Diffeomorphism on $\mathbb C$

Question

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} $

Answer 1

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.

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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 $\lvert f'\rvert^2 = u_x^2 + v_x^2$.