Zeros of the derivative of a polynomial. If all the zeros of a polynomial $f: \mathbb{C} \rightarrow \mathbb{C}$ are real, does this tell us that the zeros of the derivative are also all real valued? i.e, if $f(z) = 0$ only has real roots, will this imply that $f'(z)=0$ only has real roots? Assume that $f$ is monic.
It seems like this should be a really basic question, but attempts at a counter example have proved futile so far. 
One thing I have been trying to use is the fact that the coefficients of $f'$ can be written in terms of the symmetric functions on the zeros of $f$ itself. 
 A: this should be true. w.l.o.g let the polynomial $f$ be a monic polynomial. First, if a polynomial has only real roots, its coefficients are also only real, hence you can view the polynomial as $\mathbb{R} \to \mathbb{R}$.
in the real case, this is true: if $f$ has double zeros , $f'$ has a zero there, and the other zeroes of $f'$ must lie somewhere in between, because there need to be a local maximum or minimum somewhere.
So we have exactly $n-1$ zeros of $f'$, where n is the number of zeros of $f$
A: Yes: first, note that the derivative of a polynomial $p$ of degree $n$ has degree $n-1$.Let the roots of $p$ be $\alpha_1, \dotsc, \alpha_n$ in increasing order, not necessarily distinct. Further, the factor theorem shows that
$$ p(x) = a \prod_k (x-\alpha_k), $$
so $p(x)/a$ is a real-valued polynomial. Hence we may take $p: \mathbb{R} \to \mathbb{R}$ with no loss of generality.
Suppose first that all of the roots of $p$ are distinct. Now apply Rolle's theorem to the intervals $[\alpha_i,\alpha_{i+1}]$. This gives you $n-1$ roots, which must be all of them.
The next problem is what happens if two of the roots, $\alpha_k$, $\alpha_{k+1}$ coincide. Then it is easy to show that $p'(\alpha_k)=0$, by, say, formally differentiating the product. You can then use induction to examine the case where $k$ roots coincide to finish the proof.
