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If z is purely imaginary and $I m (z) < 0,$ then $a r g (i z) + a r g (z)$ is equal to

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$π$

$\frac{π}{2}$

$- \frac{π}{2}$

answer is C.

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Detailed Solution

Let $z = - i t$ where $t > 0,$ then

$i \bar{z} = i (i t) = - t$

$∴ \arg (i \bar{z}) + \arg (z) = π - π / 2 = π / 2$

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