Logarithm of $ei$ 
Find the logarithm of $ei$.

Firstly we know that the log of a nonzero complex number $re^{i\theta} = r\cos(\theta) + ir\cos(\theta)$ is $\log(re^{i\theta}) = \log r + i\theta$. However, I do not know how to necessarily isolate the $i$ in this given context. Can anyone help me?
 A: Since $e^z$ has period $2\pi i$ in $\mathbb{C}$, we have that for $k\in\mathbb{Z}$, $$\text{Log}(e\cdot i)=\text{Log}(e)+\text{Log}(e^{i(2k\pi+\pi/2)})=1+i\pi\left(2k+\frac12\right)$$ 
A: $\ln ei = \ln e + \ln i$
$\ln e = 1$
And $\ln i = \ln re^{i\theta}$ where $re^{i\theta} = i$.
$re^{i\theta} = r(\cos \theta) + i \sin \theta) = 1(0 + i*1)$ so
$r = 1; \cos \theta = 0$ and $\sin \theta = 1$.  So $\theta = \frac \pi 2 + 2k\pi$.
So $e*i = e*e^{i(\frac \pi 2 + 2k\pi)}$
And $\ln e*i = 1 + \frac {\pi i}2 + 2ki\pi$
A: Hint: There is no necessity to isolate $e$ and $i$. We recall the general form \begin{align*}
z=a+ib=re^{i\theta}
\end{align*}

Here we have
  \begin{align*}
z=e\cdot i=0+ie
\end{align*}
  with $a=0$ and $b=e$.

This implies that $e\cdot i$ is an imaginary number with real part $0$ and so $\theta=\frac{\pi}{2}$ .

We recall $\log(z)=\log(r)+i\theta$ and conclude
  \begin{align*}
\color{blue}{\log (ei)}=\log(e)+i\frac{\pi}{2}\color{blue}{=1+i\frac{\pi}{2}}.
\end{align*}

A: $$e.i=e.e^{i \pi/2}$$
Then
$$
\log\left(e.i\right)=1+i\frac{\pi}{2}
$$
