complex equation with exponential

**tenderline** · November 27th 2012, 01:48 PM

How can i solve $exp(\frac{1}{z})=z_0$ , where $z_0$ is any complex number?

**hollywood** · November 27th 2012, 05:54 PM

$z=\frac{1}{\log{z_0}}$ .

The log is the complex log, which is multiple-valued: $\log(re^{i\theta})=\log{r}+i(\theta+2n\pi)$ , where n is any integer.

And $\frac{1}{z}=\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}$ .

Is that what you were asking?

- Hollywood

**tenderline** · November 27th 2012, 11:11 PM

Originally Posted by hollywood

$z=\frac{1}{\log{z_0}}$ .

The log is the complex log, which is multiple-valued: $\log(re^{i\theta})=\log{r}+i(\theta+2n\pi)$ , where n is any integer.

And $\frac{1}{z}=\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}$ .

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 $e^\frac{1}{z}-3$ . So i look for solutions of $e^\frac{1}{z}=3$ . I put $e^\frac{1}{z}=e^{Log(3)}$ where $Log$ is the principal branch of complex logarithm and by periodicity of complex exp i get $\frac{1}{z}=log(3)+2n\pi i$ , where log in RHS is the real log. So $z=(log(3)+2n\pi i)^{-1}$ and i was looking for an easy way to write the RHS in order to compute residues at those points

**hollywood** · November 28th 2012, 06:48 AM

Since $\log{3}$ and $2n\pi$ are both real, you can do what I suggested in my previous post:

$\frac{1}{\log{3}+2n\pi{i}}=\frac{\log{3}-2n\pi{i}}{(\log{3})^2+4n^2\pi^2}$

Other than that, I don't really see how to simplify it.

- Hollywood

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