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Consider the function $f(z)=\frac{1}{e^z-e^{-z}}$ find the singularities and classify $0$ as a singularity.

Solution.

singularities at $$e^z=e{-z} \iff\sin y=0 \iff y=kπ \iff z_κ=ikπ $$ where $z=x+yi$ and k is integer.

Now at $k=0 ,z=0$ if its a pole of order $m$ there exists $A$ non zero such that $$\lim_{z\to\ 0} \frac{z^m}{e^z-e^{-z}}=A $$ using L'Hospital rule i get $A=0$ so since it is not a pole it is an essential singularity. Am i correct? Maybe a more elementary proof using Laurent series you can show me? Have i used something that it is not true maybe L'hospital rule is not true in Complex limits?

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  • $\begingroup$ Hint: If $f$ has an essential singularity, then so does $1/f$. $\endgroup$ – Arthur May 29 '17 at 9:35
  • $\begingroup$ ohh where can i see this result . $\endgroup$ – Manolis Lyviakis May 29 '17 at 9:39
  • $\begingroup$ An essential singularity is a singularity which is neither removable nor a pole. You can go through the different possible cases for what singularity / zero / nothing special that $f$ and $1/f$ have at a specific $z$, and which combinations can't be realised. You will find that essential singularity on one cannot correspond to anything other than an essential singularity on the other. It will be helpful to use that $1/(1/f))=f$ when doing this. $\endgroup$ – Arthur May 29 '17 at 9:51
  • $\begingroup$ @Arthur $\displaystyle \lim_{z \rightarrow z_o} f(z) = \infty$ does that mean it must be a pole? $\endgroup$ – Manolis Lyviakis May 29 '17 at 10:04
  • $\begingroup$ Yes. It's a pretty deep theorem, but it's known that around essential singularities, a function takes on any complex value (except possibly at most one) arbitrarily close to the singularity, so if $|f|$ is large everywhere close to $z_0$, then $z_0$ cannot be an essential singularity. $\endgroup$ – Arthur May 29 '17 at 10:45
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I don't know how you used L'hopital and obtained $A=0$. Because: \begin{align} \lim\limits_{z\to 0} \frac{z}{e^z-e^{-z}} = \lim\limits_{z\to 0} \frac{1}{e^z+e^{-z}} = \frac{1}{2} \end{align} So it is a pole of order 1.

You can also do it with Laurent series. For $\vert z \vert>0$ we have: \begin{align} \frac1 2 \cdot\frac{2}{e^z-e{-z}} = \frac1 2 \cdot\frac{1}{z+\frac{z^3}{3!}+\frac{z^5}{5!}+...} =\frac{1}{2z} \cdot \frac{1}{1+\frac{z^2}{3!}+\frac{z^4}{5!}+...} = \frac{1}{2z} \cdot \frac{1}{1-(-\frac{z^2}{3!}-\frac{z^4}{5!}-...)} \end{align} You can see the term on the right as a geometric series and get: \begin{align} \frac{1}{2z} \cdot \left (1+ \left(-\frac{z^2}{3!} - \frac{z^4}{5!}-..\right) + \left(-\frac{z^2}{3!} - \frac{z^4}{5!}-..\right) ^2 + ...\right)=\frac {1 }{2z }+ O(z) \end{align} So again you see that is a pole of order 1.

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We have $\lim_{z\to\ 0}zf(z)= \lim_{z\to\ 0} \frac{z}{e^z-e^{-z}}=1/2$, hence $0$ is a pole of order $1$.

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  • $\begingroup$ how did you calculate that limiti since it is 0/0 $\endgroup$ – Manolis Lyviakis May 29 '17 at 9:42
  • $\begingroup$ ohh d' hospital haha didnt thought of the case m=1 $\endgroup$ – Manolis Lyviakis May 29 '17 at 9:43

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