How can i solve $\displaystyle exp(\frac{1}{z})=z_0$, where $\displaystyle z_0$ is any complex number?
$\displaystyle z=\frac{1}{\log{z_0}}$.
The log is the complex log, which is multiple-valued: $\displaystyle \log(re^{i\theta})=\log{r}+i(\theta+2n\pi)$, where n is any integer.
And $\displaystyle \frac{1}{z}=\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}$.
Is that what you were asking?
- Hollywood
my problem, in its completeness, is the following: i'm looking for singularities of a function whose denominator is $\displaystyle e^\frac{1}{z}-3$. So i look for solutions of $\displaystyle e^\frac{1}{z}=3$. I put $\displaystyle e^\frac{1}{z}=e^{Log(3)}$ where $\displaystyle Log$ is the principal branch of complex logarithm and by periodicity of complex exp i get $\displaystyle \frac{1}{z}=log(3)+2n\pi i$, where log in RHS is the real log. So $\displaystyle z=(log(3)+2n\pi i)^{-1}$ and i was looking for an easy way to write the RHS in order to compute residues at those points
Since $\displaystyle \log{3}$ and $\displaystyle 2n\pi$ are both real, you can do what I suggested in my previous post:
$\displaystyle \frac{1}{\log{3}+2n\pi{i}}=\frac{\log{3}-2n\pi{i}}{(\log{3})^2+4n^2\pi^2}$
Other than that, I don't really see how to simplify it.
- Hollywood