The roots of the derivative $P'(z)$ of the polynomial $P(z)\in\mathbb C[x]$ lie in the convex hull of the set of roots of $P(z)$. Assume $S=\{z_1,z_2,...,z_k\}, z_i\in \mathbb C$$, C(S)$  and define $$C(S):=\{z=a_1z_1+a_2z_2+...+a_kz_k | a_i\ge0 ,a_1+a_2+...+a_k=1\}$$ where  $$A:=\{z\in \mathbb C:f(z)=0 \},~~~\text{and}~~~A':=\{z\in \mathbb C:f'(z)=0\}$$ such that    $f$  is polynomial ($f\in P[\mathbb C]$) and $\deg(f)\ge2$.
How to prove that $$ C(A')\subset C(A)$$ 
Thanks in advance! 
 A: This is the Gauss–Lucas theorem and you can find a proof here. It says that the roots 
of the derivative $P'(z)$ of the polynomial $P(z)\in\mathbb C[x]$ lie in the convex hull of the set of roots of $P(z)$.
In case $\deg P=3$ there is a stronger result ( someone might say the most marvelous theorem) called Marden's theorem but actually is Jörg Siebeck's( see this). The theorem is the following: 

Let $z_1,z_2,z_3$ be the roots of the polynomial $P(z)$ (assume that are nonlinear) and $T$ the triangle with vertices $z_1,z_2,z_3$. The Gauss–Lucas theorem says that the roots of $P'(z)$, say $\alpha_1,\alpha_2$, lie in $T$. Let $E$ be the ellipse inscribed in $T$ and tangent to the sides of $T$ at their midpoints. Marden's theorem says that $\alpha_1,\alpha_2$ are the foci of $E$.

It was named Marden's theorem by Dan Kalman because he first read the theorem at Marden's wonderful book Geometry of Polynomials.  For more on the history and a proof of this theorem see this and this.
A: Edit: following Arkamis comment, I will stress out that $C(S)$ is a very natural object called the convex hull of $S$. In the real case, this yields segments and the result follows therefore from Rolle's theorem. 
You just have to show that the roots of $f'$ are in the convex hull of the roots of $f$. Then $C(f')$ will automatically be contained in $C(f)$.
Let $f(z)$ be a degree $n$ polynomial with possibly repeated roots $(z_1,\ldots,z_n)$. We have the partial fraction decomposition:
$$
\frac{f'(z)}{f(z)}=\sum_{j=1}^n\frac{1}{z-z_j}.
$$
Pick any root $z$ of $f'$. If $z$ is a root of $f$, there is nothing to prove. If not, the lhs above is $0$ and the rhs is well-defined.
Now manipulate this equation to prove the claim, using 
$$
\frac{1}{z-z_j}=\frac{\bar{z}-\bar{z_j}}{|z-z_j|^2}.
$$
