Is it true that $log(i) = \frac\pi2i$ ? If so, are both of these legitimate proofs? They seem too beautiful not to be... Sorry if this is a naive question. I have not yet taken any upper level math courses involving complex numbers. However, in preparation for those courses, together with utilizing the knowledge that mathematicians such as Euler would just play around with infinite series, led me to this result. This seems too beautiful not to be true. If it is true, could someone explain the general framework from which this is derived?

Statement: $$log(i) = \frac\pi2i$$
Proof One: By Euler's Identity $e^{ix} = cos{x} + isin{x}$

So that $$e^{\frac\pi2i} = cos\frac\pi2 + isin\frac\pi2 = i$$
But $$e^{log(i)} = i$$
Thus $$e^{log(i)} = e^{\frac\pi2i}$$
So that $$log(i) = \frac\pi2i$$
Proof Two: Using Power Series
$$log(i) = log(\frac{2i}2) = log\left(\frac{(1+i)(1+i)}{(1+i)(1-i)}\right) = log\left(\frac{1+i}{1-i}\right) = log(1+i) - log(1-i)$$
But $$ log(1+i) = i - \frac{i^2}2 + \frac{i^3}3 - \frac{i^4}4 + ...$$
And $$-log(1-i) = i + \frac{i^2}2 + \frac{i^3}3 + \frac{i^4}4 + ...$$
Thus $$log(1+i) - log(1-i) = 2\{i + \frac{i^3}3 + \frac{i^5}5 + \frac{i^7}7 + ...\} = 2i\{1 - \frac13 + \frac15 - \frac17 + ...\} = 2i\frac\pi4 = \frac\pi2i$$
Therefore $$log(i) = \frac\pi2i$$

Added June 4, 2016: I found a delightful identity for logarithms of complex quantites in general as opposed to the case $a+bi = i$. Since the proof is essentially a generalization of my proof that $log(i) = \frac{\pi}{2}i$ as well as being aesthetically pleasing I thought it was worth posting.

Statement: $$log\left(a+bi\right) = \frac{1}{2}log\left(a^{2}+b^{2}\right) + itan^{-1}\frac{b}{a}$$

Proof: $$log\left(a+bi\right) = log\left({\sqrt{a^{2}+b^{2}}}{\frac{a+bi} {\sqrt{a^{2}+b^{2}}}}\right) = \frac{1}{2}log\left(a^{2}+b^{2}\right) + log\left({\frac{a+bi}{\sqrt{a^{2}+b^{2}}}}\right)$$

Now, $$log\left({\frac{a+bi}{\sqrt{a^{2}+b^{2}}}}\right) = log\left({\frac{1+i\frac{b}{a}}{\sqrt{1+\frac{b^{2}}{a^2}}}}\right) = log\left({1+i\frac{b}{a}}\right) - \frac{1}{2}log\left(1+\frac{b^{2}}{a^2}\right)$$

But, $$log\left(1+i\frac{b}{a}\right) = \left(i\frac{b}{a}\right)-\frac{1}{2}{\left(i\frac{b}{a}\right)}^{2} + \frac{1}{3}{\left(i\frac{b}{a}\right)}^{3} - \frac{1}{4}{\left(i\frac{b}{a}\right)}^{4} +... = 
i\left( \left(\frac{b}{a}\right)-\frac{1}{3}{\left(\frac{b}{a}\right)}^{3} + \frac{1}{5}{\left(\frac{b}{a}\right)}^{5} - \frac{1}{7}{\left(\frac{b}{a}\right)}^{7} +... \right) +
\\ \frac{1}{2}\left(\frac{b}{a}\right)-\frac{1}{4}{\left(\frac{b}{a}\right)}^{4} + \frac{1}{6}{\left(\frac{b}{a}\right)}^{6} - \frac{1}{8}{\left(\frac{b}{a}\right)}^{8} +...$$
And,$$ \frac{1}{2}log\left(1+\frac{b^2}{a^2}\right) = \frac{1}{2}\left(\left({\frac{b^{2}}{a^{2}}}\right) - \frac{1}{2}\left({\frac{b^{2}}{a^{2}}}\right)^{2} + \frac{1}{3}\left({\frac{b^{2}}{a^{2}}}\right)^{3} - \frac{1}{4}\left({\frac{b^{2}}{a^{2}}}\right)^{4} +...\right) = \\ \frac{1}{2}\left(\frac{b}{a}\right)^{2}-\frac{1}{4}{\left(\frac{b}{a}\right)}^{4} + \frac{1}{6}{\left(\frac{b}{a}\right)}^{6} - \frac{1}{8}{\left(\frac{b}{a}\right)}^{8} +...$$
So that, $$log\left({1+i\frac{b}{a}}\right) - \frac{1}{2}log\left(1+\frac{b^{2}}{a^2}\right) = i\left( \left(\frac{b}{a}\right)-\frac{1}{3}{\left(\frac{b}{a}\right)}^{3} + \frac{1}{5}{\left(\frac{b}{a}\right)}^{5} - \frac{1}{7}{\left(\frac{b}{a}\right)}^{7} +... \right) = itan^{-1}{\frac{b}{a}}$$
Therefore, $$log(a+bi) = \frac{1}{2}log(a^{2}+b^{2}) + itan^{-1}\frac{b}{a}$$ 
Example: Let a=0 and b=1 and we have 
$$log(i) = itan^{-1}\infty = \frac{\pi}{2}i + 2\pi mi$$
It depends on how you define "logarithm" in the first place. But the simplest and most straightforward approach would be to define


  Definition. $\log x$ means the number $y$ such that $e^y=x$.


With this definition the first two lines of your first proof is a complete, direct proof that $\log i = \frac12\pi i$.

The only problem is that with this reasoning $\log i$ can also be shown to be $\frac52\pi i$ and $\frac 92\pi i$ and $\frac{13}2\pi i\ldots$ because the exponential function of those numbers are also $i$.

This is not just a minor deficiency of the definition, though -- the complex logarithm is inherently multivalued in the precise sense that it is not possible to define a single function that is continuous on all of $\mathbb C\setminus\{0\}$ such that $e^{f(z)}=z$ everywhere.

The usual fudging one does in order to overcome this is to speak of the "principal logarithm", notated $\operatorname{Log}$ with a capital L, and define


  $\operatorname{Log} z$ means the number $y$ such that $e^y=x$ and the imaginary part of $y$ is in $(-\pi,\pi]$.


This is, unfortunately, discontinuous on the negative real axis, but we can't really do better.



Your second proof is an ingenious manipulation, but of a kind that will easily lead to nonsense if you're not careful. Basically you're taking the power series for $\log(1+z)$ which will only work for $|z|\le 1, z\ne -1$ and then just assuming that this extends to a function on all of $\mathbb C\setminus 0$ that satisfies the usual logarithm rules everywhere. But this is not the case -- with the same reasoning we could say
$$ 1 = \Bigl(\frac{1+i}{\sqrt2}\Bigr)^8 $$
so
$$ 0 = \log 1 = 8\log\Bigl(\frac{1+i}{\sqrt2}\Bigr) $$
which requires the logarithm on the right to be $0$ -- but actually the series will yield $\frac14\pi i$.

This is the multi-valuedness of the logarithm that strikes again -- it is not possible for any nontrivial differentiable function on $\mathbb C\setminus\{0\}$ to satisfy $f(ab)=f(a)+f(b)$ everywhere.
From Euler's identity $$e^{i\pi/2} = cos(\pi/2) + i sin(\pi/2) = i
$$

Take the natural log of both sides. 

The left $\ln(e^{i\pi/2}) = i\pi/2$ and the right is simply $\ln i$. 

Since both are equal,  $i\pi/2 = \ln i.$ 

So, it is not arbitrary log but ln. Otherwise, your identity is correct.