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Question Let $f(z)$ be meromorphic function given by $f(z)=\frac{z}{(1-e^z)\sin{z}}$ then which of the following are correct?

(1) $z=0$ is pole of order $2$

(2) for every $k$ in $\mathbb{Z}$, $z=2kπi$ is simple pole.

(3) for every $k$ in $\mathbb{Z}-\{0\}$, $z=kπ$ is a simple pole.

(4) $z=π+2πi$ is a pole.

My attempt: I solved and saw that, (1) is false, (2) is true, and (4) is false.But I am not sure about (3). Here is my work for (3).

Consider, $$\lim_{z\to kπ} (z-kπ)\frac{z}{(1-e^z)(\sin{z})}$$

$$=\lim_{z\to kπ}\frac{z^2-zkπ}{\sin{z}-e^z\sin{z}}$$

Which is ($0/0$ form) so using LH rule we get,

$$=\lim_{z\to kπ}\frac{2z-kπ}{\cos{z}-(e^z\cos{z}+e^z\sin{z})}$$

$$=\frac{kπ}{\cos{kπ}-(e^{kπ}\cos{kπ}-0)}$$

$$=\frac{kπ}{\cos{kπ}-(\cos{kπ}+i\sin{kπ})\cos{kπ}}$$

$$=\frac{kπ}{\cos{kπ}-(\cos{kπ})^2}$$

But this not always nonzero and finite(as we required by definition of pole) ( further can we check this by Laurent series too?)

So how can be (3) is true? (In key it was given (3) is true!!)

Please help....

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You've made an error when you went from $e^{k\pi}$ to $\cos{k\pi}+i\sin{k\pi}$: that would be $e^{ik\pi}$. You should end up with $$\frac{k\pi}{(1-e^{k\pi})\cos{k\pi}}, $$ which is finite and nonzero for $k \in \mathbb{Z} \setminus \{ 0\}$.

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  • $\begingroup$ Hmmm... thank you so much sir, for pointing it out. $\endgroup$ – Akash Patalwanshi May 28 '18 at 13:36

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