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Determine the number of zeros and poles inside (counted with multiplicity) of the function \begin{equation} f(z)=\frac{z^{6}\sin(\pi z)}{(1-z^{2})(2-z^{2})(3-z^{7})} \end{equation} inside the disc $\{z:|z|<4.5\}$

We have that the zeros of the divisor are given as $z=0$ (multiplicity 7), $z=\pm 1$ (multiplicity 1), $z=\pm 2$ (multiplicity 1), $z=\pm 3$, and $z=\pm 4$. Nevertheless, when $z=\pm 1$, we have that both the dividend and divisor are equal to zero - and so those two points are neither zeros nor poles.

As such, I reckon the number of zeros and poles (counted with multiplicity) inside the disc $\{z:|z|<4.5\}$ are $7+3\cdot 2=13$ and $2+7=9$. Is this a fair argument?

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  • $\begingroup$ Seems ok to me. The points $1$ and $-1$ are called removable singularities. $\endgroup$ – Harto Saarinen Oct 5 '14 at 10:56
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In general we have the following:

If $\alpha$ is zero of order $m$ of $f$ and zero of order $n$ of $g$ and $h=f/g$ then $\alpha$ is (for function $h$)

$i)$ removable singularity if $m>n$. Here one can define $h(\alpha)=0$ to "remove" the singularity and then $\alpha$ becomes zero of order $(m-n)$.

$ii)$ pole of order $(n-m)$ if $n>m$.

$iii)$ removable singularity if $m=n$.

Hope this helps.

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