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 $\frac{z^2\sin z}{\cos z -1}$?

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
$(k+\frac12)\pi$.
