Compute the first three terms in the power series of $f$ at 0.
$$ f(z) = \frac{1-cos(z)}{sin(z^2)} $$
How should I do this since the function is undefined at $f(0)$? I've seen some similar power series solved by polynomial division, would that be possible for this function?
Any help is useful, I'm very new to complex analysis.
$$f(z) = \frac{1-\cos(z)}{\sin(z^2)}$$ is undefined at $z=0$ but its limit does exist since, using equivalents, close to $z=0$, $\cos(z)\sim 1-\frac{z^2} 2$ and $\sin(z^2)\sim z^2$. So, use the standard expansions to get $$f(z)=\frac {1-\left(1-\frac{z^2}{2}+\frac{z^4}{24}-\frac{z^6}{720}+\frac{z^8}{40320}+O\left(z^{10}\right) \right)} {z^2-\frac{z^6}{6}+\frac{z^{10}}{120}+O\left(z^{14}\right) }$$
Simplify and use the long division.