Q1)

$$f(z) = {4z-5\over z(z-1)}.$$

To determine the residue of $|z|=2,$ we sum the residues of interior
poles ($z=0$ and $z=1.$)

Starting with $z=0,$
$$f(z) = -{4z-5\over z}\cdot{1\over 1-z} = (5/z-4)(1+z+z^2+\ldots),$$
giving a residue of $b_1 = 5.$

Then, $z=1,$
$$f(z) = {4(z-1)-1\over z-1}\cdot{1\over 1-(1-z)} = (4 -
1/(z-1))(1-(z-1)+(z-1)^2+\ldots),$$
giving a residue of $b_1 = -1.$

The sum of these residues is $4,$ so the value of the given integral is
$2\pi i\cdot4 = 8\pi i.$

Q2)

$$f(z) = {z^3 + 2z\over (z-i)^3}.$$

The residue at $z=i$ is $g''(i)/2!$ where $g(z) = z^3 + 2z,$ giving
$g''(i)/2! = (6i)/2! = 3i.$

Q3)

I'm going to sleep in, and I think I'll be able to actually relax on
Wednesday.

\bye