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Tasked with locating each of the isolated singularities of this function , telling whether it is removable, pole or an essential and if it's removable giving the value at the singularity and if it's a pole the order.

The function: $$\frac{z^2}{sin(z)}$$

So to check what kind of singularity it is, my understanding is we evaluate the limit.

Clearly there are singularities at $0$ and $k\pi$ for $k$ integer.

How do I evaluate the limit? Do I use L'Hospital's Rule?

Taking the derivative of the numerator and denominator gives

$$\frac{2z}{cos(z)}$$

which evaluating at 0 gives us 0. Does that mean it's a removable singularity at $z_0 = 0$ with a value of $0$? What do I do for all of the singularities at integer multiples of $\pi$?

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    $\begingroup$ correct for $0$, simple poles with residues that are easy to compute at $k\pi, k \ne 0$ $\endgroup$
    – Conrad
    Commented Apr 24, 2019 at 17:31

1 Answer 1

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You do not need L'Hospital's rule. All you have to know that the zeroes $z_k = k\pi$ of $\sin z$ have order $1$ because $\sin' z_k = \cos z_k \ne 0$. This means that $\sin z = (z - z_k) \cdot f_k(z)$ with a holomorphic function $f_k : \mathbb C \to \mathbb C$ such that $f_k(z_k) \ne 0$.

This show that $\lim_{z \to 0} \dfrac{z^2}{\sin z} = \lim_{z \to 0} \dfrac{z}{f_0(z)} = 0$. Thus you have a removable singularity at $0$.

For $k \ne 0$ you have $\dfrac{z^2}{\sin z} = \dfrac{z^2}{f_k(z)} \cdot \dfrac{1}{z - z_k}$ which shows that there is pole of order $1$ at $z_k$.

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