Find real number a such that
$\oint_c \frac{dz}{z^2-z+a}=π $
where c is the closed contour |z-i|=1 taken in the counter clockwise direction.
This is a question that has been asked in the 2021 NBHM PhD exam
I tried to apply the Cauchy integral formula here, but couldn't find anything meaningful that will lead to a solution. Please help me with your suggestions.
A useful fact here is that: $$\oint_{\gamma} \frac{{\mathrm{d} z}}{(z-r_+)(z-r_-)} = \frac{2 \pi i}{r_+-r_-}$$ assuming your contour $\gamma$ contains $r_+$ and not $r_-$.
We'll suppose for now that $r_+$ and $r_-$ are the roots of your quadratic $z^2 - z + a$, namely $r_\pm = \frac{1 \pm \sqrt{1 - 4a}}{2}$. For this integral to be $\pi$, we need $r_+ - r_- = \sqrt{1-4a} = 2i$. Hence a candidate for $a$ is $\frac{5}{4}$.
Notice $r_+ = \frac{1}{2} + i$ lies within the interior of $\left | z-i \right | = 1$ whereas $r_ = \frac{1}{2} - i$ does not, hence our assumption holds and $a = \frac{5}{4}$ works.