Let \(p\) be a nonconstant complex polynomial whose zeros all lie in the half-plane \(\operatorname{Im}z>0\).
Prove that \[ \operatorname{Im}\frac{p'(x)}{p(x)}>0 \] for real \(x\).
Find a relation between \(\deg p\) and \[ \int_{-\infty}^{\infty} \operatorname{Im}\frac{p'(x)}{p(x)}\,dx. \]
Solution
Let \(d=\deg p\), and list the zeros of \(p\), with multiplicity, as \[ \alpha_j=a_j+ib_j, \qquad b_j>0, \qquad 1\le j\le d. \]
1. For every real \(x\),\[ \frac{p'(x)}{p(x)} = \sum_{j=1}^d\frac{1}{x-\alpha_j}. \]
Write \[ p(z)=c\prod_{j=1}^d(z-\alpha_j), \qquad c\ne0. \] Since no zero \(\alpha_j\) is real, \(p(x)\ne0\) for \(x\in\RR\). Taking the logarithmic derivative of the finite product gives the displayed identity.
2. For every real \(x\) and every \(j\),\[ \operatorname{Im}\frac{1}{x-\alpha_j} = \frac{b_j}{(x-a_j)^2+b_j^2}>0. \]
Since \[ x-\alpha_j=(x-a_j)-ib_j, \] one has \[ \frac{1}{x-\alpha_j} = \frac{(x-a_j)+ib_j}{(x-a_j)^2+b_j^2}. \] The asserted imaginary part follows, and it is positive because \(b_j>0\).
3. For every real \(x\),\[ \boxed{\operatorname{Im}\frac{p'(x)}{p(x)}>0}. \]
By steps 1 and 2, \[ \operatorname{Im}\frac{p'(x)}{p(x)} = \sum_{j=1}^d \frac{b_j}{(x-a_j)^2+b_j^2}, \] a finite sum of strictly positive terms.
4. For each \(j\),\[ \int_{-\infty}^{\infty} \frac{b_j}{(x-a_j)^2+b_j^2}\,dx = \pi. \]
With \[ u=\frac{x-a_j}{b_j}, \] the integral becomes \[ \int_{-\infty}^{\infty}\frac{du}{1+u^2} = \pi. \]
5. Therefore\[ \boxed{ \int_{-\infty}^{\infty} \operatorname{Im}\frac{p'(x)}{p(x)}\,dx = \pi\deg p}. \]
4. There are exactly \(d=\deg p\) roots counted with multiplicity.
6. Q.E.D.
Step 3 proves part 1, and step 5 gives the requested relation in part 2.