Complex Numbers: Speed-Solving Techniques (4)

Question

If $z = \frac{1 + i}{1 - i}$, find $z^{2024}$.

Solution — Step by Step

Why This Works

Multiplying by the conjugate over itself rationalises any complex denominator. Once $z$ is in the form $a + bi$, all powers reduce to a periodic computation if $|z| = 1$ — the four cases $i^0, i^1, i^2, i^3$ cover everything.

The polar form gives the same answer faster: $z = i = e^{i\pi/2}$, so $z^{2024} = e^{i \cdot 1012\pi} = e^{i \cdot 0} = 1$ (since $1012\pi$ is an even multiple of $\pi$).

Alternative Method

Polar form: $1 + i = \sqrt{2} e^{i\pi/4}$ and $1 - i = \sqrt{2} e^{-i\pi/4}$. So:

$z = \frac{\sqrt{2} e^{i\pi/4}}{\sqrt{2} e^{-i\pi/4}} = e^{i\pi/2} = i$

Then $z^{2024} = e^{i \cdot 506\pi} = e^{i \cdot 0} = 1$.

Common Mistake