ (the sum of all $n$th roots of unity is always $0$ for $n \gt 1$).

Worked Example: Fourth Roots of $-16$

Find all solutions to $z^4 = -16$.

Write $-16 = 16e^{i\pi}$:

$$z = 16^{1/4} \cdot e^{i(\pi + 2k\pi)/4} = 2 \cdot e^{i(2k+1)\pi/4}, \quad k = 0, 1, 2, 3$$

• $k = 0$: $z_0 = 2e^{i\pi/4} = 2\left(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right) = \sqrt{2} + i\sqrt{2}$
• $k = 1$: $z_1 = 2e^{i3\pi/4} = -\sqrt{2} + i\sqrt{2}$
• $k = 2$: $z_2 = 2e^{i5\pi/4} = -\sqrt{2} - i\sqrt{2}$
• $k = 3$: $z_3 = 2e^{i7\pi/4} = \sqrt{2} - i\sqrt{2}$

Solving Polynomial Equations

Complex numbers allow us to solve polynomial equations that have no real solutions.

Complex Roots of Quadratics

For $az^2 + bz + c = 0$ with discriminant $\Delta = b^2 - 4ac \lt 0$:

$$z = \frac{-b \pm \sqrt{-\Delta}\, i}{2a}$$

Conjugate Root Theorem

If a polynomial has real coefficients, then any non-real complex roots occur in conjugate pairs. That is, if $a + bi$ is a root, then $a - bi$ is also a root.

Worked Example: Finding a Polynomial from Roots

Problem: A cubic polynomial has real coefficients and roots $2$, $1 + i$, and $1 - i$. Find the polynomial.

Solution:

\begin`\{aligned}` P(z) &= (z - 2)(z - (1+i))(z - (1-i))\\ &= (z - 2)\left((z - 1)^2 + 1\right)\\ &= (z - 2)(z^2 - 2z + 2)\\ &= z^3 - 2z^2 + 2z - 2z^2 + 4z - 4\\ &= z^3 - 4z^2 + 6z - 4 \end`\{aligned}`

Summary of Forms

Form	Expression	Best For
Cartesian	$z = a + bi$	Addition, subtraction
Polar	$z = r(\cos\theta + i\sin\theta)$	Multiplication, division
Euler	$z = re^{i\theta}$	Powers, roots (De Moivre)

tip

Converting between forms is essential. A good strategy: always convert to polar/Euler form before raising to a power or finding roots, and convert back to Cartesian for the final answer.

Modulus-Argument Form Operations

When complex numbers are expressed in modulus-argument form, operations on moduli and arguments follow simple, elegant rules.

Multiplication

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

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

Proof:

\begin`\{aligned}` z_1 z_2 &= r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2}\\ &= r_1 r_2 \cdot e^{i\theta_1 + i\theta_2}\\ &= r_1 r_2 \cdot e^{i(\theta_1 + \theta_2)} \end`\{aligned}`

Therefore:

$$|z_1 z_2| = |z_1||z_2|, \qquad \arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$$

Note that the argument sum may exceed $\pi$ or fall below $-\pi$, so normalise to the principal argument by adding or subtracting $2\pi$ as needed.

Division

$$\frac{z_1}{z_2} = \frac{r_1}{r_2} \cdot e^{i(\theta_1 - \theta_2)}, \qquad z_2 \neq 0$$

Proof:

\begin`\{aligned}` \frac{z_1}{z_2} &= \frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}}\\ &= \frac{r_1}{r_2} \cdot e^{i\theta_1} \cdot e^{-i\theta_2}\\ &= \frac{r_1}{r_2} \cdot e^{i(\theta_1 - \theta_2)} \end`\{aligned}`

Therefore:

$$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}, \qquad \arg\!\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$$

Powers

Applying multiplication repeatedly:

$$z^n = r^n e^{in\theta}, \qquad n \in \mathbb{'\{'}Z{'\}'}$$

Therefore:

$$|z^n| = |z|^n, \qquad \arg(z^n) = n\arg(z)$$

Worked Example: Product and Quotient

Problem: Given $z_1 = 2e^{i\pi/3}$ and $z_2 = 3e^{-i\pi/6}$, find $z_1 z_2$ and $z_1/z_2$ in Cartesian form.

Solution:

\begin`\{aligned}` z_1 z_2 &= (2)(3) \cdot e^{i(\pi/3 - \pi/6)} = 6e^{i\pi/6}\\ &= 6\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\\ &= 6\left(\frac{\sqrt{3}}{2} + i \cdot \frac{1}{2}\right) = 3\sqrt{3} + 3i \end`\{aligned}` \begin`\{aligned}` \frac{z_1}{z_2} &= \frac{2}{3} \cdot e^{i(\pi/3 - (-\pi/6))} = \frac{2}{3} e^{i\pi/2}\\ &= \frac{2}{3}\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = \frac{2}{3}i \end`\{aligned}`

Worked Example: Argument of a Product

Problem: Given $z_1 = 1 - i$ and $z_2 = -\sqrt{3} - i$, find $\arg(z_1 z_2)$.

Solution:

Convert each to modulus-argument form:

• $z_1 = \sqrt{2}\, e^{-i\pi/4}$, so $\arg(z_1) = -\pi/4$
• $z_2 = 2e^{-i5\pi/6}$, so $\arg(z_2) = -5\pi/6$

Then:

$$\arg(z_1 z_2) = -\frac{\pi}{4} + \left(-\frac{5\pi}{6}\right) = -\frac{3\pi + 10\pi}{12} = -\frac{13\pi}{12}$$

Normalise to the principal value: $-\frac{13\pi}{12} + 2\pi = \frac{11\pi}{12}$.

Complex Plane Geometry

The Argand diagram allows geometric problems to be expressed algebraically using complex numbers.

Distance Between Two Points

The distance between points representing $z_1$ and $z_2$ is:

$$d = |z_1 - z_2|$$

Proof: If $z_1 = a + bi$ and $z_2 = c + di$, then $z_1 - z_2 = (a - c) + (b - d)i$, and:

$$|z_1 - z_2| = \sqrt{(a-c)^2 + (b-d)^2}$$

This is exactly the distance formula in $\mathbb{'\{'}R{'\}'}^2$.

Midpoint of a Line Segment

The midpoint of the segment joining $z_1$ and $z_2$ is:

$$m = \frac{z_1 + z_2}{2}$$

Perpendicular Bisector (Locus)

The perpendicular bisector of the segment joining $z_1$ and $z_2$ is the set of all points $z$ satisfying:

$$|z - z_1| = |z - z_2|$$

Expanding in Cartesian form: if $z_1 = a + bi$ and $z_2 = c + di$, then

$$|z - z_1|^2 = |z - z_2|^2 \implies (x - a)^2 + (y - b)^2 = (x - c)^2 + (y - d)^2$$

which simplifies to a linear equation in $x$ and $y$.

Circle as a Locus

The set of all points at a fixed distance $r$ from $z_0$ is the circle with centre $z_0$ and radius $r$:

$$|z - z_0| = r$$

In Cartesian form, with $z_0 = x_0 + iy_0$:

$$(x - x_0)^2 + (y - y_0)^2 = r^2$$

Half-Planes and Lines

The condition $|z - z_1| \lt |z - z_2|$ describes the half-plane of points closer to $z_1$ than to $z_2$. The boundary $|z - z_1| = |z - z_2|$ is the perpendicular bisector.

Equation of a Line in the Complex Plane

A line through $z_1$ and $z_2$ can be parameterised as:

$$z = z_1 + t(z_2 - z_1), \quad t \in \mathbb{'\{'}R{'\}'}$$

Equivalently, a line with equation $\mathrm{Re}(\bar{a}\, z) = c$ for some complex constant $a \neq 0$ and real constant $c$.

Perpendicular from the Origin to a Line

The shortest distance from the origin to the line $\mathrm{Re}(\bar{a}\, z) = c$ is:

$$d = \frac{|c|}{|a|}$$

Worked Example: Locus as a Circle

Problem: Describe and sketch the locus $|z - 3 - 4i| = 5$.

Solution: This is a circle with centre $z_0 = 3 + 4i$ and radius $r = 5$. The centre is at $(3, 4)$ on the Argand diagram, and every point $z$ satisfying the equation is exactly 5 units from this centre.

Worked Example: Perpendicular Bisector

Problem: Find the Cartesian equation of the perpendicular bisector of the segment joining $z_1 = 1 + 2i$ and $z_2 = 5 - 2i$.

Solution:

$$|z - z_1| = |z - z_2| \implies |z - 1 - 2i|^2 = |z - 5 + 2i|^2$$

Let $z = x + iy$:

\begin`\{aligned}` (x - 1)^2 + (y - 2)^2 &= (x - 5)^2 + (y + 2)^2\\ x^2 - 2x + 1 + y^2 - 4y + 4 &= x^2 - 10x + 25 + y^2 + 4y + 4\\ -2x - 4y + 5 &= -10x + 4y + 29\\ 8x - 8y &= 24\\ x - y &= 3 \end`\{aligned}`

The perpendicular bisector is the line $x - y = 3$.

Worked Example: Intersection of Loci

Problem: Find the points of intersection of $|z| = 5$ and $|z - 3| = 4$.

Solution: Let $z = x + iy$:

\begin`\{aligned}` x^2 + y^2 &= 25 \tag{1}\\ (x - 3)^2 + y^2 &= 16 \tag{2} \end`\{aligned}`

Subtracting $(2)$ from $(1)$:

$$x^2 - (x - 3)^2 = 9 \implies x^2 - x^2 + 6x - 9 = 9 \implies x = 3$$

Substituting into $(1)$: $9 + y^2 = 25 \implies y^2 = 16 \implies y = \pm 4$.

The points of intersection are $z = 3 + 4i$ and $z = 3 - 4i$.

Proof of De Moivre's Theorem

Theorem: For all integers $n$,

$$\left(re^{i\theta}\right)^n = r^n e^{in\theta}$$

Proof by Mathematical Induction

Base case ($n = 1$): Trivially, $\left(re^{i\theta}\right)^1 = r^1 e^{i \cdot 1 \cdot \theta}$. Verified.

Inductive hypothesis: Assume the result holds for $n = k$, i.e.

$$\left(re^{i\theta}\right)^k = r^k e^{ik\theta}$$

Inductive step ($n = k + 1$):

\begin`\{aligned}` \left(re^{i\theta}\right)^{k+1} &= \left(re^{i\theta}\right)^k \cdot re^{i\theta}\\ &= r^k e^{ik\theta} \cdot r e^{i\theta} \quad \mathrm{(by hypothesis)}\\ &= r^{k+1} \cdot e^{i(k\theta + \theta)}\\ &= r^{k+1} e^{i(k+1)\theta} \end`\{aligned}`

So the result holds for $n = k + 1$. By induction, the theorem is true for all $n \in \mathbb{'\{'}N{'\}'}$.

Negative integers: For $n = -m$ where $m \in \mathbb{'\{'}N{'\}'}$:

$$\left(re^{i\theta}\right)^{-m} = \frac{1}{\left(re^{i\theta}\right)^m} = \frac{1}{r^m e^{im\theta}} = r^{-m} e^{-im\theta} = r^n e^{in\theta}$$

Connection to the Binomial Theorem

Expanding $(\cos\theta + i\sin\theta)^n$ using the binomial theorem and equating with De Moivre's result gives trigonometric identities. For $n = 3$:

\begin`\{aligned}` \cos 3\theta + i\sin 3\theta &= (\cos\theta + i\sin\theta)^3\\ &= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta \end`\{aligned}`

Equating real parts:

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

Equating imaginary parts:

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

Additional Worked Examples

Worked Example: Solving Higher-Degree Equations

Problem: Solve $z^5 + z^3 + z = 0$.

Solution: Factor:

$$z(z^4 + z^2 + 1) = 0$$

So $z = 0$ is one root. For $z^4 + z^2 + 1 = 0$, substitute $w = z^2$:

$$w^2 + w + 1 = 0 \implies w = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$$

So $z^2 = e^{\pm 2\pi i/3}$. Taking square roots:

\begin`\{aligned}` z^2 &= e^{2\pi i/3} \implies z = \pm\, e^{\pi i/3} = \pm\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)\\ z^2 &= e^{-2\pi i/3} \implies z = \pm\, e^{-\pi i/3} = \pm\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) \end`\{aligned}`

The five roots are $0,\; \pm\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$.

Worked Example: Geometric Problem on the Argand Diagram

Problem: The points $A$, $B$, $C$ on the Argand diagram represent $z_A = 1 + i$, $z_B = 5 + 3i$, and $z_C = 3 + 7i$. Show that triangle $ABC$ is isosceles and find its area.

Solution:

\begin`\{aligned}` AB &= |z_B - z_A| = |4 + 2i| = \sqrt{16 + 4} = 2\sqrt{5}\\ BC &= |z_C - z_B| = |-2 + 4i| = \sqrt{4 + 16} = 2\sqrt{5}\\ CA &= |z_A - z_C| = |-2 - 6i| = \sqrt{4 + 36} = 2\sqrt{10} \end`\{aligned}`

Since $AB = BC = 2\sqrt{5}$, the triangle is isosceles with $AB = BC$.

Using the formula for the area of a triangle with vertices at complex numbers:

\begin`\{aligned}` \mathrm{Area} &= \frac{1}{2}\left|\mathrm{Im}\!\left(\bar{z}_`\{AB}` \cdot z_`\{AC}`\right)\right|\\ &= \frac{1}{2}\left|\mathrm{Im}\!\left((4 - 2i)(2 + 6i)\right)\right|\\ &= \frac{1}{2}\left|\mathrm{Im}(8 + 24i - 4i + 12)\right|\\ &= \frac{1}{2}\left|\mathrm{Im}(20 + 20i)\right|\\ &= \frac{1}{2} \cdot 20 = 10 \end`\{aligned}`

Worked Example: Trigonometric Expansion via De Moivre

Problem: Express $\cos^5\theta$ in terms of $\cos(n\theta)$ for $n = 1, 3, 5$.

Solution: Using the binomial expansion of $\cos^5\theta = \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^5$:

\begin`\{aligned}` 2^5 \cos^5\theta &= (e^{i\theta} + e^{-i\theta})^5\\ &= e^{5i\theta} + 5e^{3i\theta} + 10e^{i\theta} + 10e^{-i\theta} + 5e^{-3i\theta} + e^{-5i\theta}\\ &= 2\cos 5\theta + 10\cos 3\theta + 20\cos\theta \end`\{aligned}`

Therefore:

$$\cos^5\theta = \frac{1}{16}\cos 5\theta + \frac{5}{16}\cos 3\theta + \frac{5}{8}\cos\theta$$

Worked Example: Converting Between All Three Forms

Problem: Express $z = -1 + i\sqrt{3}$ in all three standard forms and verify the conversions.

Solution:

Cartesian: $z = -1 + i\sqrt{3}$.

Polar:

$$r = |z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2$$

The point $(-1, \sqrt{3})$ lies in the second quadrant, so:

$$\theta = \pi - \arctan\frac{\sqrt{3}}{1} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$

Therefore $z = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$.

Euler: $z = 2e^{2\pi i/3}$.

Verification: $2e^{2\pi i/3} = 2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -1 + i\sqrt{3}$. Confirmed.

Common Pitfalls

Wrong Quadrant for the Argument

The function $\arctan(b/a)$ always returns a value in $(-\pi/2, \pi/2)$. If the complex number lies in the second or third quadrant, you must adjust:

\begin`\{aligned}` \mathrm{Quadrant II:}\quad &\theta = \pi - \arctan\!\left|\frac{b}{a}\right|\\ \mathrm{Quadrant III:}\quad &\theta = -\pi + \arctan\frac{b}{a} \end`\{aligned}`

Always sketch the Argand diagram before computing the argument.

Forgetting to Add $2k\pi$ When Finding Roots

When solving $z^n = w$, you must write $w = re^{i(\theta + 2k\pi)}$ with the $2k\pi$ term before dividing the argument by $n$. Omitting this term yields only one root instead of all $n$ distinct roots.

Confusing $|z|$ with $z$

$|z|$ is a real, non-negative scalar (the modulus), whereas $z$ is a complex number. They behave differently under operations. For example:

$$|z_1 + z_2| \neq |z_1| + |z_2| \quad \mathrm{in general}$$

The triangle inequality gives $|z_1 + z_2| \leq |z_1| + |z_2|$, with equality only when $z_1$ and $z_2$ have the same argument.

Calculator Mode Errors

Ensure your calculator is in radian mode when working with arguments and complex exponentials. Using degree mode by mistake is a very common source of error, since $\pi/3$ radians is $60^\circ$ but entering $60$ in radian mode gives a completely different value.

Argument Normalisation

After multiplying or dividing complex numbers, the resulting argument may lie outside $(-\pi, \pi]$. Always check whether you need to add or subtract $2\pi$ to express the answer in principal argument form.

Problem Set

Problem 1: Modulus-Argument Operations

Given $z_1 = 4e^{i\pi/4}$ and $z_2 = 2e^{-i\pi/3}$, find $|z_1 z_2|$ and $\arg(z_1/z_2)$ in exact form.

Solution:

$$|z_1 z_2| = |z_1| \cdot |z_2| = 4 \cdot 2 = 8$$$$\arg\!\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2) = \frac{\pi}{4} - \left(-\frac{\pi}{3}\right) = \frac{7\pi}{12}$$

Problem 2: Locus Description

Describe the locus defined by $|z - 1 - i| = |z + 1 + i|$ and give its Cartesian equation.

Solution: This represents the perpendicular bisector of the segment joining $z_1 = 1 + i$ and $z_2 = -1 - i$.

Let $z = x + iy$:

\begin`\{aligned}` (x - 1)^2 + (y - 1)^2 &= (x + 1)^2 + (y + 1)^2\\ x^2 - 2x + 1 + y^2 - 2y + 1 &= x^2 + 2x + 1 + y^2 + 2y + 1\\ -4x - 4y &= 0\\ y &= -x \end`\{aligned}`

The locus is the line $y = -x$.

Problem 3: Roots of a Complex Number

Find all cube roots of $z = 8i$ and express them in Cartesian form.

Solution: Write $8i = 8e^{i\pi/2}$. The cube roots are:

$$z_k = 8^{1/3} \cdot e^{i(\pi/2 + 2k\pi)/3} = 2 \cdot e^{i(\pi + 4k\pi)/6}, \quad k = 0, 1, 2$$

• $k = 0$: $z_0 = 2e^{i\pi/6} = 2\left(\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = \sqrt{3} + i$
• $k = 1$: $z_1 = 2e^{i5\pi/6} = 2\left(-\frac{\sqrt{3}}{2} + i\frac{1}{2}\right) = -\sqrt{3} + i$
• $k = 2$: $z_2 = 2e^{i3\pi/2} = 2(0 - i) = -2i$

Problem 4: De Moivre Application

Use De Moivre's theorem to find $\cos 4\theta$ in terms of powers of $\cos\theta$.

Solution:

\begin`\{aligned}` \cos 4\theta + i\sin 4\theta &= (\cos\theta + i\sin\theta)^4\\ &= \cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4i\cos\theta\sin^3\theta + \sin^4\theta \end`\{aligned}`

Equating real parts and using $\sin^2\theta = 1 - \cos^2\theta$:

\begin`\{aligned}` \cos 4\theta &= \cos^4\theta - 6\cos^2\theta(1 - \cos^2\theta) + (1 - \cos^2\theta)^2\\ &= \cos^4\theta - 6\cos^2\theta + 6\cos^4\theta + 1 - 2\cos^2\theta + \cos^4\theta\\ &= 8\cos^4\theta - 8\cos^2\theta + 1 \end`\{aligned}`

Problem 5: Geometric Locus Intersection

Find all complex numbers $z$ satisfying both $|z - 2| = 2$ and $\arg(z) = \pi/4$.

Solution: The condition $\arg(z) = \pi/4$ means $z$ lies on the ray from the origin at $45^\circ$, so $z = t(1 + i)$ for $t \gt{} 0$.

Substituting into $|z - 2| = 2$:

\begin`\{aligned}` |t(1 + i) - 2| &= 2\\ |(t - 2) + ti| &= 2\\ (t - 2)^2 + t^2 &= 4\\ t^2 - 4t + 4 + t^2 &= 4\\ 2t^2 - 4t &= 0\\ 2t(t - 2) &= 0 \end`\{aligned}`

Since $t \gt{} 0$, we get $t = 2$. Therefore $z = 2(1 + i) = 2 + 2i$.

Problem 6: Polynomial with Complex Roots

The quartic $P(z) = z^4 - 4z^3 + 14z^2 - 20z + 25$ has a root at $z = 1 + 2i$. Find all roots.

Solution: By the conjugate root theorem, $z = 1 - 2i$ is also a root. The quadratic factor is:

$$(z - (1 + 2i))(z - (1 - 2i)) = (z - 1)^2 + 4 = z^2 - 2z + 5$$

Divide $P(z)$ by $z^2 - 2z + 5$:

$$P(z) = (z^2 - 2z + 5)(z^2 - 2z + 5) = (z^2 - 2z + 5)^2$$

Solving $z^2 - 2z + 5 = 0$:

$$z = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i$$

The roots are $1 + 2i$ and $1 - 2i$, each with multiplicity 2.

Problem 7: Proof Using Modulus Properties

Prove that $|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2$ for all $z_1, z_2 \in \mathbb{'\{'}C{'\}'}$.

Solution: Let $z_1 = a + bi$ and $z_2 = c + di$.

\begin`\{aligned}` |z_1 + z_2|^2 &= (a + c)^2 + (b + d)^2 = a^2 + 2ac + c^2 + b^2 + 2bd + d^2\\ |z_1 - z_2|^2 &= (a - c)^2 + (b - d)^2 = a^2 - 2ac + c^2 + b^2 - 2bd + d^2 \end`\{aligned}`

Adding:

$$|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2a^2 + 2c^2 + 2b^2 + 2d^2 = 2(a^2 + b^2) + 2(c^2 + d^2) = 2|z_1|^2 + 2|z_2|^2$$

This is the parallelogram law: the sum of the squares of the diagonals of a parallelogram equals the sum of the squares of all four sides.

Problem 8: Fifth Roots of Unity

Find all fifth roots of unity and verify that their sum is zero.

Solution: The fifth roots of unity are:

$$z_k = e^{2k\pi i/5}, \quad k = 0, 1, 2, 3, 4$$

In Cartesian form:

\begin`\{aligned}` z_0 &= 1\\ z_1 &= \cos\frac{2\pi}{5} + i\sin\frac{2\pi}{5}\\ z_2 &= \cos\frac{4\pi}{5} + i\sin\frac{4\pi}{5}\\ z_3 &= \cos\frac{6\pi}{5} + i\sin\frac{6\pi}{5}\\ z_4 &= \cos\frac{8\pi}{5} + i\sin\frac{8\pi}{5} \end`\{aligned}`

These are the roots of $z^5 - 1 = 0$. By Vieta's formulas, the sum of the roots equals the negative coefficient of $z^4$, which is $0$.

Alternatively, using the geometric series:

$$\sum_{k=0}^{4} z_k = \sum_{k=0}^{4} \left(e^{2\pi i/5}\right)^k = \frac{1 - \left(e^{2\pi i/5}\right)^5}{1 - e^{2\pi i/5}} = \frac{1 - 1}{1 - e^{2\pi i/5}} = 0$$

Problem 9: Area of a Triangle on the Argand Diagram

The vertices of a triangle on the Argand diagram are $z_1 = 0$, $z_2 = 4 + 2i$, and $z_3 = 1 + 5i$. Find the area of the triangle.

Solution: Using the determinant formula for the area of a triangle with vertices at complex numbers:

\begin`\{aligned}` \mathrm{Area} &= \frac{1}{2}\left|\mathrm{Im}\!\left(\bar{z}_2 \cdot z_3\right)\right|\\ &= \frac{1}{2}\left|\mathrm{Im}\!\left((4 - 2i)(1 + 5i)\right)\right|\\ &= \frac{1}{2}\left|\mathrm{Im}(4 + 20i - 2i + 10)\right|\\ &= \frac{1}{2}\left|\mathrm{Im}(14 + 18i)\right|\\ &= \frac{1}{2} \cdot 18 = 9 \end`\{aligned}`

Problem 10: Argument Equations

Solve the equation $\arg(z - 1) = \arg(z + i)$ for $z \in \mathbb{'\{'}C{'\}'}$, where $z \neq 1$ and $z \neq -i$.

Solution: The condition $\arg(z - 1) = \arg(z + i)$ means the vectors from $1$ and from $-i$ to $z$ have the same direction. This occurs when $z$ lies on the line through $1$ and $-i$, excluding those two points.

Algebraically, set $\mathrm{Im}\!\left(\frac{z + i}{z - 1}\right) = 0$. Let $z = x + iy$:

$$\frac{z + i}{z - 1} = \frac{x + i(y + 1)}{(x - 1) + iy}$$

Multiplying numerator and denominator by the conjugate of the denominator:

$$\frac{(x + i(y+1))((x-1) - iy)}{(x-1)^2 + y^2}$$

The imaginary part of the numerator:

$$x(-y) + (y + 1)(x - 1) = -xy + xy + x - y - 1 = x - y - 1$$

Setting this to zero: $x - y - 1 = 0$, i.e. $y = x - 1$.

The solution set is $\{z = x + i(x - 1) : x \in \mathbb{'\{'}R{'\}'},\; z \neq 1,\; z \neq -i\}$.

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