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Q. The argument of the complex number $\left(\frac{i}{2}-\frac{2}{i}\right)$ is equal to

KEAMKEAM 2011

Solution:

Let $ z=\frac{i}{2}-\frac{2}{i} =\frac{i}{2}-\frac{2 i}{i^{2}}$
$z =\frac{i}{2}+2 i=\frac{5}{2} i=0+\frac{5}{2} i $
$\arg (z)=\tan ^{-1}\left(\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right)$
$=\tan ^{-1}\left(\frac{5 / 2}{0}\right)=\tan ^{-1}(\infty) $
$=\tan ^{-1}\left(\tan \frac{\pi}{2}\right)=\frac{\pi}{2} $