Question:
The amplitude of $\frac{1}{i}$ is equal to
(a) 0
(b) $\frac{\pi}{2}$
(c) $-\frac{\pi}{2}$
(d) $\pi$
Solution:
(c) $-\frac{\pi}{2}$
Let $z=\frac{1}{i}$
$\Rightarrow z=\frac{1}{i} \times \frac{i}{i}$
$\Rightarrow z=\frac{i}{i^{2}}$
$\Rightarrow z=-i$
Since, $z(0,-1)$ lies on the negative imaginary axis.
Therefore, $\arg (z)=\frac{-\pi}{2}$
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