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I intend to characterize the isolated singularities of $f(z) := \frac{z}{e^z - z + 1}$ which is defined on some open subset $\mathbb{C} \backslash f^{-1}({0}) \subset \mathbb{C}$. The possible singularities are only the zeros of $g(z) := e^z - z + 1$, so the approach should be to find those zeros, which is actually the difficult task.

At this point, what I know is the following:

  • Since no zeros of $g$ lie on $2\pi i \mathbb{Z}$, if we suppse that $z_{0} \in \mathbb{C}$ is a zero of $g$, then we will of course have $\lim\limits_{z \to z_{0}} |f(z)| = \infty$ and so $z_{0}$ will be a pole of $f$. Using then L'Hôpital's rule applied to $\lim\limits_{z \to z_{0}} (z-z_{0})^n f(z)$, we check that the pole will have to be of order $1$.

  • $g$ has zeros, which I found out by separating the function $g(z) = 0$ in two equations, concerning the real and imaginary parts of $g$ and then by ploting the resulting functions to see that they intersect.

My question is then: Can we see analytically that $g$ has zeros?

This is an exercise from a Reinhold Remmert's book, "Theory of Complex Functions" (line c) in exercise 1 on page 309) and I think I'm supposed to solve it analytically. Also, keep in mind that, at this point in the book, there are a lot of tools in complex analysis still not available to use, namely residue calculus and Laurent series.

Thank you in advance for all the help!

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3 Answers 3

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The function g has infinite number of zeros.
Assume otherwise, and list all its zeros ( $z_{1}$, $z_{2}$, ...$z_{n}$) with multiplicity.
Set $$P(z) = \prod_{k=1}^{k=n}(z-z_{k})$$ (if g has no zeros at all set $P(z) = 1$) and $$h(z) = \frac { g(z) } {P(z)}$$ Then h(z) is holomophic on the whole plain and has no zeros, hence it can be lifted.
In words, there is entire function, say f(z), such that h(z) = $e^{f(z)}$
Now note that $$|h(z)| <= e^{2|z|}$$ for |z| large enough and this implies that $$ |f(z)| <= 2|z| $$ for |z| large enough. In words f(z) must be a polynomial of degree at most 1, say $f(z) = az + b $.
Taking it all together, we showed that if g(z) has finite number of zeros then, for some $z_{1},z_{2}, ..., z_{n}, a, b$ we have $$g(z) = e^{az+b}\prod_{k=1}^{k=n}(z-z_{k})$$ or $$e^{z} - z + 1 = e^{az+b}\prod_{k=1}^{k=n}(z-z_{k})$$ or $$ P(z) = \prod_{k=1}^{k=n}(z-z_{k}) = \frac {e^{az+b}} {e^{z} - z + 1} $$ The last identity can not be true.
That is one can see that the coefficient a must be equal to 1 (otherwise a polynomial would grow as exponential at infinity) but if a = 1 then P(z) is bounded but not constant.

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  • $\begingroup$ The argument above is correct but maybe too complex. In fact, it shows that if g(z) is not of the form $e^{P(z)}$ where P(z) is a polynomial and yet $|g(z)| <= e^{|z|^{r}}$ for some r and |z| large enough then g(z) has infinite many zeros. But here we know formula for g(z). That is maybe one can just compute $\frac{1}{2i\pi} \int_{R}\frac{g^{'}}{g}$ for suitable region boundary R and show that this is not zero. R of the form |Re z| <= k, |Img z| <= k comes to mind. $\endgroup$
    – Salcio
    Jan 2, 2021 at 13:19
  • $\begingroup$ Why do we have that $|h(z)| \leq e^{2|z|}$? $\endgroup$
    – Daàvid
    Jan 2, 2021 at 16:00
  • $\begingroup$ h(z) = g(z)/P(z); g(z) = $e^{z} - z + 1$, and P(z) is a polynomial ... I am NOT claiming that 2 is the best constant or that we need 2 there, just that $|h(z)| <= e^{2|z|}$ for |z| large enough And this is pretty obvious calulus. $\endgroup$
    – Salcio
    Jan 2, 2021 at 20:55
  • $\begingroup$ I understood that part, I’m just not getting why that is valid for large enough |z|, sorry $\endgroup$
    – Daàvid
    Jan 2, 2021 at 21:00
  • $\begingroup$ How about this: |g(z)| = $|e^{z} - z + 1|$ <= $|e^{z}|$ + $|z +1|$ <= $e^{|z|}$ + $e^{|z|}$ <= $2*e^{|z|}$ if |z| is large enough and P(z) as a polynomial is |P(z)} > 1 for |z| large enough. Hence $|h(z)|$ <= 2*$e^{|z|}$ for |z| large enough and this is <= $e^{2*|z|}$ $\endgroup$
    – Salcio
    Jan 2, 2021 at 21:05
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If $g(z)$ had no zeroes, by Weierstrass factorization, we’d have to contend that there exists an entire function $h(z)$ such that $$e^{h(z)}=e^z-z+1\,,$$ or equivalently $$e^z(e^{h(z)-z}-1)=1-z\,.$$ It follows that $h(z)\ne z$ (that is, $h$ is not the identity function) and $h(1)-1=2\pi in_0$ for some integer $n_0$; furthermore, the entire non-constant function $h_1(z):=h(z)-z$ must never assume any value in the discrete set $\{2\pi in:n\in\mathbb{Z}\setminus\{n_0\}\}$ otherwise the left-hand side above will have at least two zeroes while the right-hand side has only one zero, which contradicts the Little Picard Theorem.

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Actually, the exercise is only about classifying those isolated singularities and, in the case of poles, to find their order. No reference is made to actually seeing them.

If $z_0$ is a zero of $g$, can it be a multiple zero? If it was, then it would be a zero of $g'$ too. But $g'(z_0)=e^{z_0}-1$. And $g(z_0)=0\iff e^{z_0}-z_0+1=0$. So, $e^{z_0}$ is equal to $1$ and also to $z_0-1$, which means that $z_0=2$. But $g(2)=e^2-1\ne0$. Therefore, $g$ has no multiple zeros and so each isolated singularity of $f$ is a simple pole.

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  • $\begingroup$ Yes, that is the kind of reasoning I did, assuming that g has zeros. Although it is a simplified way to do it (for which I thank you), it assumes that g has zeros, which is what I am trying to find out analytically $\endgroup$
    – Daàvid
    Jan 2, 2021 at 12:10

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