Finding The Zeros Of $\frac{z^2\sin z}{\cos z -1}$

Question

$$f(z)=\frac{z^2\sin z}{\cos z -1}$$

$$\frac{z^2\sin z}{\cos z -1}=0\iff z^2\sin z=0$$ so $z=\pi k$ and $z=0$ are zeros, to find the order we must derive $\frac{z^2\sin z}{\cos z -1}$?

Answer 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 $(k+\frac12)\pi$.

Answer 2

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