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I had a complex analysis exam today. One question was

$$ f(z)= \frac{z(\pi -z)^2}{\sin^2{z}}$$

What are the singularities? What are the removable ones? What are the poles of order 1? What are the poles of order 2?

If I remember correctly I just took limits for $$\lim_{z \to z_0}(z-z_0)^mf(z)$$ to check whether $k\pi$ with $k \in $ {$...,-2,-1,0,.1,2,...$} is a pole of order $m$. I used that removable singularity is a pole of order $m$.

I also tried the Taylor expansion of $\sin(z)^2$ but that did not work...

Can you explain how this works? And what the answers are?

Thank you for helping me out

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I had an exam today with the exact same question, only the numerator was $z(z-\pi)^2$ instead of $z(\pi-z)^2$ – Teun Verstraaten Jan 22 at 22:17
I guess that's the same exam :P – Joyeuse Saint Valentin Jan 23 at 11:32

1 Answer

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$\sin z$ has single zeros at $z=n\pi$ for every integer $n$. The easiest way to see this is to check that the derivative of $\sin$ is non-zero at these points. Hence $\sin^2 z$ has double zeros at $z=n\pi$.

On the other hand $z(\pi-z)^2$ clearly has a single zero at $z=0$ and a double zero at $z=\pi$.

Cancel all the possible common zeros. What do you end up with?

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Double zeros? Dont know about that. Im lookibg for poles – Joyeuse Saint Valentin Jan 22 at 23:07
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@JoyeuseSaintValentin Poles for your $f$ come from zeros to the denominator. – mrf Jan 22 at 23:13
But I guess you want to say that $z=0$, and $z=\pi$ are removable? – Joyeuse Saint Valentin Jan 23 at 13:50
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@JoyeuseSaintValentin Not quite, $z=0$ is not a removable singularity. You can only cancel one of the zeros in the denominator. (But $z=\pi$ is indeed a removable singularity.) – mrf Jan 23 at 13:54
Thank you. I just used the the limit approach. So I calculated limits for $z \to z_0$ for $z_0= 0, \pi$ and $k\pi$ for $|k|>1$, assuming (for each $z_0$) that it was a removable singularity, pole of order 1, and pole of order 2. After some algebra and manipulation I think I had the same answers. For example $z=0$ is pole of order $1$ because $\lim_{z \to 0} (z-0)^1 f(z)$ exists. For the order points I used the same method, concluding the same as you I believe. Thank you for your help :) – Joyeuse Saint Valentin Jan 23 at 14:35

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