Finding the first three terms of the Laurent series near $z=0$

Question

Let $$f(z) = \frac{1-e^z}{\sin^2z}$$ Find the first three terms of the Laurent series near $z=0$ and then evaluate $$I =\oint_{|z| = 1}f(z)dz$$

If we write $\sin^2 z = \frac{1-\cos2z}{2}$ and then try to use Taylor expansions of $e^z$ and $\cos z$, we should then divide two power series and this is really difficult by hand(this question was in an exam). Also if we do this, what's the radius of convergence? Is there any easier method for finding the first three terms of the Laurent series?

Answer 1

$$e^z=1+z+\frac{z^2}2+\frac{z^3}6+\ldots\,,\;\sin z=z-\frac{z^3}6+\frac{z^5}{120}-\ldots\implies$$

$$\implies\sin^2z=z^2-\frac13z^4+\frac{2z^4}{45}\ldots\implies$$

$$\frac1{\sin^2z}=\frac1{z^2\left(1-\frac13z^2+\ldots\right)}=\frac1{z^2}\left(1+\frac{z^2}3+\frac{z^4}9+\ldots\right)\implies$$

$$\frac{1-e^z}{\sin^2z}=\left(-z-\frac{z^2}2-\frac{z^3}6-\ldots\right)\frac1{z^2}\left(1+\frac{z^2}3+\frac{z^4}{49}\ldots\right)=$$

$$=\frac1{z^2}\left(-z-\frac{z^2}2-\frac{z^3}3+\ldots\right)=-\frac1z-\frac12-\frac z2+\ldots$$