Complex Numbers

Some equations have no real solution. The simplest is $x^2 = -1$, because the square of any real number is never negative. To repair this gap we introduce a single new symbol, the imaginary unit $i$, defined by the one rule that does all the work:

A complex number is any expression of the form $z = a + ib$, where $a$ and $b$ are real numbers. Here $a$ is the real part, written $\operatorname{Re}(z)$, and $b$ is the imaginary part, written $\operatorname{Im}(z)$. Notice that $\operatorname{Im}(z) = b$ is the coefficient of $i$, not $ib$ itself. When $b = 0$ the number is purely real, and when $a = 0$ (with $b \ne 0$) it is purely imaginary.

Equality: two complex numbers are equal exactly when their real parts match and their imaginary parts match — a single equation in complex numbers is really two equations in disguise.

Powers of $i$ cycle every four steps. From $i^2 = -1$ we get $i^3 = i^2 \cdot i = -i$ and $i^4 = (i^2)^2 = 1$. After that the pattern repeats, so to find any power of $i$ you only need the remainder of the exponent on division by $4$.

Arithmetic. Addition and subtraction act part by part. Multiplication uses ordinary distribution, after which every $i^2$ is replaced by $-1$ and like terms are gathered:

Division is the one operation that needs a trick: multiply the top and bottom by the conjugate of the denominator, $c - id$, which turns the denominator into the real number $c^2 + d^2$.

A caution with square roots of negatives: the schoolbook rule $\sqrt{x}\,\sqrt{y} = \sqrt{xy}$ fails when both numbers are negative. Always convert first: $\sqrt{-4} = 2i$, so $\sqrt{-4}\,\sqrt{-9} = (2i)(3i) = 6i^2 = -6$, never $\sqrt{36} = 6$.

Deeper Insight — one symbol, and the whole number system closes up: The entire chapter rests on a single act of definition: declare $i^2 = -1$ and agree that $i$ obeys the ordinary laws of algebra. Everything else is forced. Once you accept $a + ib$, addition and multiplication are not new rules to memorise but the familiar distributive law applied while remembering to swap $i^2$ for $-1$. The deeper pay-off is closure: in the real numbers, polynomial equations could run off the edge of the system and have no answer, but the moment $i$ exists, every polynomial equation has a full set of roots — this is the Fundamental Theorem of Algebra, and it is why complex numbers feel less like an invention and more like a discovery. Treat $i$ as an ordinary quantity you can add and multiply, never as something mystical, and the algebra behaves exactly as you expect.

Every complex number $z = a + ib$ can be pictured as the point $(a, b)$ in a plane: the horizontal axis carries the real part and the vertical axis carries the imaginary part. This picture is the Argand plane (or complex plane), and it turns algebra into geometry.

Conjugate. The conjugate of $z = a + ib$ is obtained by flipping the sign of the imaginary part:

Geometrically, $\overline{z}$ is the mirror image of $z$ across the real axis. A key product collapses to a real number:

Modulus. The modulus $|z|$ is the distance of the point from the origin, found by Pythagoras:

Because $z\,\overline{z} = |z|^2$, the multiplicative inverse of a non-zero $z$ has a clean closed form — there is no need to guess:

Polar form. Instead of the coordinates $(a, b)$, a point can be located by its distance $r = |z|$ from the origin and the angle $\theta$ its line makes with the positive real axis. Then $a = r\cos\theta$ and $b = r\sin\theta$, which gives the polar (or trigonometric) form:

The angle $\theta$ is the argument of $z$. The value taken in $(-\pi, \pi]$ is the principal argument, written $\arg(z)$. The quadrant of the point decides the correct angle, so you must check signs of $a$ and $b$ rather than trust $\tan^{-1}(b/a)$ blindly.

Deeper Insight — the modulus and conjugate are two halves of one idea: The conjugate and the modulus are not separate tricks; they are bound together by the single identity $z\,\overline{z} = |z|^2$. That identity is exactly why division and inversion of complex numbers work at all: multiplying by the conjugate is the move that converts a complex denominator into a real one, and the real number it produces is precisely $|z|^2$. Geometry makes the rest intuitive — $|z|$ is a length, so $|z_1 z_2| = |z_1||z_2|$ (lengths multiply) and the modulus obeys the triangle inequality $|z_1 + z_2| \le |z_1| + |z_2|$ just as distances do. The polar form then reveals the deepest fact of all: multiplying complex numbers adds their arguments and multiplies their moduli, so multiplication by a complex number is a rotation combined with a scaling. Seeing complex numbers as points with a length and a direction, rather than as $a + ib$ alone, is what makes the next chapters on rotations and roots fall into place.

A quadratic equation has the form $ax^2 + bx + c = 0$ with $a \ne 0$. Its roots are given by the quadratic formula, and the quantity under the square root — the discriminant $D = b^2 - 4ac$ — decides the nature of those roots.

Over the real numbers, a negative discriminant means "no solution". Once complex numbers exist, that gap closes: a negative $D$ simply produces a square root of a negative number, which we write using $i$. The result is a pair of complex roots in the form $a + ib$.

When $D < 0$, write $\sqrt{D} = \sqrt{-(4ac - b^2)} = i\sqrt{4ac - b^2}$, so the roots become:

For an equation with real coefficients, complex roots always arrive in conjugate pairs: if $p + iq$ is a root, then $p - iq$ is the other. This is why a quadratic over the reals can never have exactly one non-real root.

Relations between roots and coefficients still hold, and they hold even when the roots are complex:

These let you check an answer quickly: for conjugate roots $p \pm iq$, the sum is the real number $2p$ and the product is $p^2 + q^2$, both real, exactly as $-b/a$ and $c/a$ must be.

Deeper Insight — why complex roots are a feature, not a failure: A negative discriminant used to mark the end of the road, but it is really a signpost that the answers live one dimension over, off the real line. The reason this works so cleanly is the conjugate pairing forced by real coefficients: the imaginary parts of $p + iq$ and $p - iq$ cancel in the sum and combine into the real quantity $p^2 + q^2$ in the product, so the coefficients $-b/a$ and $c/a$ stay perfectly real even though the roots are not. This is a small instance of the Fundamental Theorem of Algebra, which promises that a degree-$n$ polynomial has exactly $n$ roots in the complex numbers, counted with multiplicity — no equation is ever truly unsolvable. The practical lesson is to stop treating $D < 0$ as an error and start reading it as an instruction: factor out the $i$, simplify the surd, and report the conjugate pair.