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 
$$log(i) = itan^{-1}\infty = \frac{\pi}{2}i + 2\pi mi$$

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.
