Complex Numbers | ChatWhole Learn

Why It Matters

ℹ️ Why It Matters

Complex numbers are far more than an abstract extension of the real number system. They are the natural language for describing oscillations, rotations, and wave phenomena. Every alternating current in an electrical circuit, every quantum mechanical wavefunction, every signal transmitted through fiber optics relies on complex arithmetic. The fundamental theorem of algebra — that every polynomial of degree n has exactly n roots in the complex plane — guarantees that complex numbers are algebraically complete. Without them, Fourier analysis, control theory, fluid dynamics, and much of modern physics would be impossible. Complex numbers unify exponential and trigonometric functions through Euler's formula, transforming difficult calculus problems into elegant algebra. Mastering complex numbers is the essential first step toward contour integration, residue calculus, and the powerful applications that follow.

Core Definitions

DfComplex Number

A complex number is an ordered pair $(a, b)$ of real numbers, written $z = a + bi$ , where $i$ is the imaginary unit satisfying $i^2 = -1$ . The real part is $\operatorname{Re}(z) = a$ and the imaginary part is $\operatorname{Im}(z) = b$ . The set of all complex numbers is denoted $\mathbb{C}$ .

DfConjugate

The complex conjugate of $z = a + bi$ is $\overline{z} = a - bi$ . The conjugate reflects $z$ across the real axis in the complex plane.

DfModulus (Absolute Value)

The modulus of $z = a + bi$ is $|z| = \sqrt{a^2 + b^2}$ , representing the distance from the origin to $z$ in the complex plane.

DfArgument

The argument of $z = a + bi$ is the angle $\theta$ such that $z = |z|(\cos\theta + i\sin\theta)$ . The principal argument $\operatorname{Arg}(z) \in (-\pi, \pi]$ is the unique value in this range. The general argument is $\arg(z) = \operatorname{Arg}(z) + 2k\pi$ for any integer $k$ .

DfPolar Form

Any nonzero complex number can be written as $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$ , where $r = |z|$ and $\theta = \arg(z)$ . This representation is called the polar form or exponential form.

DfRoots of Unity

The $n$ -th roots of unity are the solutions to $z^n = 1$ . They are given by $z_k = e^{2\pi ik/n}$ for $k = 0, 1, 2, \ldots, n-1$ . These are equally spaced on the unit circle at angles $2\pi k/n$ .

Key Formulas

Complex Number Arithmetic

z_1 + z_2 = (a+c) + (b+d)i, \quad z_1 \cdot z_2 = (ac - bd) + (ad + bc)i

Here,

$z_1 = a + bi$ =First complex number
$z_2 = c + di$ =Second complex number

Modulus

|z| = \sqrt{a^2 + b^2}, \quad |z|^2 = z \cdot \overline{z}

Here,

$z = a + bi$ =Complex number
$|z|$ =Distance from origin

Division of Complex Numbers

\frac{z_1}{z_2} = \frac{(a+bi)(c-di)}{c^2 + d^2} = \frac{ac+bd}{c^2+d^2} + \frac{bc-ad}{c^2+d^2}i

Here,

$z_1 = a+bi$ =Numerator
$z_2 = c+di$ =Denominator (nonzero)

Polar Form

z = re^{i\theta}, \quad r = |z|, \quad \theta = \arg(z)

Here,

$r$ =Modulus (magnitude) of z
$\theta$ =Argument (angle) of z in radians

Euler's Formula

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

Here,

$\theta$ =Angle in radians
$e^{i\theta}$ =Point on unit circle at angle \theta

De Moivre's Theorem

(re^{i\theta})^n = r^n e^{in\theta} = r^n(\cos n\theta + i\sin n\theta)

Here,

$r$ =Modulus of original complex number
$\theta$ =Argument of original complex number
$n$ =Integer exponent

N-th Roots

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

Here,

$r^{1/n}$ =Real n-th root of the modulus
$\theta + 2k\pi$ =Argument offset by full rotations

Trigonometric Identities via Euler

\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}

Here,

$e^{i\theta}$ =Exponential form of rotation
$e^{-i\theta}$ =Conjugate rotation

Polar Multiplication

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

Here,

$r_1, r_2$ =Moduli of the two numbers
$\theta_1, \theta_2$ =Arguments of the two numbers

Important Theorems

ThEuler's Formula

For any real number $\theta$ , $e^{i\theta} = \cos\theta + i\sin\theta$ .

Proof Sketch: Start with the Taylor series for $e^x$ , $\cos x$ , and $\sin x$ . Substitute $x = i\theta$ into the exponential series. Since $i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$ , the series splits into real (cosine) and imaginary (sine) parts, yielding the result. $\square$

Key Consequence: $e^{i\pi} + 1 = 0$ (Euler's identity), connecting the five most important constants in mathematics.

ThFundamental Theorem of Algebra

Every non-constant polynomial $p(z) = a_n z^n + a_{n-1} z^{n-1} + \cdots + a_0$ with complex coefficients has at least one root in $\mathbb{C}$ . Equivalently, $p(z)$ factors completely as $p(z) = a_n(z - z_1)(z - z_2) \cdots (z - z_n)$ where $z_1, \ldots, z_n$ are the roots (counted with multiplicity).

Implication: $\mathbb{C}$ is algebraically closed — no polynomial equation lacks solutions. This is the essential reason complex numbers are so powerful in analysis.

ThDe Moivre's Theorem

For any real number $r > 0$ , any angle $\theta$ , and any integer $n$ :

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

Proof Sketch: Use induction on $n$ . For $n = 1$ , trivial. Assume true for $n$ . Then $(re^{i\theta})^{n+1} = (re^{i\theta})^n \cdot re^{i\theta} = r^n e^{in\theta} \cdot re^{i\theta} = r^{n+1} e^{i(n+1)\theta}$ . $\square$

ThProperties of Modulus and Argument

For $z_1, z_2 \in \mathbb{C}$ :

$|z_1 z_2| = |z_1| \cdot |z_2|$
$|z_1 + z_2| \leq |z_1| + |z_2|$ (Triangle Inequality)
$\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) \pmod{2\pi}$
$\arg(z_1 / z_2) = \arg(z_1) - \arg(z_2) \pmod{2\pi}$
$z \cdot \overline{z} = |z|^2$

ThRoots of Unity Form a Group

The $n$ -th roots of unity $\{1, \omega, \omega^2, \ldots, \omega^{n-1}\}$ where $\omega = e^{2\pi i/n}$ form a cyclic group under multiplication. They are vertices of a regular $n$ -gon inscribed in the unit circle. Any polynomial $z^n - 1$ factors as $z^n - 1 = \prod_{k=0}^{n-1}(z - \omega^k)$ .

Worked Examples

📝Example 1: Basic Complex Arithmetic

Let $z_1 = 3 + 4i$ and $z_2 = 1 - 2i$ .

Addition: $z_1 + z_2 = (3+1) + (4-2)i = 4 + 2i$

Multiplication: $z_1 z_2 = (3)(1) - (4)(-2) + [(3)(-2) + (4)(1)]i = 3 + 8 + (-6 + 4)i = 11 - 2i$

Division: $\frac{z_1}{z_2} = \frac{(3+4i)(1+2i)}{(1-2i)(1+2i)} = \frac{3 + 6i + 4i + 8i^2}{1 + 4} = \frac{3 - 8 + 10i}{5} = \frac{-5 + 10i}{5} = -1 + 2i$

Modulus: $|z_1| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$

Conjugate: $\overline{z_2} = 1 + 2i$

Verification: $z_2 \cdot \overline{z_2} = (1-2i)(1+2i) = 1 + 4 = 5 = |z_2|^2$ ✓

📝Example 2: Converting to Polar Form

Convert $z = -\sqrt{3} + i$ to polar form.

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

Step 2: $\tan\theta = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}}$ . Since $z$ is in the second quadrant ( $\operatorname{Re}(z) < 0$ , $\operatorname{Im}(z) > 0$ ):

\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}

Result: $z = 2e^{i \cdot 5\pi/6} = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)$

Verification: $2\cos(5\pi/6) = 2 \cdot (-\sqrt{3}/2) = -\sqrt{3}$ ✓ and $2\sin(5\pi/6) = 2 \cdot (1/2) = 1$ ✓

📝Example 3: Computing Powers with De Moivre's Theorem

Compute $(1 + i)^8$ .

Step 1: Convert to polar form. $r = \sqrt{1^2 + 1^2} = \sqrt{2}$ , $\theta = \pi/4$ (first quadrant, $\tan\theta = 1$ ).

So $1 + i = \sqrt{2} \cdot e^{i\pi/4}$ .

Step 2: Apply De Moivre's Theorem:

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

Alternative (brute force): $(1+i)^2 = 2i$ , $(2i)^2 = -4$ , $(-4)^2 = 16$ . Same result. ✓

📝Example 4: Finding N-th Roots of a Complex Number

Find all cube roots of $z = 8i$ .

Step 1: Write $z$ in polar form. $z = 8i = 8e^{i\pi/2}$ (modulus 8, argument $\pi/2$ ).

Step 2: Apply the $n$ -th root formula with $n = 3$ : $z^{1/3} = 8^{1/3} \cdot e^{i(\pi/2 + 2k\pi)/3} = 2 \cdot e^{i(\pi/6 + 2k\pi/3)}$ for $k = 0, 1, 2$ .

Step 3: Compute each root:

$k = 0$ : $w_0 = 2e^{i\pi/6} = 2\left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = \sqrt{3} + i$
$k = 1$ : $w_1 = 2e^{i(\pi/6 + 2\pi/3)} = 2e^{i5\pi/6} = 2\left(-\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = -\sqrt{3} + i$
$k = 2$ : $w_2 = 2e^{i(\pi/6 + 4\pi/3)} = 2e^{i3\pi/2} = 2(0 - i) = -2i$

Check: $w_0^3 = (\sqrt{3}+i)^3 = 8i$ ✓, $w_1^3 = (-\sqrt{3}+i)^3 = 8i$ ✓, $w_2^3 = (-2i)^3 = 8i$ ✓

These three roots form an equilateral triangle inscribed in a circle of radius 2.

📝Example 5: Euler's Formula and Trigonometric Identities

Prove that $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$ .

Step 1: By De Moivre's Theorem, $(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$ .

Step 2: Expand the left side using the binomial theorem:

(\cos\theta + i\sin\theta)^3 = \cos^3\theta + 3i\cos^2\theta\sin\theta + 3i^2\cos\theta\sin^2\theta + i^3\sin^3\theta

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

= (\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta)

Step 3: Equate real parts: $\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = 4\cos^3\theta - 3\cos\theta$ ✓

📝Example 6: Roots of Unity and Polynomial Factorization

Factor $z^4 - 1$ over $\mathbb{C}$ .

Step 1: The 4th roots of unity are $z_k = e^{2\pi ik/4}$ for $k = 0, 1, 2, 3$ :

$z_0 = 1$
$z_1 = e^{i\pi/2} = i$
$z_2 = e^{i\pi} = -1$
$z_3 = e^{i3\pi/2} = -i$

Step 2: Factor:

z^4 - 1 = (z - 1)(z - i)(z + 1)(z + i)

Step 3: Group into real quadratics: $(z-1)(z+1) = z^2 - 1$ and $(z-i)(z+i) = z^2 + 1$

So $z^4 - 1 = (z^2 - 1)(z^2 + 1) = (z-1)(z+1)(z^2+1)$ .

This demonstrates how complex roots enable complete factorization of polynomials.

Practice Problems

📝Problem 1: Complex Division

Compute $\frac{(2 + 3i)(1 - i)}{3 + 4i}$ .

💡Solution

Step 1: Numerator: $(2+3i)(1-i) = 2 - 2i + 3i - 3i^2 = 2 + i + 3 = 5 + i$

Step 2: Divide by $3 + 4i$ by multiplying by conjugate:

\frac{5+i}{3+4i} = \frac{(5+i)(3-4i)}{(3+4i)(3-4i)} = \frac{15 - 20i + 3i - 4i^2}{9 + 16} = \frac{15 + 4 - 17i}{25} = \frac{19 - 17i}{25}

Result: $\frac{19}{25} - \frac{17}{25}i$

📝Problem 2: Polar Form Conversion

Express $z = -1 - i\sqrt{3}$ in polar form and find $z^6$ .

💡Solution

Step 1: $r = \sqrt{1 + 3} = 2$

Step 2: $\tan\theta = \frac{-\sqrt{3}}{-1} = \sqrt{3}$ . Third quadrant: $\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$

Polar form: $z = 2e^{i4\pi/3}$

Step 3: $z^6 = 2^6 \cdot e^{i \cdot 6 \cdot 4\pi/3} = 64 \cdot e^{i \cdot 8\pi} = 64 \cdot e^{i \cdot 0} = 64$

(Since $8\pi$ is a multiple of $2\pi$ , the exponential equals 1.)

📝Problem 3: Finding All Fifth Roots of Unity

Find all 5th roots of unity and express them in both exponential and rectangular form.

💡Solution

The 5th roots of unity are $z_k = e^{2\pi ik/5}$ for $k = 0, 1, 2, 3, 4$ .

Let $\omega = e^{2\pi i/5} = \cos(2\pi/5) + i\sin(2\pi/5)$ . Using the exact values $\cos(2\pi/5) = \frac{\sqrt{5}-1}{4}$ and $\sin(2\pi/5) = \frac{\sqrt{10+2\sqrt{5}}}{4}$ :

$z_0 = 1$
$z_1 = \omega = \frac{\sqrt{5}-1}{4} + \frac{\sqrt{10+2\sqrt{5}}}{4}i$
$z_2 = \omega^2 = e^{4\pi i/5} = \frac{-\sqrt{5}-1}{4} + \frac{\sqrt{10-2\sqrt{5}}}{4}i$
$z_3 = \omega^3 = e^{6\pi i/5} = \frac{-\sqrt{5}-1}{4} - \frac{\sqrt{10-2\sqrt{5}}}{4}i$
$z_4 = \omega^4 = e^{8\pi i/5} = \frac{\sqrt{5}-1}{4} - \frac{\sqrt{10+2\sqrt{5}}}{4}i$

Properties: These form a regular pentagon on the unit circle. They satisfy $z^5 = 1$ and $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$ .

📝Problem 4: Modulus Inequalities

Prove that $|z_1 + z_2| \leq |z_1| + |z_2|$ for all $z_1, z_2 \in \mathbb{C}$ .

💡Solution

Step 1: Compute $|z_1 + z_2|^2 = (z_1 + z_2)\overline{(z_1 + z_2)} = (z_1 + z_2)(\overline{z_1} + \overline{z_2})$

= z_1\overline{z_1} + z_1\overline{z_2} + z_2\overline{z_1} + z_2\overline{z_2} = |z_1|^2 + z_1\overline{z_2} + \overline{z_1\overline{z_2}} + |z_2|^2

Step 2: Note $z_1\overline{z_2} + \overline{z_1\overline{z_2}} = 2\operatorname{Re}(z_1\overline{z_2})$ .

So $|z_1 + z_2|^2 = |z_1|^2 + 2\operatorname{Re}(z_1\overline{z_2}) + |z_2|^2$ .

Step 3: Since $\operatorname{Re}(w) \leq |w|$ for any $w$ , we have $2\operatorname{Re}(z_1\overline{z_2}) \leq 2|z_1||\overline{z_2}| = 2|z_1||z_2|$ .

Step 4: Therefore $|z_1 + z_2|^2 \leq |z_1|^2 + 2|z_1||z_2| + |z_2|^2 = (|z_1| + |z_2|)^2$ .

Taking square roots: $|z_1 + z_2| \leq |z_1| + |z_2|$ . $\square$

📝Problem 5: Complex Equation Solving

Solve $z^2 + 2z + 5 = 0$ for $z \in \mathbb{C}$ .

💡Solution

Step 1: Apply the quadratic formula: $z = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2}$

Step 2: $z_1 = -1 + 2i$ and $z_2 = -1 - 2i$

Verification: $(z+1)^2 = z^2 + 2z + 1$ , so $z^2 + 2z + 5 = (z+1)^2 + 4 = 0$ implies $(z+1)^2 = -4$ , so $z+1 = \pm 2i$ , giving $z = -1 \pm 2i$ . ✓

Geometric interpretation: These roots are reflections of each other across the real axis (they are complex conjugates). They lie at distance $\sqrt{1+4} = \sqrt{5}$ from the origin.

Common Mistakes

Mistake	Correction	Example
Forgetting that $\arg(z)$ is multi-valued	The principal argument $\operatorname{Arg}(z) \in (-\pi, \pi]$ , but $\arg(z) = \operatorname{Arg}(z) + 2k\pi$	$\arg(-1) = \pi + 2k\pi$
Multiplying conjugates incorrectly	$\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$ , NOT $\overline{z_1} + \overline{z_2}$	$\overline{(2+3i)(1+i)} = (2-3i)(1-i)$
Confusing $\|z\|^2$ with $z^2$	$\|z\|^2 = z\overline{z} = a^2+b^2$ (real), while $z^2 = a^2 - b^2 + 2abi$ (complex)	$\|3+4i\|^2 = 25 \neq (3+4i)^2 = -7+24i$
Wrong quadrant for argument	Use $\operatorname{atan2}(y, x)$ , not just $\arctan(y/x)$	$\arg(-1-i) = -3\pi/4$ , not $\pi/4$
Assuming $e^{i\theta} = \cos\theta + \sin\theta$	It's $e^{i\theta} = \cos\theta + i\sin\theta$ — don't forget the $i$	$e^{i\pi/2} = i$ , not $1$
Applying De Moivre to negative $r$	De Moivre requires $r > 0$ ; rewrite negative modulus first	$(-2e^{i\theta})^n = 2^n e^{in(\theta + \pi)}$
Forgetting that roots come in conjugate pairs	If $z_0$ is a root of a real polynomial, so is $\overline{z_0}$	If $2+i$ is a root, so is $2-i$

Interview / Exam Questions

Q1: What is Euler's formula, and why is it significant?

A1: Euler's formula states $e^{i\theta} = \cos\theta + i\sin\theta$ . Its significance is threefold: (1) it unifies exponential and trigonometric functions, (2) it provides a compact representation of rotations in the complex plane ( $e^{i\theta}$ rotates a point by angle $\theta$ ), and (3) it yields Euler's identity $e^{i\pi} + 1 = 0$ , connecting five fundamental constants. It is the foundation of Fourier analysis, phasor representation in electrical engineering, and much of complex analysis.

Q2: Why can't $|z|^2$ equal $z^2$ for a nonzero complex number?

A2: $|z|^2 = a^2 + b^2$ is always a non-negative real number, while $z^2 = a^2 - b^2 + 2abi$ is real only if $ab = 0$ . For $z = 1+i$ : $|z|^2 = 2$ but $z^2 = 2i$ . The modulus squared is a geometric quantity (distance squared from origin), while $z^2$ is an algebraic operation (squaring the complex number).

Q3: What are the $n$ -th roots of unity, and what geometric figure do they form?

A3: The $n$ -th roots of unity are the solutions $z^n = 1$ , given by $z_k = e^{2\pi ik/n}$ for $k = 0, 1, \ldots, n-1$ . They are equally spaced on the unit circle at angles $0, 2\pi/n, 4\pi/n, \ldots, 2\pi(n-1)/n$ . Geometrically, they are the vertices of a regular $n$ -gon inscribed in the unit circle. For $n = 3$ they form an equilateral triangle; for $n = 4$ , a square; for $n = 6$ , a regular hexagon.

Q4: If $z_1 z_2 = 0$ , must $z_1 = 0$ or $z_2 = 0$ ? Prove or disprove.

A4: Yes. If $z_1 z_2 = 0$ , then $|z_1 z_2| = |z_1| \cdot |z_2| = 0$ . Since $|z_1|$ and $|z_2|$ are non-negative reals, one of them must be zero. If $|z_1| = 0$ then $z_1 = 0$ (since $|z| = 0$ iff $z = 0$ ). Similarly for $z_2$ . This is the zero-product property, which holds in $\mathbb{C}$ just as in $\mathbb{R}$ .

Q5: How do you compute $\arg(z_1 / z_2)$ and what subtleties arise?

A5: $\arg(z_1/z_2) = \arg(z_1) - \arg(z_2) \pmod{2\pi}$ . The subtlety is that if you use principal arguments, the result may not be a principal argument. For example, $\operatorname{Arg}(-1) = \pi$ and $\operatorname{Arg}(i) = \pi/2$ , but $\operatorname{Arg}(-1/i) = \operatorname{Arg}(i) = \pi/2 \neq \pi - \pi/2 = \pi/2$ (happens to work here). But $\operatorname{Arg}((-1)/(1+i))$ needs adjustment. Always reduce modulo $2\pi$ to the interval $(-\pi, \pi]$ .

Q6: Prove that if $|z| = 1$ , then $\overline{z} = 1/z$ .

A6: If $|z| = 1$ , then $z\overline{z} = |z|^2 = 1$ . Dividing both sides by $z$ (which is nonzero since $|z|=1$ ): $\overline{z} = 1/z$ . Geometrically, this means the conjugate of a point on the unit circle is its reciprocal, which is also on the unit circle.

Quick Reference

📋Formula Summary

Formula	Expression	Notes
Modulus	$\\|z\\| = \sqrt{a^2 + b^2}$	Distance from origin
Conjugate	$\overline{a+bi} = a - bi$	Reflection across real axis
Polar Form	$z = re^{i\theta}$	$r = \\|z\\|$ , $\theta = \arg(z)$
Euler's Formula	$e^{i\theta} = \cos\theta + i\sin\theta$	Fundamental identity
De Moivre	$(re^{i\theta})^n = r^n e^{in\theta}$	Powers in polar form
N-th Root	$z^{1/n} = r^{1/n}e^{i(\theta+2k\pi)/n}$	$n$ distinct roots
Product	$z_1z_2 = r_1r_2 e^{i(\theta_1+\theta_2)}$	Multiply moduli, add arguments
Quotient	$z_1/z_2 = (r_1/r_2)e^{i(\theta_1-\theta_2)}$	Divide moduli, subtract arguments
Triangle Inequality	$\\|z_1+z_2\\| \leq \\|z_1\\| + \\|z_2\\|$	Equality iff $z_1, z_2$ are collinear
Modulus Squared	$z\overline{z} = \\|z\\|^2$	Always real, non-negative
Cosine Formula	$\cos\theta = (e^{i\theta} + e^{-i\theta})/2$	From Euler's formula
Sine Formula	$\sin\theta = (e^{i\theta} - e^{-i\theta})/(2i)$	From Euler's formula

Cross-References

092 - Complex Functions — Analyticity, Cauchy-Riemann equations, and conformal mappings build on the algebraic foundations of complex numbers.
093 - Contour Integration — Contour integrals use polar form and De Moivre's theorem to parameterize paths in the complex plane.
094 - Residue Theory — Finding poles and computing residues requires fluency with complex arithmetic and roots of unity.
095 - Applications — Signal processing (Fourier transforms) and control theory (Z-transforms) use complex numbers as their fundamental language.
Linear Algebra (Topic 14) — Eigenvalues of real matrices may be complex; the characteristic polynomial roots live in $\mathbb{C}$ .
Differential Equations — Complex exponentials $e^{(a+bi)t}$ arise in solutions to linear ODEs with complex characteristic roots.