One value of ${(1 + i)}^{1 / 2}$ is $2^{1 / 4} e^{iπ / 8} .$ Then the other value is

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$2^{1 / 2} e^{i \frac{9 π}{8}}$

$2^{1 / 2} e^{i \frac{5 π}{8}}$

$2^{1 / 2} e^{i \frac{π}{8}}$

$2^{1 / 4} e^{i \frac{9 π}{8}}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} {(1 + i)}^{1 / 2} & = & {\sqrt{2}}^{1 / 2} {(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}^{1 / 2} \end{array}$

$\begin{array}{rcl} = & 2^{1 / 4} {(\cos \frac{π}{4} + i \sin \frac{π}{4})}^{1 / 2} \end{array}$

$\begin{array}{rcl} = & 2^{1 / 4} (\cos \frac{2 kπ + \frac{π}{4}}{2} + i \sin \frac{2 kπ + \frac{π}{4}}{2}) : k = 0, 1 \end{array}$

$\begin{array}{rcl} {(1 + i)}^{1 / 2} & = & 2^{1 / 4} cis \frac{π}{8}, 2^{1 / 4} cis \frac{9 π}{8} \end{array}$

$\begin{array}{rcl} = & 2^{1 / 4} e^{iπ / 8}, 2^{1 / 4} e^{i 9 π / 8} \end{array}$

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