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Q.

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

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a

$2^{1 / 4} e^{- i (π / 8)}$

b

$2^{1 / 4} e^{- i (5 π / 8)}$

c

$2^{1 / 4} e^{i (5 π / 8)}$

d

$2^{1 / 4} \cdot e^{i (9 π / 8)}$

answer is D.

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Detailed Solution

$\begin{array}{l} 1 + i = \sqrt{2} (\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) = \sqrt{2} c i s \frac{π}{4} \\ {(1 + i)}^{\frac{1}{2}} = {(\sqrt{2} c i s \frac{π}{4})}^{\frac{1}{2}} \\ {(\sqrt{2} cis \frac{π}{4})}^{1 / 2} = 2^{1 / 4} cis \frac{(2 kπ + \frac{π}{4})}{2}; k = 0, 1 \\ = 2^{1 / 4} e \frac{iπ}{8}; 2^{1 / 4} e \frac{i 9 π}{8} \end{array}$

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