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Question

(B) To simplify $\frac{1 - i}{1 + i}$,we multiply the numerator and denominator by the conjugate of the denominator,which is $(1 - i)$:$\frac{1 - i}{1

Vedclass · Accepted Answer

$\cos \frac{\pi}{2} - i \sin \frac{\pi}{2}$

Vedclass · Answer

(B) To simplify $\frac{1 - i}{1 + i}$,we multiply the numerator and denominator by the conjugate of the denominator,which is $(1 - i)$:$\frac{1 - i}{1 + i} = \frac{(1 - i)(1 - i)}{(1 + i)(1 - i)}$$= \frac{1 - 2i + i^2}{1^2 - i^2}$$= \frac{1 - 2i - 1}{1 + 1}$$= \frac{-2i}{2} = -i$Now,expressing $-i$ in polar form $\cos 	heta + i \sin 	heta$:Since $\cos \frac{\pi}{2} = 0$ and $\sin \frac{\pi}{2} = 1$,we have:$-i = 0 - i(1) = \cos \frac{\pi}{2} - i \sin \frac{\pi}{2}$

$\frac{1 - i}{1 + i}$ is equal to

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