First slide

De-moivre's theorem

Question

The value of ${(\frac{1 + \sin \frac{π}{8} + i \cos \frac{π}{8}}{1 + \sin \frac{π}{8} - i \cos \frac{π}{8}})}^{8}$

Easy

A

-1
B

0
C

1
D

$2 \cos \frac{3 π}{8}$

Solution

$L e t \sin \frac{π}{8} + i \cos \frac{π}{8} = z$
${(\frac{1 + \sin \frac{π}{8} + i \cos \frac{π}{8}}{1 + \sin \frac{π}{8} - i \cos \frac{π}{8}})}^{8} = {(\frac{1 + z}{1 + \frac{1}{z}})}^{8} = z^{8} = {[c os (\frac{π}{2} - \frac{π}{8}) + i \sin (\frac{π}{2} - \frac{π}{8})]}^{8}$
$= {[\cos \frac{3 π}{8} + i \sin \frac{3 π}{8}]}^{8} = \cos 3 π + i \sin 3 π = - 1 (by Demovier' s theorem)$

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