MDoubt No 000012 - GATE

GATE [EE, 2013, 1 mark]
Question : Square roots of $-i$, where $i=\sqrt{-1}$, are

(a) $i, -i$
(b) $\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right),$ $\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right)$
(c) $\cos\left(\dfrac{\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right),$ $\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{\pi}{4}\right)$
(d) $\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(-\dfrac{3\pi}{4}\right),$ $\cos\left(-\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right)$

Solution : As we know that
$-i=i^{3}$
$=e^{i3\pi/2}$
$\therefore \sqrt{-i}=e^{i3\pi/2\times 1/2}$
$=\cos\left(\dfrac{3\pi}{4}\right)+i\sin\left(\dfrac{3\pi}{4}\right)$
Now $,-i=\dfrac{1}{i}=e^{-i\pi/2}$
$\implies (-i)^{1/2}=e^{-i\pi/2\times 1/2}$
$=e^{-i\pi/4}$
$=\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right)$
CORRECT ANSWER : B