Roots of a polynomial and its derivative

Question

All roots of a complex polynomial have positive imaginary part. Prove that all roots of its derivative also have positive imaginary part.

It's not a homework. This issue has been proposed in the materials to prepare for exams.

Answer 1

If I'm not mistaken, the roots of $f'$ are in the convex hull of the roots of $f$.

Answer 2

A well known result :

We consider the logarithmic derivative of $f$:

$\displaystyle\frac{f'}{f}(x)=\displaystyle \sum \frac 1 {x-x_i}$ where $x_i$ are roots of $f$ (you can show this result by derivating $\ln(f(x))$ wrt $x$)

Then if $Z$ a root of $f'$, it's also one of the logarithmic derivative.

So $\Im\left(\displaystyle\frac{f'}{f}(Z)\right)=0=\Im\left(\displaystyle\sum \frac 1 {Z-x_i}\right)=\displaystyle\sum \Im\left(\frac 1 {Z-x_i}\right)=\displaystyle\sum \frac {\Im(x_i)-\Im(Z)}{|Z-x_i|^2}$. If we suppose that $\Im(Z)<0$, then the contradiction is blatant.