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Complex Numbers

The value of ...

Question

The value of $(s i n \frac{π}{8} + i cos \frac{π}{8})^{8} (s i n \frac{π}{8} - i c o s \frac{π}{8})^{8}$ is

A

-1

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B

0

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C

1

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D

2i

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Solution

The correct option is C 1
${(sin π / 8 + i cos π / 8)}^{8} {(sin \frac{π}{8} - 2 cos \frac{π}{8})}^{8}$
$\Rightarrow (sin π + i cos π) (sin π - i cos π)$
$\Rightarrow - i^{2} (cos π - i sin π) (cos π + 2 sin π)$
$\Rightarrow 1 \times e^{- π i} \cdot e^{π 2} [∵ i^{- 2} = - i]$
$\Rightarrow e^{0} = 1 A n s .$

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Similar questions

{⎛ ⎜ ⎜ ⎝ \frac{sin \frac{π}{8} + i cos \frac{π}{8}}{sin \frac{π}{8} - i cos \frac{π}{8}} ⎞ ⎟ ⎟ ⎠}^{8}

\frac{{(sin π / 8 + i cos π / 8)}^{8}}{{(sin π / 8 - i cos π / 8)}^{8}}

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{(sin \frac{π}{8} + i cos \frac{π}{8})}^{8}

Q. Show that one value of

{⎛ ⎜ ⎜ ⎝ \frac{1 + sin \frac{π}{8} + i cos \frac{π}{8}}{1 + sin \frac{π}{8} - i cos \frac{π}{8}} ⎞ ⎟ ⎟ ⎠}^{\frac{8}{3}}

- 1

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{⎡ ⎢ ⎢ ⎣ \frac{1 + sin \frac{π}{8} + i cos \frac{π}{8}}{1 + sin \frac{π}{8} - i cos \frac{π}{8}} ⎤ ⎥ ⎥ ⎦}^{8} =

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