Let \(p(z)\) be a nonconstant complex polynomial all of whose roots satisfy \[ \operatorname{Re}z>0. \] Prove that every root of \(p'\) also satisfies \[ \operatorname{Re}z>0. \]
Solution
Write \[ p(z) = c\prod_{j=1}^n(z-\lambda_j), \] where the roots are listed with multiplicity. By hypothesis, \[ \operatorname{Re}\lambda_j>0 \] for every \(j\).
1. Any root of \(p'\) that is also a root of \(p\) lies in the half-plane\(\operatorname{Re}z>0\).
Every root of \(p\) has positive real part by hypothesis. Hence any point that is simultaneously a root of \(p\) and \(p'\) already lies in the required half-plane.
2. If \(w\) is not a root of \(p\), then\[ \frac{p'(w)}{p(w)} = \sum_{j=1}^n\frac{1}{w-\lambda_j}. \]
Differentiating the factored expression for \(p\) gives \[ p'(z) = p(z) \sum_{j=1}^n\frac{1}{z-\lambda_j} \] whenever \(p(z)\ne0\). Dividing by \(p(w)\) yields the formula.
3. If \(\operatorname{Re}w\leq0\), then for every \(j\),\[ \operatorname{Re}\frac{1}{w-\lambda_j}<0. \]
Since \(\operatorname{Re}\lambda_j>0\), \[ \operatorname{Re}(w-\lambda_j) = \operatorname{Re}w-\operatorname{Re}\lambda_j <0. \] For any nonzero complex number \(u\), \[ \operatorname{Re}\frac1u = \frac{\operatorname{Re}u}{\abs{u}^2}. \] Applying this with \(u=w-\lambda_j\) gives the claim.
4. Every root \(w\) of \(p'\) satisfies\[ \boxed{\operatorname{Re}w>0}. \]
Let \(p'(w)=0\). If \(p(w)=0\), step 1 applies. Suppose instead that \(p(w)\ne0\). Then step 2 gives \[ 0 = \frac{p'(w)}{p(w)} = \sum_{j=1}^n\frac{1}{w-\lambda_j}. \] If \(\operatorname{Re}w\leq0\), every summand on the right has strictly negative real part by step 3, so their sum has strictly negative real part and cannot equal \(0\). Therefore \(\operatorname{Re}w>0\).
5. Q.E.D.
Step 4 proves the assertion for every zero of \(p'\).