T-11 · Imaginary numbers

A quarter turn.

Descartes called them imaginary, and the name stuck. But multiplying by i is a quarter turn of the plane, so i² = −1 says only that two quarter turns make a half turn. Here are the equations that forced these numbers on mathematicians, among them a cubic whose real roots can be reached only by passing through them, and Euler's e^(iπ) = −1, built up from small turns.

i¹ = i

Multiplying by i

i·(a + bi) = ai + bi² = −b + ai: the coordinates swap and one changes sign. That is a quarter turn about 0, counter-clockwise, and the triangle under z turns with it.

z	3 + 2i
iz	−2 + 3i
iz ⟂ z	3·(−2) + 2·3 = 0 at right angles
\|iz\|² = \|z\|²	(−2)² + 3² = 13 the legs swap, the hypotenuse stays

Two quarter turns make a half

So i² = −1 is no mystery: two quarter turns are a half turn, which is what multiplying by −1 does. The powers of i repeat every four, and negative powers turn the other way, since 1/i = −i.

i·i	−1 i² = −1
i¹, from the cycle	i
by 1 multiplication	i the same
1/i	−i
cycle	1 → i → −1 → −i → 1

As a matrix

On coordinates (a, b), multiplying by i is the matrix J. Its square is −I, the half turn. A quarter turn leaves no real direction where it was, so J has no real eigenvector: its eigenvalues are the roots of λ² + 1 = 0, which are ±i.

0−110

J·(3, 2) = (−2, 3)

J² = −I	[−1, 0, 0, −1] a half turn
J⁴ = I	a full turn
det, trace	1, 0 so λ² − 0·λ + 1 = λ² + 1
eigenvalues	±i: no real ones at all

Halfway: √i

A quarter turn has a half: an eighth of a turn, (1 + i)/√2. Squared in exact arithmetic with √2 it gives i. And −1, like every number but 0, has two square roots: ±i.

((1 + i)/√2)²	= i, exactly
(−i)²	−1 so √−1 = ±i