Complex Number in a logarithm I am having a little trouble understanding complex numbers with logarithms. How would I do the log of $e$ ($\log_{i}{e}$)? What I did firstly was to do $\frac{\log{e}}{\log{i}}$. I don’t have any idea how to simplify this. 
 A: $i = e^{i \frac{\pi}{2}}$
So $\ln i = i\frac{\pi}{2}$.
Hence $$\frac{\ln e }{\ln i} = \frac{2}{i \pi}$$
Although do note we took the principle branch and complex logarithms are multivalued functions.
A: The complex logarithm is multi-valued, so
$$\log i = \log(e^{i\pi/2 + 2\pi i k}) = \frac{i\pi}{2} + 2\pi i k, k \in \mathbb{Z}$$
$$\log_i(e) = \frac{1}{i\pi/2 + 2\pi i k} = -\frac{2i}{\pi + 4\pi k}$$
A: All you need to know is $log(z)=log(x)+i\theta$,where $z$ is a complex number and $x=|z|$ and $\theta=amplitude$
A: $\log_i e$ is a number $z$ such that $i^z=e$; in turn, because $i=e^{i\pi/2}$, $i^z$ can be written as $(e^{\pi i/2})^z=\exp(i\pi z/2)$.  Because $|\exp(i\pi t)|=1$ for $t$ real, we see that $z$ cannot be real.  Let $z=u+iv$ with $u$, $v$ real.  Then $\exp(i\pi z/2)=\exp(i\pi u/2-\pi v/2)$.  To have that equal $e$, we must have $i\pi u/2=2n\pi i$ for some integral $n$, since $e$ is real, and $\pi v/2=-1$.
A: Notice:


*

*$$i=e^{\frac{\pi i}{2}}=e^{\left(2\pi k+\frac{\pi}{2}\right)i}\space\space\space\space\space\space\text{with}\space k\in\mathbb{Z}$$

*$$\ln(e)=\log_{e}(e)=\frac{\ln(e)}{\ln(e)}=1$$

*$$\ln(i)=\ln\left(e^{\left(2\pi k+\frac{\pi}{2}\right)i}\right)=\ln\left(ie^{2i\pi k}\right)=\frac{\pi i}{2}\space\space\space\space\space\space\text{with}\space k\in\mathbb{Z}$$


So:
$$\log_{i}(e)=\frac{\ln(e)}{\ln(i)}=\frac{1}{\ln\left(ie^{2i\pi k}\right)}=\frac{1}{\frac{\pi i}{2}}=\frac{2}{\pi i}=-\frac{2i}{\pi}\space\space\space\space\space\space\text{with}\space k\in\mathbb{Z}$$
