Why is $\frac{2ln(i)}{i}=\pi$ true?

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.

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.

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