Use Laurent series to evaluate

Question

Use Laurent series to evaluate $\int_{|z|=2}dz/(4z-z^2)$. I've proved that $1/(4z-z^2)=1/4z+\sum_{n=0}^\infty z^n/4^{n+2}$. And I know the following formula:

Is the answer to this integration: $ln(2)/2+2\pi i/64$?

Answer 1

By inspection, the only pole of the function relevant in $\;|z|\le2\;$ is zero, so we need a Laurent series in an annulus of the form $\;0<|z|<r\le2\;$ :

$$\frac1{4z-z^2}=\frac1z\cdot\frac1{4-z}=\frac1{4z}\cdot\frac1{1-\frac z4}\stackrel{\text{Why can we?}}=\frac1{4z}\sum_{n=0}^\infty\frac{z^n}{4^n}=\sum_{n=0}^\infty\frac{z^{n-1}}{4^{n+1}}\implies$$

$$\implies a_{-1}\stackrel{\text{with}\;n=0\;}=\frac14$$

and then

$$\oint_{|z|=2}\frac{dz}{4z-z^2}=2\pi i\cdot\frac14=\frac{\pi i}2$$