Zeros of $z^2 \text{cos}z^2$

Question

Is there an easy way to find the zeros of the function $z^2 \text{cos}z^2$, $z\in \mathbb{C}$ and the respective orders (multiplicities)? All I can think of is to find $f^{(1)},f^{(2)},...$ but then again that did not take me anywhere. Any help will be much appreciated. Thanks

Answer 1

Clearly this function has a zero of multiplicity $2$ at $z=0$.

Consider now $g(z)=\cos(z^2)$.

Since $g'(z)=-2z\sin(z^2)$, there is no point $z\in\Bbb C$ where $g(z)=g'(z)=0$. Then, other zeros have multiplicity $1$