Complex Numbers

Ask a beginner to solve $x^2 = -1$ and they will tell you, correctly, that no real number squared gives a negative. So mathematicians did something audacious: they invented a new number that does. That single act — declaring a symbol $i$ with the property $i^2 = -1$ — opens a whole new landscape where every polynomial has roots, where rotation becomes multiplication, and where the equations governing electricity, quantum mechanics, and signal processing suddenly become simple.

Complex numbers are not a mathematical curiosity or a "fake" number system. They are as concrete and useful as the real numbers, and they are the natural completion of arithmetic. This page will show you what they are, how to compute with them fluently, and — just as important — why anyone ever needed them.

Learning Objectives

• Understand the imaginary unit $i$ and why it was introduced.
• Add, subtract, multiply, and divide complex numbers with confidence.
• Plot complex numbers on the complex (Argand) plane.
• Compute the modulus and conjugate, and use conjugates to divide.
• Convert between rectangular form $a + bi$ and polar form $r(\cos\theta + i\sin\theta)$.
• Appreciate the historical struggle that made complex numbers "respectable."

Quick Answer

A complex number has the form $a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit defined by $i^2 = -1$. Here $a$ is the real part and $b$ is the imaginary part. You add and multiply them just like binomials, replacing $i^2$ with $-1$ whenever it appears. Every complex number corresponds to a point $(a, b)$ in the complex plane. Its modulus $|a + bi| = \sqrt{a^2 + b^2}$ is its distance from the origin, and its conjugate $a - bi$ is its mirror image across the real axis. In polar form, a complex number is described by its modulus $r$ and angle $\theta$, which makes multiplication and taking powers remarkably easy.

Where It Came From

The story of complex numbers is a story of reluctant acceptance. They did not arrive to solve $x^2 = -1$ — nobody in the 1500s cared about equations with no solutions. They arrived, unexpectedly, from equations that clearly did have real solutions.

In 1545 the Italian mathematician Gerolamo Cardano, in his book Ars Magna, published a formula for solving cubic equations. When he applied it to certain cubics, the formula demanded that he take the square root of a negative number in the middle of the calculation — even though the final answer turned out to be a perfectly ordinary real number. Cardano was baffled. He famously worked out $5 + \sqrt{-15}$ and $5 - \sqrt{-15}$ as solutions to a problem, remarked that the reasoning was "as subtle as it is useless," and moved on.

The breakthrough came from Rafael Bombelli around 1572. Studying the cubic $x^3 = 15x + 4$, whose obvious real solution is $x = 4$, Bombelli noticed that Cardano's formula produced $\sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}}$. Rather than dismiss the negative roots, Bombelli boldly assumed they obeyed ordinary algebra, computed with them, watched the "impossible" parts cancel, and recovered $x = 4$. This was the crucial insight: even if you did not know what $\sqrt{-1}$ was, you could compute with it consistently and get true results.

For two more centuries these quantities were treated with suspicion — René Descartes coined the dismissive term "imaginary" in 1637. Rehabilitation came from two giants. Leonhard Euler (1700s) introduced the symbol $i$ for $\sqrt{-1}$ and discovered the astonishing formula $e^{i\theta} = \cos\theta + i\sin\theta$, linking complex numbers to trigonometry and exponentials. Finally Carl Friedrich Gauss (early 1800s) gave them a concrete geometric home as points in a plane and proved the Fundamental Theorem of Algebra: every polynomial of degree $n$ has exactly $n$ roots in the complex numbers. What had been "impossible" was now the setting in which algebra becomes complete.

The Imaginary Unit and the Form a + bi

Everything begins with one definition:

$i^2 = -1, \qquad \text{equivalently} \qquad i = \sqrt{-1}.$

From this, powers of $i$ cycle with period four:

$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^5 = i, \ldots$

To simplify a high power of $i$, divide the exponent by 4 and keep the remainder. For example, $i^{35}$: since $35 = 4 \times 8 + 3$, we have $i^{35} = i^3 = -i$.

A complex number is any expression $z = a + bi$ with $a, b$ real. We call $a = \operatorname{Re}(z)$ the real part and $b = \operatorname{Im}(z)$ the imaginary part (note: the imaginary part is the real number $b$, not $bi$). Real numbers are just complex numbers with $b = 0$, and pure imaginary numbers have $a = 0$.

Worked example. Simplify $\sqrt{-49}$. $\sqrt{-49} = \sqrt{49}\cdot\sqrt{-1} = 7i.$

Arithmetic: Add, Subtract, Multiply, Divide

Addition and subtraction combine real parts with real parts and imaginary with imaginary:

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

Worked example. $(3 + 2i) + (1 - 5i) = (3 + 1) + (2 - 5)i = 4 - 3i.$

Multiplication uses the distributive law (FOIL), then replaces $i^2$ with $-1$:

$(a + bi)(c + di) = ac + adi + bci + bd\,i^2 = (ac - bd) + (ad + bc)i.$

Worked example. Multiply $(3 + 2i)(1 - 5i)$. $= 3(1) + 3(-5i) + 2i(1) + 2i(-5i) = 3 - 15i + 2i - 10i^2.$ Since $i^2 = -1$, the last term is $-10(-1) = +10$: $= (3 + 10) + (-15 + 2)i = 13 - 13i.$

Division is the clever step. To divide, multiply the top and bottom by the conjugate of the denominator, which turns the denominator into a real number.

Worked example. Compute $\dfrac{2 + 3i}{1 - 2i}$. Multiply by the conjugate $1 + 2i$: $\frac{2 + 3i}{1 - 2i}\cdot\frac{1 + 2i}{1 + 2i} = \frac{(2 + 3i)(1 + 2i)}{(1 - 2i)(1 + 2i)}.$ Numerator: $2 + 4i + 3i + 6i^2 = 2 + 7i - 6 = -4 + 7i$. Denominator: $1^2 - (2i)^2 = 1 - 4i^2 = 1 + 4 = 5$. $= \frac{-4 + 7i}{5} = -\frac{4}{5} + \frac{7}{5}i.$

The Complex Plane, Modulus, and Conjugate

Gauss's key idea was to picture $z = a + bi$ as the point $(a, b)$ — with the horizontal real axis and the vertical imaginary axis. This picture is called the complex plane or Argand diagram (after Jean-Robert Argand, who published it in 1806).

The modulus (or absolute value) is the distance from the origin to that point, by the Pythagorean theorem:

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

The conjugate, written $\bar{z}$, reflects the point across the real axis by flipping the sign of the imaginary part:

$\overline{a + bi} = a - bi.$

A beautiful identity ties them together: $z\bar{z} = a^2 + b^2 = |z|^2$. This is exactly why multiplying by the conjugate clears the denominator in division — it produces a real number.

Worked example. For $z = 3 - 4i$:

• Conjugate: $\bar{z} = 3 + 4i$.
• Modulus: $|z| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
• Check: $z\bar{z} = (3 - 4i)(3 + 4i) = 9 + 16 = 25 = |z|^2.$ ✓

Polar Form and Multiplication as Rotation

Because a complex number is a point in the plane, we can also locate it by its distance $r = |z|$ from the origin and the angle $\theta$ it makes with the positive real axis (the argument). Then:

$a = r\cos\theta, \quad b = r\sin\theta, \qquad z = r(\cos\theta + i\sin\theta).$

Using Euler's formula this is written compactly as $z = r e^{i\theta}$. To find $r$ and $\theta$ from $a + bi$: $r = \sqrt{a^2 + b^2}$ and $\theta = \arctan(b/a)$, adjusted for the correct quadrant.

The payoff is enormous. To multiply two complex numbers in polar form, you multiply the moduli and add the angles:

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

So multiplication by a complex number rotates and scales. Raising to a power (De Moivre's theorem) becomes $z^n = r^n e^{in\theta}$.

Worked example. Convert $z = 1 + i$ to polar form and compute $z^8$.

• Modulus: $r = \sqrt{1^2 + 1^2} = \sqrt{2}$.
• Argument: $\theta = \arctan(1/1) = 45^\circ = \pi/4$ (first quadrant).
• Polar form: $z = \sqrt{2}\,(\cos 45^\circ + i\sin 45^\circ)$.
• Power: $z^8 = (\sqrt{2})^8\,(\cos(8\cdot 45^\circ) + i\sin(8\cdot 45^\circ)) = 16(\cos 360^\circ + i\sin 360^\circ) = 16(1 + 0i) = 16.$

Doing $(1+i)^8$ by hand in rectangular form would be painful; polar form makes it a one-liner.

Real-World Applications

• Electrical engineering. Alternating-current circuits use complex numbers (engineers write $j$ instead of $i$) to track both the magnitude and phase of voltages and currents. Impedance combines resistance and reactance into a single complex quantity, turning calculus problems into algebra.
• Signal processing. The Fourier transform decomposes signals into complex exponentials $e^{i\omega t}$. This underlies audio compression (MP3), image formats (JPEG), and wireless communication.
• Quantum mechanics. The wavefunction that describes a particle is complex-valued; complex amplitudes are not optional here but fundamental to the theory.
• Control systems and stability. Whether a bridge, aircraft, or feedback loop is stable depends on where certain roots sit in the complex plane.
• Fractals and computer graphics. The Mandelbrot set is defined by iterating $z \to z^2 + c$ over complex numbers; 2D rotations in graphics can be encoded as complex multiplication.

Common Mistakes

Mistake 1: Writing $\sqrt{-4}\cdot\sqrt{-9} = \sqrt{36} = 6$. Why it is wrong: the rule $\sqrt{x}\sqrt{y} = \sqrt{xy}$ only holds for non-negative reals. Correction: convert to $i$ first. $\sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6$. The correct answer is $-6$, not $6$.

Mistake 2: Confusing the imaginary part with $bi$. Why it is wrong: for $z = 3 + 5i$, students say the imaginary part is $5i$. By definition, $\operatorname{Im}(z) = 5$, a real number — not $5i$. The $i$ is a label, not part of the value.

Mistake 3: Forgetting that $i^2 = -1$ during multiplication. Why it is wrong: treating $i$ like an ordinary variable and leaving $i^2$ in the answer. Correction: always substitute $i^2 = -1$ and combine, e.g. $(2i)(4i) = 8i^2 = -8$, not $8i^2$.

Mistake 4 (bonus): Getting the argument's quadrant wrong. $\arctan(b/a)$ from a calculator only returns angles between $-90^\circ$ and $90^\circ$. For a number like $-1 - i$ (third quadrant), you must add $180^\circ$ to the raw arctan result.

Comparison and Connections

Feature	Real Numbers $\mathbb{R}$	Complex Numbers $\mathbb{C}$
Form	$a$	$a + bi$
Visualize as	Points on a line	Points in a plane
$x^2 = -1$ solvable?	No	Yes ($x = \pm i$)
Ordered ($<$, $>$)?	Yes	No meaningful order
Polynomial roots	May have none	Always has $n$ roots (degree $n$)

Complex numbers extend the reals: every real number is a complex number with zero imaginary part. The trade-off is that you lose ordering — it makes no sense to say $i < 2i$ — but you gain algebraic completeness. The modulus $|z|$ generalizes the absolute value of a real number, and the conjugate is closely tied to reflection and to the dot product in the plane. Polar form connects complex numbers directly to trigonometry and to the exponential function through Euler's formula.

Practice Questions

Recall

Simplify $i^{26}$. Answer: $26 = 4\times 6 + 2$, so $i^{26} = i^2 = -1$.

Understanding

Find the modulus and conjugate of $z = -6 + 8i$. Answer: $\bar{z} = -6 - 8i$; $|z| = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.

Application

Compute $\dfrac{5}{2 + i}$. Guidance: Multiply top and bottom by the conjugate $2 - i$. Numerator: $5(2 - i) = 10 - 5i$. Denominator: $(2 + i)(2 - i) = 4 + 1 = 5$. Result: $\dfrac{10 - 5i}{5} = 2 - i$.

Analysis

Convert $z = -1 + \sqrt{3}\,i$ to polar form, then find $z^3$. Guidance: $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. The point is in the second quadrant; raw $\arctan(\sqrt{3}/(-1)) = -60^\circ$, so add $180^\circ$ to get $\theta = 120^\circ$. Then $z^3 = 2^3(\cos 360^\circ + i\sin 360^\circ) = 8(1) = 8$. So $z^3 = 8$ — meaning $z$ is a cube root of 8 that is not real.

FAQ

Is $i$ a real thing, or just made up? It is exactly as "real" as negative numbers or zero — all were once distrusted. "Imaginary" is an unfortunate historical name. Complex numbers describe physical reality (AC circuits, quantum states) so precisely that engineers rely on them daily. They are a legitimate, rigorous number system.

Why can't complex numbers be ordered like real numbers? Any consistent ordering would have to make $i > 0$ or $i < 0$. Both lead to contradictions: for instance, if $i > 0$ then $i^2 > 0$, but $i^2 = -1 < 0$. So no order relation is compatible with the arithmetic.

What is the difference between the modulus and the argument? The modulus $r$ is how far the number is from the origin (its size). The argument $\theta$ is which direction it points (its angle). Together they pin down the number, just like $(a, b)$ coordinates do.

When should I use polar form instead of $a + bi$? Use rectangular form ($a + bi$) for adding and subtracting. Switch to polar form for multiplication, division, powers, and roots — angles add and moduli multiply, which is far easier than expanding by hand.

Does every equation have complex solutions? Every non-constant polynomial equation does — that is the Fundamental Theorem of Algebra. A degree-$n$ polynomial has exactly $n$ complex roots (counting repeats). Not all equations are polynomials, but within algebra, complex numbers close the gap that reals leave open.

Quick Revision

• Imaginary unit: $i^2 = -1$; powers cycle $i, -1, -i, 1$ with period 4.
• Standard form: $z = a + bi$; $\operatorname{Re}(z) = a$, $\operatorname{Im}(z) = b$.
• Add/subtract: combine like parts. Multiply: FOIL, then $i^2 = -1$.
• Conjugate: $\overline{a + bi} = a - bi$; divide by multiplying by the conjugate.
• Modulus: $|z| = \sqrt{a^2 + b^2}$, and $z\bar{z} = |z|^2$.
• Polar form: $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$, with $r = |z|$, $\theta = \arctan(b/a)$ (fix the quadrant).
• Multiply in polar: multiply moduli, add angles. Powers: $z^n = r^n e^{in\theta}$ (De Moivre).

Related Topics

Prerequisites

• Algebra overview
• Quadratic equations and the quadratic formula
• Basic trigonometry (sine, cosine, and angles)

Related Topics

• Polynomials and the Fundamental Theorem of Algebra
• Trigonometry and the unit circle (for polar form)

Next Topics

• De Moivre's theorem and roots of unity
• Euler's formula and exponential form
• Vectors and the geometry of the plane