Integrate
$$\int_{\vert z \vert = 2}\frac{z+4}{z(z-1)}dz$$
Attempt: I can either do partial fractions or realize that I have singularities at $z=1,0$ Thus the answer is
$$2\pi i \left(\left(\frac{z+4}{z}\right)\Bigg\vert_{z=1}+\frac{z+4}{z-1}\Bigg\vert_{z=0}\right)=2\pi i.$$
Is this correct? I did partial frations and got the same thing.
Yes. Your function has two simple poles at points $z=0$ at $z=1$, the residue of a function of the form $\frac{g(z)}{(z-z_0)}$ at the point $z_0$, where $g(z)$ is analytic in the neighborhood of $z_0$, is $g(z_0)$, the residue at a point $z_0$ is, by definition, the contour integral $$\frac{1}{2\pi i}\oint_{|z-z_0|=\varepsilon}f(z)dz$$ where $\varepsilon$ is sufficiently small for $f(z)$ to be analytic at $U_{\varepsilon}(z_0)\backslash \{z_0\}$, and your contour encompasses both poles, so you can deform it into two circles around the poles and the line connecting them, passed in both directions (which cancels out the integrals along it). Consider using \left and \right to format your brackets.