First slide

De-moivre's theorem

Question

Let z₁ and z₂ be nth roots of unity which subtend a right angle at the origin. Then n must be of the form

Moderate

A

4k + 1
B

4k + 2
C

4k + 3
D

4k

Solution

nth roots of unity are given by
$\cos (\frac{2 m π}{n}) + i \sin (\frac{2 m π}{n}) = e^{2 m π i / n} for m = 0, 1, 2, \dots, n - 1$
Let $z_{1} = e^{2 m_{1} π i / n}$ and $z_{2} = e^{2 m_{2} π i / n}$
where $0 \leq m_{1}, m_{2} < n, m_{1} \neq m_{2}$
As the join of $z_{1} a n d z_{2}$ subtend a right angle at the origin $z_{1} / z_{2}$ is purely imaginary we get
$\frac{e^{2 m_{1} π i / n}}{e^{2 m_{2} π i / n}} = i l$ for some real $l \Rightarrow e^{2 (m_{1} - m_{2}) π i / n} = i l$
$\Rightarrow \cos [\frac{2 (m_{1} - m_{2}) π}{n}] + i \sin [\frac{2 (m_{1} - m_{2}) π}{n}] = i l$
$\Rightarrow \cos [\frac{2 (m_{1} - m_{2}) π}{n}] = 0 \Rightarrow \frac{2 (m_{1} - m_{2}) π}{n} = \frac{π}{2}$
$\Rightarrow n = 4 (m_{1} - m_{2})$
Thus, $n$ must be of the form $4 k .$

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