# Complex stationary point of $\frac{z}{1-e^{-z}}+z$?

I apply the method of steepest descents I need to know the stationary points $z_0$ of the function $$p(z)=\frac{z}{1-e^{-z}}+z,$$ such that, $0 <\mathrm {Im} (z)<2 \pi$. That is, I want $z_0$ such that: $$p'(z_0)=\frac{1+2e^{2z}-e^z(3+z)}{(e^z-1)^2}=0$$

Using Mathematica I find that

FindRoot[p'[z]==0,{z,-2+5 I}]

gives: $$z_0=-1.69329+\mathbb i \space 4.93943,$$ I seek an analytical approximation to $z_0$

Now, through trial and error (i.e. guessing), I've found that $z_1=-1-\log 2 + \frac{\mathbb i \pi^2} 2$ is close ($<0.005$) to $z_0$.

Alternatively, truncating the Laurent series of $p(z)$ about $z=2 \pi \mathbb i$ gives $$p_2(z) = \frac{2 \pi \mathbb i}{z-2 \pi \mathbb i}+1+2 \pi \mathbb i +(\frac 3 2+\frac{\mathbb i \pi} 6 )(z-2 \pi \mathbb i),$$ which has stationary points $2\pi \mathbb i \pm (1.63244+\mathbb i \space 1.1514)$, in particular, $$z_2=-1.62144+5.13179 \mathbb i,$$ which is about $0.2$ to $z_0$ and taking more terms in the series gives a stationary point closer to $z_0$.

Questions:

• Can $z_1$ be obtained from $p'(z)=0$?

• Can a better estimate to $z_0$ be made?

• There's a typo in your $p'(z_0)$ expression, it should be $2e^{2z}$, not $2e^z$ May 27, 2015 at 6:24

There is even a closed form solution to the equation $$p'(z)=0$$ in terms of Lambert's W function
$$z_n=W_n(e)-1$$
where $$n$$ labels the different branches of $$W$$. With this knowledge you try to give approximate solutions using the different asymtotic formulas in the link given above!