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

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

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

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

Intuition

Complex numbers extend the number line into a plane — the Argand diagram — where every number has both a real and imaginary coordinate. Think of the imaginary unit i as a 90-degree rotation operator: multiplying by i spins a number a quarter turn counterclockwise. Polar form describes complex numbers by distance and direction rather than coordinates, making multiplication intuitive: multiply the distances and add the angles. De Moivre’s theorem is the power tool for this: raising a complex number to a power means raising the distance and multiplying the angle. Roots of unity are equally spaced points on a circle, like hours on a clock face.

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

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}` <div class="table-wrap"> |t(1 + i) - 2| &= 2\\ | --- | --- | |(t - 2) + ti| &= 2\\ </div> (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 $\{z = x + i(x - 1) : x \in \mathbb{R},\; z \neq 1,\; z \neq -i\}$.

---

See Also

• Number And Algebra
• Number and Algebra
• Proof and Logic