First slide

De-moivre's theorem

Question

The expression ${[\frac{1 + \sin \frac{π}{8} + icos \frac{π}{8}}{1 + \sin \frac{π}{8} - icos \frac{π}{8}}]}^{8} =$

Moderate

A

1
B

-1
C

i
D

-i

Solution

Let $\sin \frac{π}{8} + icos \frac{π}{8} = z$
$\Rightarrow {[\frac{1 + \sin \frac{π}{8} + icos \frac{π}{8}}{1 + \sin \frac{π}{8} - icos \frac{π}{8}}]}^{8}$
$\begin{array}{l} = {(\frac{1 + z}{1 + \frac{1}{z}})}^{8} = z^{8} \\ = {(\sin \frac{π}{8} + icos \frac{π}{8})}^{8} \\ = {(\cos (\frac{π}{2} - \frac{π}{8}) + isin (\frac{π}{2} - \frac{π}{8}))}^{8} \\ = {(\cos \frac{3 π}{8} + isin \frac{3 π}{8})}^{8} \\ = \cos 3 π = - 1 \end{array}$

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