TITLE: Why is $\frac{2ln(i)}{i}=\pi$ true?
QUESTION [8 upvotes]: I was playing with Wolfram|Alpha (just for fun, cause math is fun ), and I found that $\ln(-1)=i\pi$ which is quite obvious when you already know Euler's Identity. So, I continued playing with numbers and when I typed 
$$\frac{2\ln(i)}{i}$$
it returned "$\pi$". And I have no idea why this is true. I looked it up on the internet, but I have not found anything interesting. Do you guys have a nice explanation for this?

REPLY [2 votes]: Think of the logarithm as giving you back the angle between the positive real axis and the number you have plugged in (times a factor $i$). You were not surprised to see
$$\log(-1)=\pi i.$$
However, this statement just means that the angle between the positive axis, and the negative axis (represented by $-1$) is $180^°$ or (in radians) $\pi$. And when you look at the complex plain, you would see that $i$ is "above" the zero, hence there is a $90^°$ angle between the positive axis and the complex axis (represented by $i$), which expressed by radians is $\pi/2$. Hence
$$\log(i)=\frac\pi 2 i.$$
There is some more math and some more subtleties hidden inside. But this should suffice as a first explanation.

Maybe also this might be interesting. You probably know that logarithms "convert" multipication into addition in the sense $\log(ab)=\log(a)+\log(b)$. And you you know that $i\cdot i=-1$. So it would be quite natural to assume that 
$$\log(i)+\log(i)=\log(-1).$$
And thats exactly what you found.