# Characterizing isolated singularities of $\frac{z}{e^{z} - z + 1}$

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!

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.

• 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. Jan 2, 2021 at 13:19
• Why do we have that $|h(z)| \leq e^{2|z|}$? Jan 2, 2021 at 16:00
• 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. Jan 2, 2021 at 20:55
• I understood that part, I’m just not getting why that is valid for large enough |z|, sorry Jan 2, 2021 at 21:00
• 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 <= $e^{2*|z|}$ Jan 2, 2021 at 21:05

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.

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.

• 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 Jan 2, 2021 at 12:10