Leave lesson

Pure · Complex numbers

Chapter 1 · 3

The idea

Complex roots of polynomial equations

Solving quadratics with a negative discriminant to get a conjugate pair of complex roots; the theorem that complex roots of a real-coefficient polynomial come in conjugate pairs; using one known complex root to find the others and factorise cubics and quartics; and finding the square roots of a + bi by equating real and imaginary parts.

A full journey — read it, play with it, work it, then earn real exam marks. Everything stays on the timeline below.

In this lesson — start anywhere

Pure · Complex numbers

Complex roots of polynomial equations

Solving quadratics with a negative discriminant to get a conjugate pair of complex roots; the theorem that complex roots of a real-coefficient polynomial come in conjugate pairs; using one known complex root to find the others and factorise cubics and quartics; and finding the square roots of a + bi by equating real and imaginary parts.

Why it works

Every quadratic now solves

Once we have ii with i2=−1i^2 = -1, every quadratic can be solved. Take z2−4z+13=0z^2 - 4z + 13 = 0. The quadratic formula still applies: z=4±(−4)2−4(1)(13)2=4±−362.z = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(13)}}{2} = \frac{4 \pm \sqrt{-36}}{2}. The discriminant is negative, which is exactly the case that had no real solution. Now we handle it: −36=36 −1=6i\sqrt{-36} = \sqrt{36}\,\sqrt{-1} = 6i, so z=4±6i2=2±3i.z = \frac{4 \pm 6i}{2} = 2 \pm 3i. The two roots are 2+3i2 + 3i and 2−3i2 - 3i — the same real part, opposite imaginary parts. They are conjugates of each other. That is not a coincidence.

Keep reading — free

The rest of the explanation, plus 3 worked examples you step through move by move.

Start free

Takes a minute — no card.