Evaluate the residue of $ω(z) =\frac{\sqrt{(z−1)(z−5)}}{z} dz $ at point z = 0 for the branch $\sqrt{(z − 1)(z − 5)}$ defined by having positive limit values on the northern shore of the cut [1, 5].
The last part means : 'positive imaginary values' here. That mean that when z go to some point from (1,5) the product (z-1)(z-5)-> something negative, and square root of (z-1)(z-5) -> something purely imaginary. For example, for z->3 we have $\sqrt{(z-1)(z-5)}$-> plus or minus 2i. And by ''positive'' here we mean 2i. The northern shore in the initial formulation mean that limit is not just z->3, but $z=3+i.\epsilon$, $\epsilon->0$, $\epsilon>0$. Or the same that argument of (z-1)(z-5) on the ''upper'' or ``nothern'' side of the segment (1,5) is $\pi$ and not $-\pi$.
Does anybody can help me, I'm totally lost on how to prove it...
Your immediate task is to find which value to assume for $\sqrt{(z-1)(z-5)}$ at $z=0$ such that it comes from the right branch.
Pick a small positive real $\delta$ and imagine moving $z$ from $1+\delta$ in a small counterclockwise semicircle around $1$ to $1-\delta$ and then along the real axis to $0$. Because we're not crossing the branch cut, the select square root should be continuous along this journey.
When $z=1+\delta$ we have $(z-1)(z-5) \approx -4\delta$ since the difference between $z-5$ and $1-5$ is negligible in this context. Since we start out from the "northern shore", the square root we start our journey with is $2i\sqrt\delta$.
As we move around the semicircle, the $z-5\approx -4$ factor hardly changes, but $z-1$ factor turns counterclockwise around the origin at a constant distance of $\delta$. So $(z-1)(z-2)$ also moves counterclockwise from $-4\delta$ to $4\delta$. In order to stay continuous the square root has to move counterclockwise from $2i\sqrt\delta$ at half the angular speed -- so it ends up at $-2\sqrt\delta$.
Now moving along the real axis to $0$, the value of $(z-1)(z-5)$ stays positive all the way, so the relevant square root will not change sign during this process. It stays negative and ends up at $-\sqrt5$ when we reach $z=0$.