Multiplying by \(i\) is a rotation
Multiplication by a complex number has a picture. Once you can see it, a lot of complex number questions get shorter.
The algebra
Take any complex number \(z = a + bi\) and multiply by \(i\), using \(i^2 = -1\):
\[ iz = i(a + bi) = ai + bi^2 = -b + ai. \]So the point \((a, b)\) moves to \((-b, a)\).
Why that is a quarter-turn
- Same distance from the origin. \(|iz| = \sqrt{(-b)^2 + a^2} = \sqrt{a^2 + b^2} = |z|\).
- At right angles. The vectors \(\begin{pmatrix} a \\ b \end{pmatrix}\) and \(\begin{pmatrix} -b \\ a \end{pmatrix}\) have dot product \(-ab + ab = 0\).
- Anticlockwise. For \(z = 1\), \(iz = i\): the positive real axis turns to the positive imaginary axis.
In polar form, \(i = \operatorname{cis}\frac{\pi}{2}\). Multiplying moduli and adding arguments gives \(|iz| = |z|\) and \(\arg(iz) = \arg(z) + \frac{\pi}{2}\).
Powers of \(i\)
| Multiply by | Equals | Effect on a point |
|---|---|---|
| \(i\) | \(i\) | Rotate \(90^\circ\) anticlockwise |
| \(i^2\) | \(-1\) | Rotate \(180^\circ\) |
| \(i^3\) | \(-i\) | Rotate \(90^\circ\) clockwise |
| \(i^4\) | \(1\) | Back where you started |
The general rule
Multiplying by \(w = r\operatorname{cis}\theta\) scales distances from the origin by \(r\) and rotates anticlockwise by \(\theta\). Multiplying by \(i\) is the case \(r = 1\), \(\theta = \frac{\pi}{2}\).
Questions
- Let \(z = 2 + 5i\). Find \(iz\) and describe how the point has moved.
- Let \(z = -3 + 4i\). Find \(i^2 z\) and \(i^3 z\).
- Show that \(|iz| = |z|\) for any \(z = a + bi\).
- The point \(P\) represents \(z = 1 + 2i\). The point \(Q\) is \(P\) rotated \(90^\circ\) clockwise about the origin. Find the complex number that \(Q\) represents.
- A square has vertices at \(O\), \(z = 3 + i\) and \(iz\). Find the complex number for the fourth vertex.
- Describe the effect of multiplying a complex number by \(1 + i\).
Worked solutions
1.
\[ iz = i(2 + 5i) = 2i + 5i^2 = -5 + 2i \]\(iz = -5 + 2i\)
The point \((2, 5)\) has rotated \(90^\circ\) anticlockwise about the origin to \((-5, 2)\).
2.
\(i^2 = -1\), so \(i^2 z = -(-3 + 4i) = 3 - 4i\).
\(i^3 = -i\), so \(i^3 z = -i(-3 + 4i) = 3i - 4i^2 = 4 + 3i\).
\(i^2 z = 3 - 4i, \quad i^3 z = 4 + 3i\)
3.
\(iz = -b + ai\), so
\[ |iz| = \sqrt{(-b)^2 + a^2} = \sqrt{a^2 + b^2} = |z|. \]\(|iz| = |z|\) as required.
4.
A clockwise quarter-turn is multiplication by \(-i\) (that is, \(i^3\)).
\[ -i(1 + 2i) = -i - 2i^2 = 2 - i \]\(Q\) represents \(2 - i\)
5.
\(iz = i(3 + i) = -1 + 3i\). The sides from \(O\) to \(z\) and from \(O\) to \(iz\) are equal and perpendicular, so the fourth vertex is their sum (the parallelogram rule).
\[ z + iz = (3 + i) + (-1 + 3i) \]\(2 + 4i\)
6.
\(|1 + i| = \sqrt{2}\) and \(\arg(1 + i) = \frac{\pi}{4}\), so \(1 + i = \sqrt{2}\operatorname{cis}\frac{\pi}{4}\).
Rotate \(45^\circ\) anticlockwise about the origin and scale distances by \(\sqrt{2}\).