The order of zero at $ z=0$ of $ f(z) = \cos(z^3) - 1 $.

Question

I'm trying to find the order of the zero at $z = 0$ of the function $ f(z) = \cos(z^3) -1 $.

I know I can express $f(z) $ as $$ f(z) = - \frac{z^6}{2!} + \frac{z^{12}}{4!} - \frac{z^{18}}{6!} + \ldots $$ I am unsure how to proceed.

Answer 1

Now, factor out $z^6$. So, you get $$f(z) = z^6g(z)$$ with $g(0)\neq 0$