If $i z^{4} + 1 = 0$ , then z can take the value

\frac{1 + i}{\sqrt{2}}

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cos \frac{π}{8} + i sin \frac{π}{8}

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\frac{1}{4 i}

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i

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Solution

The correct option is B $cos \frac{π}{8} + i sin \frac{π}{8}$
$i z^{4} + 1 = 0 \Rightarrow z^{4} = - \frac{1}{i} = \frac{i^{2}}{i} = i$
Let $z^{4} = cos \frac{π}{2} + i sin \frac{π}{2}$
$∴ z = {[cos \frac{π}{2} + i sin \frac{π}{2}]}^{1 / 4}$
U\sin g De-Moivre's theorem,
$(cos θ + i sin θ)^{n} = cos n θ + i sin n θ, n ε I$
Hence, $z = cos \frac{π}{8} + i sin \frac{π}{8}$

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

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

{(\frac{1 + cos \frac{π}{8} - i sin \frac{π}{8}}{1 + cos \frac{π}{8} - i sin \frac{π}{8}})}^{2} =

{(\frac{1 + c o s \frac{π}{8} - i s i n \frac{π}{8}}{1 + c o s \frac{π}{8} + i s i n \frac{π}{8}})}^{8}

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

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