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$\begingroup$ Calculate $\arg(i)+\arg(-1)$:
(a) $\arg(-i) \quad$ (b) $\arg(1-i) \quad$ (c) $\arg(1+i) \quad$ (d) $\text{None}$
I found it $\arg(-i)$ but it's not correct, my teacher said.
$\endgroup$ 3 1 Answer
$\begingroup$ Well, $\arg(i)=\pi/2$ and $\arg(-1)=\pi,$ so $\arg(i)+\arg(-1)=?$ And which of the answers corresponds to this?
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