Show that f must be constant on C

Question

This is a problem that I have been encountered after reading about analytic functions in complex analysis.

Suppose $f(z) = f(x + iy)$ is analytic on $\mathbb{C}$. Let $u= \Re ~f$ and $v = \Im ~f$. Show that if the range of $f$ is contained in the parabola described by $v = u^2$, then f must be constant on $\mathbb{C}$.

What I am thinking:

Since it is assumed that the function is analytic, and they are breaking down into $u$ and $v$, I know I have to somehow use the Cauchy-Riemann equations.

Any insight here would be greatly appreciated.

Answer 1

Another idea, use Liouville: $${\rm Im}\ f = v = u^2\ge 0\implies\exp(if) \hbox{ bounded}\implies\exp(if) \hbox{ constant}\implies\cdots$$

Answer 2

I'll leave to you to check that this all follows from the Cauchy-Riemann equations and differentiating the identity $v=u^2$: $$ v_x=2uu_x=2uv_y=2u(-2uu_y)=(2u)^2v_x. $$ Therefore either $v_x(z)=0$ or $u(z)=\pm 1/2$ and in this last case we have necessarily $v(z)=1/4$. Therefore for every point in $\mathbb{C}$ either $v_x(z)=0$ or $v(z)=1/4$. By continuity of $v_x$ this implies that $v_x=0$ throughout. Similarly we conclude that $v_y$ is zero and by Cauchy-Riemann so are $u_x$ and $u_y$. Therefore $f$ is constant.

Answer 3

Regarding $f$ as map $f: \mathbb{R}^2 \to \mathbb{R}^2$ the derivative $f'(z_0)$ is the Jacobian matrix $df$ which take a tangent vector $v$ at $x_0$ to a tangent vector $df(v) = f'(z_0) v$ at $f(z_0)$. Here you regard the vector $v$ as a complex number. If $f'(z_0) \neq 0$ then $df(z_0)$ is surjective hence the image of $f$ can not be contained in a smooth curve $\gamma \subset \mathbb{R}^2$ (otherwise the image of $df(z_0)$ is contained in the tangent line to $\gamma$). So if you know that the image of $f$ is contained in a smooth curve $\gamma$ then $f'(z) = 0$ for all $z \in \mathbb{C}$ hence $f$ is constant.