1. The equation that breaks the reals
Start with what real numbers can do. The equation $x^2 - 1 = 0$ has two real solutions, $x = \pm 1$. The equation $x^2 = 0$ has one. And the equation
$$ x^2 + 1 = 0 $$has none — because $x^2 \geq 0$ for every real $x$, so $x^2 + 1$ is always at least $1$ and never zero. The graph of $y = x^2 + 1$ floats above the x-axis and never touches it. As far as $\mathbb{R}$ is concerned, this equation simply doesn't have an answer.
That is irritating. Every other quadratic with a discriminant problem has the same flavor: $x^2 + 4 = 0$, $x^2 - 2x + 5 = 0$, and so on. There's a whole family of perfectly reasonable-looking polynomials that the real numbers refuse to factor.
The fix is breathtakingly cheap. Invent a new symbol whose only job is to be a square root of $-1$. Call it $i$, declare
$$ i^2 = -1, $$and see what happens when you let $i$ mingle with the real numbers under the usual rules of arithmetic. What you get is not a kludge. It's a new number system — the complex numbers $\mathbb{C}$ — in which every polynomial equation has a solution, not just $x^2 + 1 = 0$. That sweeping payoff is the Fundamental Theorem of Algebra, and it's the reason the gamble was worth taking.
Descartes coined "imaginary" in 1637 as a dismissal. The name stuck even though the numbers are no more imaginary than negative numbers or fractions — all of which were once dismissed too. Quantum mechanical state vectors literally live in complex Hilbert spaces; the universe does not seem to share Descartes' opinion.
2. Defining $i$ and the complex numbers
A complex number is an expression $z = a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit satisfying $i^2 = -1$. The set of all complex numbers is written $\mathbb{C} = \{\, a + bi \mid a, b \in \mathbb{R}\,\}$.
The real part is $\operatorname{Re}(z) = a$ and the imaginary part is $\operatorname{Im}(z) = b$. Both are real numbers — the imaginary part is just $b$, not $bi$.
Some examples, with their parts laid out:
| $z$ | $\operatorname{Re}(z)$ | $\operatorname{Im}(z)$ | Notes |
|---|---|---|---|
| $3 + 4i$ | $3$ | $4$ | Generic complex |
| $-2i$ | $0$ | $-2$ | Pure imaginary |
| $7$ | $7$ | $0$ | A real number is a complex number with no imaginary part |
| $\tfrac{1}{2} - \tfrac{\sqrt{3}}{2}i$ | $\tfrac{1}{2}$ | $-\tfrac{\sqrt 3}{2}$ | Lives on the unit circle (we'll see why) |
Two complex numbers are equal iff their real parts match and their imaginary parts match. So $a + bi = c + di$ means $a = c$ and $b = d$ — one complex equation is actually two real equations.
$i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, $i^5 = i$, $\ldots$ — the pattern has period four. To compute $i^n$, reduce $n$ modulo 4. $i^{100} = i^{4 \cdot 25} = 1$; $i^{101} = i$.
3. Four operations on $a + bi$
Treat $i$ as a variable, do the usual algebra, and replace $i^2$ with $-1$ whenever it appears. That single rule — plus ordinary distributivity and commutativity — generates every formula below.
Addition and subtraction
Just combine real and imaginary parts separately:
$$ (a + bi) + (c + di) = (a + c) + (b + d)i $$ $$ (a + bi) - (c + di) = (a - c) + (b - d)i $$Identical in shape to adding 2-D vectors. Example: $(3 + 4i) + (1 - 2i) = 4 + 2i$.
Multiplication — the interesting one
Expand the product like a binomial, then collapse using $i^2 = -1$:
$$ \begin{aligned} (a + bi)(c + di) &= ac + adi + bci + bdi^2 \\ &= ac + adi + bci - bd \\ &= (ac - bd) + (ad + bc)i. \end{aligned} $$You don't need to memorize the final formula — just remember the procedure: distribute and substitute. Example: $(3 + 4i)(1 - 2i) = 3 - 6i + 4i - 8i^2 = 3 - 2i + 8 = 11 - 2i$.
This is the magical bit. The same rule that lets you square the binomial $(a + b)^2$ also encodes — once we see the polar form in a moment — rotation in the plane. A piece of high-school algebra has secretly been doing geometry the entire time.
Division
Postponed for one section, because the trick that makes it clean — multiplying by the conjugate — deserves its own subsection.
4. Conjugates, modulus, and division
The conjugate of $z = a + bi$ is $\bar{z} = a - bi$ — flip the sign of the imaginary part.
The conjugate is the closest thing in $\mathbb{C}$ to "the other one." Conjugation respects all the operations: $\overline{z + w} = \bar{z} + \bar{w}$ and $\overline{zw} = \bar{z}\bar{w}$. It also fixes the real numbers: if $z$ is real, $\bar{z} = z$.
The single fact that makes the conjugate useful is what happens when you multiply $z$ by it:
$$ z \bar{z} = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2 $$That product is a real, non-negative number. Its square root is the modulus (or absolute value):
$$ |z| = \sqrt{a^2 + b^2}, \qquad |z|^2 = z\bar{z}. $$Geometrically, $|z|$ is the distance from the origin to the point $(a, b)$ — just the Pythagorean theorem in fresh clothes. Example: $|3 + 4i| = \sqrt{9 + 16} = 5$.
The conjugate trick for division
To divide $z/w$, multiply top and bottom by $\bar{w}$. The denominator becomes $w\bar{w} = |w|^2$, which is real, and a real denominator makes life easy.
$$ \frac{a + bi}{c + di} \;=\; \frac{(a + bi)(c - di)}{(c + di)(c - di)} \;=\; \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} $$Worked example: $\dfrac{1 + 2i}{3 - i}$. Multiply by $\dfrac{3 + i}{3 + i}$:
$$ \frac{(1 + 2i)(3 + i)}{(3 - i)(3 + i)} \;=\; \frac{3 + i + 6i + 2i^2}{9 + 1} \;=\; \frac{1 + 7i}{10} \;=\; \tfrac{1}{10} + \tfrac{7}{10}i. $$Notice that this also proves $\mathbb{C}$ is a field: every nonzero complex number has a multiplicative inverse, namely $1/z = \bar{z}/|z|^2$.
The modulus is the hypotenuse of a right triangle with legs $|a|$ and $|b|$ — not the sum of the legs. $|3 + 4i| = 5$, not $7$.
5. The complex plane
Identify the complex number $a + bi$ with the point $(a, b)$ in the plane. The horizontal axis is the real axis, the vertical axis is the imaginary axis, and the whole picture is called the Argand diagram (after Jean-Robert Argand, who published it in 1806; Wessel had the same idea seven years earlier).
Under this identification:
- Addition is vector addition. Adding $z$ and $w$ slides $w$ over by $z$'s coordinates — the parallelogram rule from physics.
- Modulus is distance from the origin. $|z|$ is the length of the vector to $(a, b)$.
- Conjugation is reflection across the real axis. $z = a + bi \mapsto \bar{z} = a - bi$ flips the sign of the vertical coordinate.
- Multiplication is rotation and scaling. This is the headline fact, and it pops out cleanly only once we switch to polar form (next section).
Every complex number is two things at once: an algebraic expression $a + bi$ you can manipulate symbolically, and a point/arrow in the plane you can manipulate geometrically. The whole power of $\mathbb{C}$ comes from being able to switch between the two views at will.
6. Polar form
A point in the plane can also be specified by its distance from the origin and the angle the vector makes with the positive real axis. Call those $r$ and $\theta$:
$$ a = r\cos\theta, \qquad b = r\sin\theta. $$Substituting into $z = a + bi$ gives the polar form:
$$ z = r(\cos\theta + i\sin\theta), \qquad r = |z|, \quad \theta = \arg(z). $$Here $r \geq 0$ is the modulus and $\theta$ is the argument of $z$ — the angle, measured in radians, from the positive real axis. The argument is only defined modulo $2\pi$, since adding a full turn lands you on the same point; the principal value is the unique choice in $(-\pi, \pi]$.
Converting between forms
Rectangular to polar:
$$ r = \sqrt{a^2 + b^2}, \qquad \theta = \operatorname{atan2}(b, a) $$where $\operatorname{atan2}$ is just $\arctan(b/a)$ adjusted to give the angle in the correct quadrant. Polar to rectangular: $a = r\cos\theta$, $b = r\sin\theta$.
Example: $z = 1 + i$ has $r = \sqrt{2}$ and $\theta = \pi/4$ (it's $45^\circ$ above the real axis), so $1 + i = \sqrt{2}\bigl(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\bigr)$.
Why polar form is worth the bookkeeping
Watch what happens when you multiply two complex numbers in polar form:
$$ \begin{aligned} &r_1(\cos\theta_1 + i\sin\theta_1) \cdot r_2(\cos\theta_2 + i\sin\theta_2) \\ &= r_1 r_2 \bigl[(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) + i(\cos\theta_1\sin\theta_2 + \sin\theta_1\cos\theta_2)\bigr] \\ &= r_1 r_2 \bigl[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\bigr]. \end{aligned} $$That last step uses the angle-sum identities from trigonometry. The result is gorgeous:
Multiplying complex numbers multiplies their moduli and adds their arguments.
That's what we meant by "rotation and scaling." Multiplying by $i = \cos(\pi/2) + i\sin(\pi/2)$ doesn't change any magnitude — it just rotates by $90^\circ$. Multiplying by $2i$ rotates by $90^\circ$ and doubles the length. The complicated-looking multiplication formula $(ac - bd) + (ad + bc)i$ is, in disguise, simple rotation.
7. Euler's formula
Now the punchline. The angle-addition rule for multiplication of complex numbers in polar form looks exactly like the law of exponents — $\theta_1 + \theta_2$ in the angle, just like $e^{x_1} \cdot e^{x_2} = e^{x_1 + x_2}$. That can't be a coincidence, and it isn't.
For every real $\theta$,
$$ e^{i\theta} = \cos\theta + i\sin\theta. $$
Setting $\theta = \pi$ gives the most-quoted equation in mathematics:
$$ e^{i\pi} + 1 = 0 \qquad \text{(Euler's identity)} $$which packages $0$, $1$, $i$, $\pi$, and $e$ into a single relationship.
With Euler's formula in hand, the polar form becomes the exponential form:
$$ z = r e^{i\theta}. $$And the multiplication rule simplifies to plain exponent arithmetic:
$$ \left(r_1 e^{i\theta_1}\right)\left(r_2 e^{i\theta_2}\right) = r_1 r_2\, e^{i(\theta_1 + \theta_2)}, \qquad \frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \frac{r_1}{r_2} e^{i(\theta_1 - \theta_2)}. $$Why this is true (informally)
A full derivation belongs in calculus, where the Taylor series of $e^x$, $\cos x$, and $\sin x$ are well in hand. But the shape of the argument is worth seeing now. Each function has a power series:
$$ \begin{aligned} e^{x} &= 1 + x + \tfrac{x^2}{2!} + \tfrac{x^3}{3!} + \tfrac{x^4}{4!} + \cdots \\ \cos x &= 1 - \tfrac{x^2}{2!} + \tfrac{x^4}{4!} - \cdots \\ \sin x &= x - \tfrac{x^3}{3!} + \tfrac{x^5}{5!} - \cdots \end{aligned} $$Substitute $i\theta$ into the series for $e^x$ and group the terms by whether the power of $i$ is even or odd. Using $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, the even-power terms reassemble into $\cos\theta$ and the odd-power terms reassemble into $i\sin\theta$. Done — modulo the analytic question of whether you're allowed to substitute imaginary numbers into a power series, which is a real (and answerable) concern that calculus class will resolve.
Justifying "you're allowed to plug $i\theta$ into the series for $e^x$" requires the theory of power series convergence in the complex plane. That work is exactly what complex analysis is for. For now, take Euler's formula as a well-earned definition and enjoy what it does.
Multiplication by $e^{i\alpha}$ is rotation
Since $|e^{i\alpha}| = \sqrt{\cos^2\alpha + \sin^2\alpha} = 1$, multiplying any complex number by $e^{i\alpha}$ keeps its modulus the same and adds $\alpha$ to its argument — a pure rotation by angle $\alpha$ around the origin. This single fact is the reason engineers reach for $e^{i\omega t}$ to describe oscillating signals.
8. De Moivre's theorem
Euler's formula plus the law of exponents gives an immediate consequence. Raise $e^{i\theta}$ to the $n$-th power:
$$ \bigl(e^{i\theta}\bigr)^n = e^{in\theta}. $$Expanding both sides via Euler's formula:
For every integer $n$,
$$ (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta). $$
The theorem is named after Abraham de Moivre (1722), who stated it before Euler tied the bow with the exponential. It's the right tool whenever you want to raise a complex number to an integer power without grinding through a binomial expansion.
Example. Compute $(1 + i)^8$ in two breaths:
- Convert: $1 + i = \sqrt{2}\, e^{i\pi/4}$.
- Raise: $\bigl(\sqrt{2}\bigr)^8 \cdot e^{i \cdot 8 \cdot \pi/4} = 16\, e^{i 2\pi} = 16 \cdot 1 = 16$.
Try doing that by expanding $(1 + i)^8$ rectangularly. You can — but you won't enjoy it.
De Moivre is also a manufacturing plant for trig identities. Compare real and imaginary parts of $(\cos\theta + i\sin\theta)^2 = \cos(2\theta) + i\sin(2\theta)$ on each side and out drops
$$ \cos(2\theta) = \cos^2\theta - \sin^2\theta, \qquad \sin(2\theta) = 2\sin\theta\cos\theta $$— the double-angle formulae, derived for free from one line of algebra.
9. Roots of unity
The equation $z^n = 1$ — find all complex numbers whose $n$-th power is $1$ — looks innocent and has a beautiful answer. Write $z = re^{i\theta}$ and $1 = 1 \cdot e^{i \cdot 0}$. For $z^n = 1$ we need $r^n e^{in\theta} = e^0$, so $r = 1$ and $n\theta$ is a whole number of full turns: $n\theta = 2\pi k$ for some integer $k$.
That gives
$$ z_k = e^{2\pi i k / n}, \qquad k = 0, 1, 2, \ldots, n - 1. $$For $k \geq n$ the values just repeat, so there are exactly $n$ distinct solutions. They all sit on the unit circle (since $|z_k| = 1$), spaced at equal angles of $2\pi/n$. These are the $n$-th roots of unity.
The five fifth roots
For $n = 5$ the five solutions are $z_k = e^{2\pi i k / 5}$ for $k = 0, 1, 2, 3, 4$. They lie at angles $0$, $72^\circ$, $144^\circ$, $216^\circ$, $288^\circ$ on the unit circle — the vertices of a regular pentagon inscribed in the circle, with one vertex pinned at $z = 1$.
In general, the $n$-th roots of unity are the vertices of a regular $n$-gon inscribed in the unit circle, with one vertex always at $z = 1$. They satisfy a sum that's worth knowing: for $n \geq 2$,
$$ 1 + z_1 + z_2 + \cdots + z_{n-1} = 0 \quad (\text{the } n\text{-th roots of unity sum to zero}). $$This is just the statement that the centroid of a regular polygon centered at the origin is the origin — geometry and algebra agreeing again.
$n$-th roots of any complex number
The same method finds the $n$-th roots of any nonzero $w = \rho e^{i\phi}$. There are exactly $n$ of them, given by
$$ z_k = \rho^{1/n}\, e^{i(\phi + 2\pi k)/n}, \qquad k = 0, 1, \ldots, n - 1, $$evenly spaced around a circle of radius $\rho^{1/n}$. Example: $z^3 = 8 = 8 e^{i\cdot 0}$ has three cube roots — $2$, $2 e^{i 2\pi/3} = -1 + i\sqrt{3}$, and $2 e^{i 4\pi/3} = -1 - i\sqrt{3}$.
In the reals, $x^4 = 1$ has only two solutions — $\pm 1$. In the complexes, it has four — $\{1, i, -1, -i\}$. The "missing" solutions you couldn't find were on the imaginary axis the whole time. This is a small foretaste of what's about to happen at full scale: every polynomial gets all of its roots back, not just some.
10. Common pitfalls
$\operatorname{Im}(3 - 4i) = -4$, not $-4i$. The imaginary part is a real number — it's the coefficient of $i$, not $i$ itself.
The pleasant rule $\sqrt{x}\sqrt{y} = \sqrt{xy}$ holds for positive reals, not for negatives. For $a, b > 0$: $\sqrt{-a}\,\sqrt{-b} = (i\sqrt{a})(i\sqrt{b}) = i^2\sqrt{ab} = -\sqrt{ab}$. Don't pull negatives out of square roots before you've introduced the $i$.
Writing $\dfrac{a + bi}{c + di}$ and stopping is leaving the answer half-cooked. Multiply by $\overline{c + di}/\overline{c + di}$ so the denominator becomes real. Only then can you cleanly read off the real and imaginary parts.
They share three symbols and almost nothing else. $e^\theta$ is a positive real number that grows without bound. $e^{i\theta}$ has modulus $1$ — it lives on the unit circle and oscillates forever. Different functions, different worlds.
You cannot say "$z > 0$" or "$z_1 < z_2$" for non-real complex numbers. There is no total ordering of $\mathbb{C}$ compatible with its arithmetic — proven, not just inconvenient. Statements that look like inequalities on complex numbers are either nonsense, or really about $|z|$ (which is real and can be compared).
11. Worked examples
Open each only after you've tried it. The aim is to recognize the move, not memorize the answer.
Example 1 · Multiply $(2 + 3i)(4 - i)$
Expand and replace $i^2$ with $-1$:
$$ \begin{aligned} (2 + 3i)(4 - i) &= 8 - 2i + 12i - 3i^2 \\ &= 8 + 10i + 3 \\ &= 11 + 10i. \end{aligned} $$Example 2 · Simplify $\dfrac{5 + 2i}{1 - 3i}$
Multiply top and bottom by the conjugate of the denominator, $1 + 3i$:
$$ \frac{(5 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)} = \frac{5 + 15i + 2i + 6i^2}{1 + 9} = \frac{-1 + 17i}{10} = -\tfrac{1}{10} + \tfrac{17}{10}i. $$Example 3 · Modulus and argument of $-1 + i\sqrt{3}$
$|{-1 + i\sqrt{3}}| = \sqrt{1 + 3} = 2$. The point lies in the second quadrant (real part negative, imaginary part positive). Its reference angle has $\tan = \sqrt{3}/1$, so $\theta = \pi - \pi/3 = 2\pi/3$. Therefore
$$ -1 + i\sqrt{3} = 2\,e^{i\cdot 2\pi/3}. $$Example 4 · Compute $(1 - i)^6$
Convert: $1 - i$ has modulus $\sqrt{2}$ and argument $-\pi/4$, so $1 - i = \sqrt{2}\, e^{-i\pi/4}$. Raise to the sixth power:
$$ (1 - i)^6 = \bigl(\sqrt{2}\bigr)^6 e^{-i \cdot 6\pi/4} = 8\, e^{-i\, 3\pi/2} = 8\,(\cos\tfrac{-3\pi}{2} + i\sin\tfrac{-3\pi}{2}) = 8 \cdot (0 + i) = 8i. $$Example 5 · Find the four fourth roots of $-16$
Write $-16 = 16\, e^{i\pi}$. The four fourth roots are
$$ z_k = 16^{1/4}\, e^{i(\pi + 2\pi k)/4} = 2\, e^{i(\pi + 2\pi k)/4}, \qquad k = 0, 1, 2, 3. $$Plugging in $k = 0, 1, 2, 3$ gives angles $\pi/4$, $3\pi/4$, $5\pi/4$, $7\pi/4$, and therefore
$$ 2\,e^{i\pi/4} = \sqrt{2} + i\sqrt{2}, \quad -\sqrt{2} + i\sqrt{2}, \quad -\sqrt{2} - i\sqrt{2}, \quad \sqrt{2} - i\sqrt{2}. $$Four points on a circle of radius $2$, $90^\circ$ apart, with the first one rotated $45^\circ$ off the real axis — exactly a square.
12. Where this is heading: the fundamental theorem
We invented $i$ to solve a single equation, $x^2 + 1 = 0$. The astonishing payoff — and the reason mathematicians stopped calling $\mathbb{C}$ "imaginary" with a sneer — is that this single extension fixes everything.
Every non-constant polynomial with complex coefficients has at least one complex root. Equivalently: every polynomial of degree $n$ factors completely into $n$ linear factors over $\mathbb{C}$.
The proof — Gauss gave several, starting with his 1799 dissertation — is not algebraic. It uses topology and continuity, which is striking given how purely algebraic the statement sounds. We won't prove it here; it belongs in complex analysis. But the consequence is the one you want to carry away from this page:
By adjoining one number, we made every polynomial solvable. There is no further extension to make. $\mathbb{C}$ is, in this precise sense, the end of the line.
Once you trust that, complex numbers stop feeling like a curiosity and start feeling like the natural setting for almost every piece of algebra you'll encounter. Eigenvalues are complex. Roots of characteristic polynomials are complex. Solutions of linear ODEs are complex. Fourier analysis is complex. Quantum mechanics is complex. The geometry of rotation — the thing the multiplication rule was secretly describing — is everywhere, and the complex numbers are how it shows up algebraically.
From here you can branch in three directions. Modular arithmetic (next) zooms back out to integer number systems and a different kind of cyclical structure. Trigonometry gives you the geometric tools to push Euler's formula further. And complex analysis — eventually — takes calculus to $\mathbb{C}$, where holomorphic functions turn out to be among the most rigid, surprising objects in all of mathematics.