# Why is the argument of $i$ equal to $\pi/2$?

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 -

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

Is that the way I should be approaching it?

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You will find this wiki article very relevant. – Sasha Jan 24 '12 at 1:12
Are you sure the arctangent rule is the definition of "argument" in the presentation you're following? That would be unusual; commonly one defines the argument in some different way, and then the arctangent rule is just a computational technique that works if (and only if) the real part is positive. – Henning Makholm Jan 24 '12 at 1:38

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.