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Q.

The real part of a z that satisfies ${iz}^{4} + 1 = 0$ is

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a

sin $\frac{π}{4}$

b

0

c

-1

d

cos $\frac{π}{8}$

answer is B.

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Detailed Solution

$\begin{array}{rcl} {iz}^{4} & = & - 1 \\ {iz}^{4} & = & i^{2} \Rightarrow z^{4} = i \\ z^{4} & = & \cos \frac{π}{2} + i \sin \frac{π}{2} \\ z & = & {(\cos \frac{π}{2} + i \sin \frac{π}{2})}^{1 / 4} \\ z & = & \cos \frac{π}{8} + i \sin \frac{π}{8} \Rightarrow Re (z) = \cos \frac{π}{8} \end{array}$

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