Why is $\frac{2ln(i)}{i}=\pi$ true? 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?
 A: For the principal branch of the logarithm we have $\ln i=i\pi/2$, since $e^{i\pi/2}=i$. Hence $2\ln i=i\pi$, and upon dividing by $i$ the result follows.
A: This is equivalent to
$$\ln(i) = i\frac{\pi}{2}$$
Taking $e$ to the power of both sides gives us
$$e^{\ln(i)} = e^{i\frac \pi 2}$$
Both sides are equal to $i$ (the right side is because of Euler's formula, which states that for real $x$ we have $e^{ix} = \cos x+i\sin x$). So, everything works out. 
A: Actually, the form $e^{i\pi }+1=0$ is just a special case of Euler 's identity, which states that
$$e^{iz}=\cos{z}+i\sin{z}$$
So by Euler 's identity,
$$e^{i\frac{\pi}{2}}=i$$
So that
$$\frac{2\ln(i)}{i}=\frac{2i\frac{\pi}{2}}{i}=\pi$$
A: 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.
A: As $\operatorname{Ln}(-1)=i\pi$, $\operatorname{Ln}(i^2)=i\pi$, so $2\operatorname{Ln}(i)=i\pi$, and $\frac{2\operatorname{Ln}(i)}{i}=\pi$.
This uses the single-valued complex logarithm, $\operatorname{Ln}$. (see The complex logarithm, exponential and power functions (43), and equation (56), where we assume the principal value.)
A: That is an example of multivalued functions in Complex Analysis. Assuming you know that every complex number $z$ can be written like $z=|z|e^{Argz}$, then $Argz$ is called $z$'s argument, which can be $\alpha, \alpha+2\pi,...,\alpha+2k\pi$. Then we define the first one argunent, namely $\alpha$, to be the main value of $Arg z$, named as $argz$.
$Ln(z)$, just like $Argz$ is also a multivalued function in Complex Analysis. With simple calculation, we have $Ln(z)=ln|z|+iArgz$, where the function $ln(*)$ means the real logarithmic function. Now $Ln(i)=Ln(1*e^{\pi/2})=ln1+i(\pi/2+2k\pi)$, and we now define the main value of the multivalued function $Ln(*)$, named again as $ln(*)$, and $ln(z)=ln|z|+iarg(z)$. Generally speaking, we restrict $\alpha=argz\in (-\pi, \pi]$, and this ensures the uniqueness of the main value. So $ln(i)=0+i*(\pi/2)=i\pi/2$. So everything clear.
