First slide

De-moivre's theorem

Question

If $z + \frac{1}{z} = 2 \cos θ, z \in C$ then $z^{2 n} - 2 z^{n} \cos (n θ)$ is equal to

Moderate

A

1
B

0
C

-1
D

$- n$

Solution

$\begin{array}{l} z + \frac{1}{z} = 2 \cos θ \Rightarrow z^{2} - 2 z \cos θ + 1 = 0 \\ \Rightarrow z = \cos θ \pm i \sin θ \end{array}$
Now, $z^{2 n} - 2 z^{n} \cos (n θ)$
$\begin{array}{l} = z^{n} [z^{n} - 2 \cos (n θ)] \\ = z^{n} [\cos (n θ) \pm i \sin (n θ) - 2 \cos (n θ)] \\ = - z^{n} {\bar{z}}^{n} = - 1 \end{array}$

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