An entire function such that $if(z)= \bar f(z)$ for all $z$ is constant

Question

Let $f$ be an entire function such that $if(z)= \bar f(z)$ for all $z\in\Bbb C$. Show that $f$ is the constant function.

I didn't understand it but I took some notes which seem like gibberish to me now. Maybe someone could clarify:

• I had drawn a line on a graph that has an arg of $\pi/4$ then drew the conjugate of that so a line with arg of $-\pi/4$. Where did this come from?

• I had defined $f(z) = w$ and did $iw=\bar w$. Were supposed to solve(?) that and somehow end up with $w=$ $e^{-i\pi/4 + in\pi}$.

Can someone elaborate on these steps and clarify what's happening?

Answer 1

If $\overline f$ is also an analytic function then its components satisfy the Cauchy-Riemann equations in reverse. Together with the ordinary CR equations this puts all partial derivatives equal to 0.

Independently of these considerations, you can also argue that $(\overline f)^2=if\overline f=i|f|^2,$ which lies on the positive imaginary axis, so the image of $\overline f$ lies on the line through 0 defined by argument $\pi/4+k\pi.$

This implies that $\frac1{1+\overline f}$ is a bounded entire function so it, as well as $\overline f$ and $f$ itself, must be constant.

Answer 2

If $f$ and $\bar f$ are holomorphic, they are constant. Proof. If $f$ holomorphic then $\det f'(z_0) = \vert f'(z_0)\vert ^2 \geq 0$, where $f'(z_0)$ is a linear map from $\bf R^2$ to itself.

As the linear map $C: z\to \bar z$ has determinant $\det C=-1$, $\det (\bar f '(z_0)) \leq 0$. So if at some point $f'(z)\not = 0$ $\bar f$ is not holomorphic.