Show that the roots of P' lie in the same half plane as the roots of P The problem statement is:
Part(a)
Assume P(z) is a non-constant polynomial with all of its roots in some half plane H.  Show that the derivative P'(z) must also have all of its roots in H.
Part (b)
Show that the roots of P' always lie within the closed convex hull of the set of roots of P.
The hint given is: 
After obvious reductions of the problem, look at the behavior of $\frac{P'}{P}$ on the complementary half-plane.
My thoughts: 
So I think that, whatever half-plane H that we have to work with, we can always just look at $P(e^{i\theta}z)$, which has its zeros in the UHP, for a suitable rotation by angle theta.  
Now we can just show that $P'(e^{i\theta}z)e^{i\theta}$ also has its roots in the UHP -- and then undoing the rotations shows that both P and P' had its roots in the same half-plane H.
Using the hint regarding looking at the complementary half-plane, we then want to show that P' does not vanish in the LHP (we already know P doesn't vanish in the LHP, by assumption.)  
Again following the hint we look at P'/P in the LHP and show that it does not vanish.  I am stuck at this point.  How can I proceed?
And regarding part(b) what is a "convex hull"?  I am guessing that the answer to part (b) is immediate, after figuring out part(a).
Any ideas are welcome.
Thanks,

HINT: 
Let $$P(z) = c \prod_{1\le k \le n} (z - z_k)$$
then $$P'(z)/P(z) = \sum_{1\le k \le n}\frac{1}{ (z - z_k)}$$
Therefore, if $z$ is a root of $P'$ that is not a root of $P$ ( i. e a simple root) then we have 
$$\sum_{1\le k \le n}\frac{1}{ (z - z_k)}=0$$
Let's show that $\sum_{1\le k \le n}\frac{1}{ (z - z_k)}\ne 0$ for any point not in the relative interior of the convex hull of the $z_k$ and different from all $z_k$'s.  Indeed, consider such a point $z$. Then the nonzero complex numbers $(z-z_k)$ have all arguments inside a halfplane, and so will the numbers 
$\frac{1}{z-z_k}$, so their sum cannot be $0$. 
