# If $\,f(z)=\exp((z+1)/(z-1))\,$ then all singular points of $1/(f(z)-a)$ are simple poles

Here is a question from an old qualifying exam.

Let $f(z)=e^{\frac{z+1}{z-1}}$.

1. Show that $f$ maps the unit disc $D$ in to the unit disk.(I can show this using properties of LFT.

2. Let $0 <|a|<1$. Prove that all isolated singular points of $\frac{1}{f(z)-a}$ in the unit disc are simple poles. Enumerate the poles explicitely.

I know that this maps the unit disk into an outer disk, but how does that show that we have simple poles.

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Here's a sketch. Let me know if this is what you're looking for.

The poles can be found by solving

$$\frac{z+1}{z-1} = 2n\pi i + \log a$$

for $z$, where $\log a$ is any fixed value of the logarithm. You'll find a unique solution $z = z_n$ for each $n$, and you can translate them to the origin by letting $w = z-z_n$. You can then rewrite your function as

$$g(w) = f(w+z_n) = \left(a e^{\zeta_n w + O(w^2)}-a\right)^{-1} = (a\zeta_n w)^{-1}\Bigl(1+O(w)\Bigr)$$

as $w \to 0$ for some $\zeta_n = \zeta_n(a) \in \mathbb{C}$.

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Thanks for your answer Antonio. But still, how do I show that all isolated singular points must be simple poles. How do I know there are no others, essential, or poles of a different order? –  Jack Dawkins Jan 11 '13 at 23:36
@user54755 they are all simple because $$\lim_{w \to 0} w \,g(w) = (a\zeta_n)^{-1}.$$ In particular, the limit exists. –  Antonio Vargas Jan 11 '13 at 23:39
@user54755 Were you asking why there were no other ones? The idea is that $(z+1)/(z-1)$ is analytic for $z \neq 1$, so $1/(\exp[(z+1)/(z-1)]-a)$ is analytic when $z \neq 1$ and $\exp[(z+1)/(z-1)] \neq a$, which is exactly when $(z+1)/(z-1) \neq 2n\pi i+\log a$ for any $n$. –  Antonio Vargas Jan 12 '13 at 4:54
Antonio: Yes, I was wondering why there were no others. You cleared it up. Thanks!! –  Jack Dawkins Jan 14 '13 at 6:26