Let $P(z)$ be a monic polynomial with complex coefficients with all roots distinct and in $\{z \in C : \Im(z) \lt 0\}$.
$(a)$ Prove that the sum of all the residues of $\frac{P^{'}}{P}$ is the degree of the polynomial $P$.
$(b)$ Prove that $ P^{'}$ has no real root.
My idea was that option $(a)$ as I take $f(z)$=$\frac{p^{'}(z)}{p(z)}$
$\deg(p(z))\ge \deg(p^{'}(z))+2$
Residue theorem: If $f$ is analytic in a domain except for isolated singularities at $a_1,\dots a_k$ then for any closed contour $\gamma\in D$ on which none of the points $a_k$ lie, we have $$\frac{1}{2\pi i}\int_{\gamma}f(z)dz=\sum_{1}^{k}n(\gamma;a_k)Res[f(z);a_k].$$
Here I don't know how to proceed further.
For option $(b)$ if I take even polynomial degree that $p(x) =x^2+1$ then it will not have real roots
As I don't know the actual proof. Please assist and help me.
Thanks in advance for helping.
The first part is a residue theorem exercise. For the second part, write $p(z) = \prod (z-a_i)$, where the $a_i$ are the roots of $p$, possibly with repetition. Then $p'(z)/p(z) = \sum \frac{1}{z-a_i} $. Suppose $z$ is any complex number with $\Im(z) \ge 0$. Then $$ \Im(p'(z)/p(z)) = \Im \left(\sum \frac{1}{z-a_i}\right) = \Im\left(\sum \frac{\bar z - \bar a_i}{|z-a_i|^2}\right). $$ Now since $\Im(z) \ge 0$, it follows that $\Im(z - a_i) > 0\ $ for all $i$, hence $\Im(\bar z - \bar a_i) < 0$ for all $i$, hence $\Im(p'(z)/p(z)) \ne 0$, hence $p'(z) \ne 0$.
Note: I just took the proof of the Gauss-Lucas theorem from the wikipedia article, and took a short cut appropriate to the special case.