Showing $f(z)$ is constant when $Im(f^2(z)-f(z))=0$

Question

I don't have the full question but I'm assuming $f(z)$ must be entire for this to occur. Also note that $u>1$ where $u$ is the real part of $f(z)$.

If we solve the given eqn then we are left with $v=0$. I have the $v=0$ drawn out on an axis. Now my professor has stated that we "rotate" this to the imaginary axis by multiplying it with $i$ so we somehow end up with $|e^{if(z)}|=1$. What's going on here?

Answer 1

$\Im(f^2(z) - f(z)) =0 $ gives that $2uv = v$. This means either $u=\frac{1}{2}$ or $v=0$. Now use Cauchy-Riemann equation to deduce that $f$ is constant

Answer 2

It seems your professor is saying to let $g(z) = f(z)^2 - f(z)$, and then argue why the holomorphic function $e^{ig(z)}$ must have constant modulus equal to $1$.

Answer 3

HINT:

The result is called maximum modulus principle. Assuming that the function $f$ is defined on a connected open set, since $|e^{i g}|$ is constant you conclude that $e^{i g}$ is constant and so $g ( = f^2 - f)$ is constant, and from here $f$ constant.

Answer 4

Here is a barehanded proof for a slightly more general case, that is when $u$ is not assumed to be greater than $1$.

Let $f(z) = \sum_{n=0}^\infty a_n z^n$ be an analytic function (WLOG analytic at the origin). If we write $f(z) = u(z) + iv(z)$ then we have $$f(z)^2 = u(z) - v(z) + 2i u(z)v(z).$$

Thus $Im(f(z)^2 - f(z)) = 2u(z)v(z) - v(z) = v(z)(2u(z)-1) = 0$, and we have for every $z$ either $v(z) = 0$ and/or $u(z) = 1/2$.

Consider the sequence $z_m = 1/m$. By the Pigeon hole principle, either $v(z_m)=0$ for infinitely many $m$ or $u(z_m) = 1/2$ for infinitely many $m$. We can pass to a subsequence where either $v(z_m)$ or $u(z_m)$ is constant. Without loss of generality assume that $u(z_m) = 1/2$ for a subsequence $x_m$ of $z_m$.

For real $x$, $Re(f(x)) = u(x) = \sum_{n=0}^\infty Re(a_n) x^n$.

Note that for any real analytic function, if $x_m$ is a sequence that converges to the origin, and $u(x_m)$ is constant, then $u$ is constant. This follows from the Taylor series and this question: Show that $f^{(n)}(0)=0$ for $n=0,1,2, \dots$ where we take our function to be $u(z) - 1/2$.

Thus $u(z) = Re(f(z))$ is constant. The Cauchy-Riemann equations then tell us that $v(z)$ is constant.