What Is a Complex Number?

Solve x^2 + 1 = 0 and you hit a wall. Rearranging gives x^2 = -1, and no real number squares to a negative — squaring always lands on zero or above. The quadratic formula runs into the same wall whenever the discriminant goes negative: it asks for the square root of a negative number, which the real line simply does not contain.

So we do the boldest thing in algebra — we invent the missing number. Define a brand-new symbol i, the imaginary unit, by the single defining property

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

That one rule is the whole foundation. Everything else follows from ordinary algebra with i^2 rewritten as -1 wherever it appears.

Solving the impossible equation

Watch x^2 + 1 = 0 dissolve, one line at a time.

Step 1 — isolate the square. Subtract 1 from both sides:

x^2 = -1.

Step 2 — take the square root. A square root comes with both a positive and a negative branch, so we keep the \pm:

x = \pm\sqrt{-1}.

Step 3 — name the new number. By definition \sqrt{-1} = i, so the two solutions are simply

x = i \qquad \text{and} \qquad x = -i.

Step 4 — check. Substituting back, i^2 + 1 = -1 + 1 = 0. The equation that had no real answer has exactly two answers — once you allow i.

A complex number is a real part plus an imaginary part

Mix a real number with a multiple of i and you get a complex number:

z = a + bi.

Here a is the real part, written \operatorname{Re}(z), and b is the imaginary part, written \operatorname{Im}(z) (note: the imaginary part is the real number b, not bi). Ordinary real numbers are just the special case b = 0, so the reals sit comfortably inside the complex numbers.

Arithmetic: it's just algebra with i² = −1

Adding — combine like with like

Step 1 — group real with real, imaginary with imaginary. Addition is componentwise; the i terms gather together just like a common factor:

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

For example (3 + 2i) + (1 + 4i) = 4 + 6i — add the reals, add the imaginaries, done.

Multiplying — expand, then use i² = −1

Step 1 — expand the brackets exactly as you would for two real binomials (every term times every term):

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

Step 2 — replace i^2 with -1. This is the only new move; the last term flips sign:

= ac + adi + bci - bd.

Step 3 — collect real and imaginary parts. Group the two real terms and the two imaginary terms:

(a + bi)(c + di) = (ac - bd) + (ad + bc)i.

For example (3 + 2i)(1 + 4i) = (3 - 8) + (12 + 2)i = -5 + 14i.

The conjugate cleans things up

Flip the sign of the imaginary part and you get the complex conjugate, written \bar{z}:

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

Step 1 — multiply a number by its conjugate. Using the multiplication rule with c = a and d = -b:

z \bar{z} = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2.

Step 2 — apply i^2 = -1. The cross terms +abi and -abi cancelled, and the last term becomes +b^2:

z\bar{z} = a^2 + b^2.

The imaginary part vanished — z\bar{z} is always a non-negative real number. That single fact is what makes division by a complex number work (multiply top and bottom by the conjugate to clear the i from the denominator), and it is the seed of the modulus we meet next.

Adjoin a symbol i with i^2 = -1 to the real numbers. Then:

The quadratic formula solves ax^2 + bx + c = 0 with

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

When the discriminant b^2 - 4ac is negative there are no real roots — but there are two complex ones. Take x^2 - 2x + 5 = 0, which you could also reach by completing the square to (x - 1)^2 = -4. Here b^2 - 4ac = 4 - 20 = -16, so

x = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i.

The two roots are 1 + 2i and 1 - 2i — a conjugate pair, as complex roots of a real polynomial always are. With i in hand, every quadratic has exactly two roots.

See a product computed

Slide the four coefficients of z_1 = a + bi and z_2 = c + di. The diagram works the multiplication rule live: the real part ac - bd and the imaginary part ad + bc update as you move the knobs, so you can watch i^2 = -1 doing its job in the -bd term.