Question:
The argument of $\frac{1-i}{1+i}$ is
(a) $-\frac{\pi}{2}$
(b) $\frac{\pi}{2}$
(c) $\frac{3 \pi}{2}$
(d) $\frac{5 \pi}{2}$
Solution:
(a) $-\frac{\pi}{2}$
Let $z=\frac{1-i}{1+i}$
$\Rightarrow z=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$
$\Rightarrow z=\frac{1+i^{2}-2 i}{1-i^{2}}$
$\Rightarrow z=\frac{1-1-2 i}{1+1}$
$\Rightarrow z=\frac{-2 i}{2}$
$\Rightarrow z=-i$
Since, $z$ lies on negative direction of imaginary axis.
Therefore, $\arg (z)=\frac{-\pi}{2}$
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