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?
On
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.
On
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.
On
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.
On
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.)
On
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.
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 $$\frac{2\ln(i)}{i}=\frac{2i\frac{\pi}{2}}{i}=\pi$$