I have a question regarding to complex functions:
Let there be $\Omega \subseteq C$ and let be $f=u_1 +iv_1$ an analytic function in $\Omega$. And suppose that a function $g=u_2 +iv_2$ is non-analytic for every $z \in \Omega$ but does have continuous partial derivatives.
Prove that $h=f(g)$ is analytic in $\Omega$ if and only if, $f$ is a constant function in $\Omega$.
To prove that $h$ is analytic given that $f$ is constant is trivial. But for the other direction I'm stuck. I've tried Cauchy–Riemann equations and the chain rule to prove that $f'(z_0) =0 | z_0\in\Omega$, but it has not gotten me anywhere, would like some advise with explanation if possible, thanks!
Best Answer
Inverse $f^{-1}$ exists near any point $z_0$ at which $f^\prime(z_0) \neq 0$. Assuming such $z_0$ exists, near it we would have $g = f^{-1}(h)$. This would be a contradiction since $f^{-1}$ is also analytic due to Cauchy–Riemann equations and because we are given that at each point $g$ is non-analytic.
Or you can directly apply Cauchy–Riemann equations to $h = u_1(u_2, v_2) + v_1(u_2, v_2) i$, assuming $h$ is analytic.
You will obtain $$\frac{\partial u_1}{\partial x} \cdot \frac{\partial u_2}{\partial x} + \frac{\partial u_1}{\partial y} \cdot \frac{\partial v_2}{\partial x} = \frac{\partial v_1}{\partial x} \cdot \frac{\partial u_2}{\partial y} + \frac{\partial v_1}{\partial y} \cdot \frac{\partial v_2}{\partial y}, $$
$$\frac{\partial u_1}{\partial x} \cdot \frac{\partial u_2}{\partial y} + \frac{\partial u_1}{\partial y} \cdot \frac{\partial v_2}{\partial y} = -\frac{\partial v_1}{\partial x} \cdot \frac{\partial u_2}{\partial x} - \frac{\partial v_1}{\partial y} \cdot \frac{\partial v_2}{\partial x}. $$
Using Cauchy–Riemann for $f$, i.e $\frac{\partial u_1}{\partial x} = \frac{\partial v_1}{\partial y}, \frac{\partial u_1}{\partial y} = -\frac{\partial v_1}{\partial x}$, you can rewrite this as follows $$\begin{pmatrix} \frac{\partial u_1}{\partial x} & -\frac{\partial v_1}{\partial x}\\ \frac{\partial v_1}{\partial x} & \frac{\partial u_1}{\partial x} \end{pmatrix} \cdot \begin{pmatrix}\frac{\partial u_2}{\partial x} - \frac{\partial v_2}{\partial y} \\ \frac{\partial u_2}{\partial y} + \frac{\partial v_2}{\partial x} \end{pmatrix} = 0.$$
Assuming $f^\prime = \frac{\partial u_1}{\partial x} + i\frac{\partial v_1}{\partial x} \neq 0$ at some point of $\Omega$. we will have that determinant of the above matrix is non-zero. This would mean that at this point $g$ satisfies Cauchy-Riemann equaions.