Diffeomorphism on $\mathbb C$ Let $P= a_{0} z^{n}  +a_{1} z^{n-1}+  \cdots +a_{n}$, with $ a_{0} \neq 0,z \in \mathbb C$.  
I don't know why $P$ fails to be a local diffeomorphism only at the zeros of the derivative polynomial $P'(z)= \sum  a_{n-j}  j z^{j-1} $
 A: If we have a real-differentiable function $f \colon U \to \mathbb{C}$, where $U\subset \mathbb{C}$, with real part $u$ and imaginary part $v$, then its Jacobian matrix in real form is
$$J_f^{\mathbb{R}} = \begin{pmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{pmatrix},$$
and in complex form, with the Wirtinger derivatives, it is
$$J_f^{\mathbb{C}} = \begin{pmatrix}
\frac{\partial f}{\partial z} & \frac{\partial f}{\partial \overline{z}}\\
\frac{\partial\overline{f}}{\partial z} & \frac{\partial\overline{f}}{\partial\overline{z}}
\end{pmatrix}.$$
For a holomorphic $f$, the Cauchy-Riemann equations yield
$$\det J_f^{\mathbb{R}} = \det J_f^{\mathbb{C}} = \lvert f'\rvert^2,$$
so a holomorphic function is a local diffeomorphism at all points where the derivative does not vanish by the (real) inverse function theorem.

Computation of the determinants:


*

*Complex form: The Cauchy-Riemann equations have the form $\frac{\partial f}{\partial\overline{z}} = 0$. Since we always have $\frac{\partial \overline{g}}{\partial z} = \overline{\frac{\partial g}{\partial\overline{z}}}$ and $\frac{\partial \overline{g}}{\partial \overline{z}} = \overline{\frac{\partial g}{\partial z}}$, we have
$$J_f^{\mathbb{C}} = \begin{pmatrix}f' & 0\\ 0 & \overline{f'}\end{pmatrix}$$
and $\det J_f^{\mathbb{C}} = \lvert f'\rvert^2$.

*Real form: The Cauchy-Riemann equations are $u_x = v_y$ and $u_y = -v_x$ in real form, and that gives $\det J_f^{\mathbb{R}} = u_xv_y - u_yv_x = u_x^2 + v_x^2$. Now, $$\begin{align}
\frac{\partial f}{\partial z} &= \frac12\left(\frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y}\right)\\
&= \frac12\left(u_x+iv_x - i(u_y+iv_y)\right)\\
&= \frac12\left((u_x+v_y) + i(v_x-u_y)\right)\\
&= u_x + iv_x,
\end{align}$$
so $\lvert f'\rvert^2 = u_x^2 + v_x^2$.
