TITLE: Why is the argument of $i$ equal to $\pi/2$?
QUESTION [7 upvotes]: So it's obvious geometrically that the argument of $z=i$ is $\pi/2$.
However the method of getting the argument is $\arctan(y/x)$. And when in the case of $z=i$, $y/x = 1/0$ which is undefined...
So when you want to find the argument of a complex number is this the correct process -

$\operatorname{Argument}(z) = \arctan(y/x)$ if $x\neq0$.
If $x = 0$, then $\operatorname{Argument}(z) =  \pi/2\text{ or }-\pi/2$

Is that the way I should be approaching it?

REPLY [2 votes]: Although 1/0 is undefined, $\lim_{x \to 0^+}\operatorname{tan}(y/x)=\pi/2$ for any y. This corresponds with the geometric justification. One considers the limit as x goes to 0 from above because this is what happens when x rotates counterclockwise from 1 to i. To get the argument of -i, one still considers the counterclockwise rotation, which leads to the limit from below, which gives instead $-\pi/2$. The arguments of multiples of i and -i work out the same way.