Hey Rui,
Can you please help me with question 3 and 4?
Sketch solution for Q3: That region isn't even gonna be a region. It's literally just gonna be a tiny segment of the ray. First, verify that the ray \( \arg z = \frac\pi3\) and the circle \( |z|=2\) intersect at \(z = 2 \text{cis} \frac\pi3\). (Recall from my lecture that we only need to consider the boundary, which is why we don't care too much about the interior of the circle; \(|z|<2\).)
Then, because you know the locus of \(z\), you also know the locus of \(z-i\). Reason being that \(-i\) is just a translation
downwards by one unit.
Since the endpoints of the locus of \(z\) were at \(0\) with an open circle and \(2\text{cis}\frac\pi3\) with a closed circle, the endpoints of the locus of \(z-i\) are at \(-i\) and \(2\text{cis} \frac\pi3-i\). So you just have to find the arguments of these two complex numbers to know the extremities. Of course, you will need to expand out that cis into Cartesian form beforehand.
I will look at Q4 later, but my guess would to be start with \(U_n = \int \frac{dx}{(x^2+1)^{n-1}} \) and then take the IBP suggestion, differentiating \( \frac{1}{(x^2+1)^{n-1}} \) and integrating \(1\).