Finding The Zeros Of $\frac{z^2\sin z}{\cos z -1}$ $$f(z)=\frac{z^2\sin z}{\cos z -1}$$
$$\frac{z^2\sin z}{\cos z -1}=0\iff z^2\sin z=0$$ so $z=\pi k$ and $z=0$ are zeros, to find the order we must derive $\frac{z^2\sin z}{\cos z -1}$?
 A: Since
$\sin(2x)=2\sin(x)\cos(x)$
and
$\cos(2x)=\cos^2(x)-\sin^2(x)
=1-2\sin^2(x)$
so
$\cos(2x)-1
=-2\sin^2(x)
$,
$f(2z)
=\frac{4z^2\sin 2z}{\cos 2z -1}
=\frac{4z^22\sin(z)\cos(z)}{-2\sin^2(z)}
=\frac{-2z^2\cos(z)}{\sin(z)}
=-2\frac{z}{\sin(z)}z\cos(z)
$.
The zeros are
$z=0$
(taking the limit
since it is undefined there)
and the zeros of
$\cos(z)$
which are
$(k+\frac12)\pi$.
A: Note that the denominator $\cos z - 1$ has zeroes at $z = 2m\pi$, so not all zeroes of $\sin z$ are zeroes of $f(z)$. To show this, apply L'Hopital's rule
$$ \lim_{z\to 2m\pi \ne 0} \frac{z^2\sin z}{\cos z - 1} = \lim_{z\to 2m\pi \ne 0}\frac{2z\sin z + z^2\cos z}{-\sin z} = \text{DNE} $$
Hence, the zeroes of $\sin z$ where $z=n\pi = 2m\pi$ are actually poles. This leaves the odd multiples, $z = (2m+1)\pi$, which are indeed zeroes of $f(z)$. 
$z = 0$ is also a zero, which you can show by applying L'Hopital once more
To simplify computations, we rewrite
$$ f(z) = \frac{z^2\sin z}{\cos z - 1} = -\frac{2z^2 \sin(\frac{z}{2})\cos(\frac{z}{2})}{2\sin^2(\frac{z}{2})} = -z^2\cot\left(\frac{z}{2}\right) $$
which confirms $z = (2m+1)\pi$ as the zeroes of $\cot(\frac{z}{2})$. Futhermore
$$ f'(z) = -2z\cot\left(\frac{z}{2}\right) + \frac{z^2}{2}\csc^2\left(\frac{z}{2}\right) $$
You can check that
$$ \lim_{z\to 0} f'(z) = -4\lim_{z\to0} \frac{\frac{z}{2}}{\sin\left(\frac{z}{2}\right)}\cos\left(\frac{z}{2}\right) + 2 \lim_{z\to 0} \frac{\left(\frac{z}{2}\right)^2}{\sin^2\left(\frac{z}{2}\right)} = -2 $$
and
$$ f'\big((2m+1)\pi\big) = \frac{(2m+1)^2\pi^2}{2} $$
Proving that these are all first-order zeroes
