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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?

Edit: Error in p'(z) leading to erroneous answer

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

1 Answer 1

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There is even a closed form solution to the equation $p'(z)=0$ in terms of Lambert's W function

They are given by:

$z_n=W_n(e)-1$

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!

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  • $\begingroup$ Really? W_1(e)-1 = -1.53+4.59i $\endgroup$
    – pdmclean
    May 27, 2015 at 0:17
  • $\begingroup$ Sorry, there was a typo. $\endgroup$
    – pdmclean
    May 27, 2015 at 11:16

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