0

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?

Cayde
  • 87
  • 1
    You probably know the theorem that a holomorphic function (defined on a connected open set) with constant imaginary part must be constant, right? – David Dec 14 '15 at 18:18
  • Actually I know: If $f$ is analytic on D (open and connected) and $|f(z_0)|$ is the max of $f(z)$ where $z_0 \in D$ then $f(z)$ is constant. What's your theorem called? I've taken note of it but can you try to apply the theorem I've stated before (also) applying yours? By that I mean walk me through it. Sorry I am asking for a lot, I just really want to understand this. Thanks friend! – Cayde Dec 14 '15 at 18:30
  • I don't know a name for it. It's a direct consequence of the open mapping theorem, but in most books something similar is proved before that. Do you have any other theorems that look similar to this one? – David Dec 14 '15 at 18:32
  • Oh, my Doc said to not use open mapping theorem because we won't approach it in time (for our exam). He did mention that it would be very simple to use what you stated but unfortunately, for our exam we aren't allowed. I have written how my Doc has done it above, if you can take a glance at it. – Cayde Dec 14 '15 at 18:34

4 Answers4

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

  • $u>1$ from the question though. – Cayde Dec 14 '15 at 18:41
  • @Cayde, if $u > 1$ then this means that $v(z) = 0$ everywhere in the domain of $f$. Thus the Cauchy-Reimann equations tell you that $u$ is constant. – Joel Dec 14 '15 at 19:16
  • After C-R, can we deduce that $u(z)=c & v(z)=0$ so $g(z)$ is a function that maps $\Bbb C \rightarrow \Bbb R$? – Cayde Dec 14 '15 at 19:59
  • Assuming $g$ is the function defined by David in his answer, you can show that without using C-R equations. It is given that $\Im(g(z)) = 0$ –  Dec 14 '15 at 20:03
  • Yeah, sorry I should have specified. And yeah, so by what was given I could just tell. Silly me. Thanks for the help! – Cayde Dec 14 '15 at 20:10
0

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 $1$.

David
  • 6,306
  • My biggest concern is: how does he go from having $v=0$ on a graph, multiplying it with $i$(?) and getting the imaginary axis. He mentioned to me that the "trick" is to orient it so it becomes the imaginary axis and then you can tell. I haven't had a chance to ask him either unfortunately. – Cayde Dec 14 '15 at 18:40
  • Orangeskid's answer is what I was hinting at. – David Dec 14 '15 at 18:43
  • Oh, sorry, I sometimes miss the more obvious things. Let me take a gander and give you feedback. – Cayde Dec 14 '15 at 18:46
0

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.

orangeskid
  • 53,909
  • Hello friend, yes this chapter goes by that name, what a coincidence! Jokes aside, how does he equate $|e^{ig}|$ to 1? Where does the 1 come from? – Cayde Dec 14 '15 at 18:43
  • Cayde, what kind of number is $g(z)$? – David Dec 14 '15 at 18:44
  • $g$ is real of course! – Cayde Dec 14 '15 at 18:46
  • @ Cayde: $e^{i ( a+ i b)} = e^{-b} \cdot e^{i a}$, and the first factor is the absolute value . If $b=0$ then you get $e^{i a}$, of absolute value $1$. – orangeskid Dec 14 '15 at 18:47
  • Is this what my doc meant by "multiply by $i$ to get the imaginary axis"? when he then took the modulus of the exp? – Cayde Dec 14 '15 at 18:49
0

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.

Community
  • 1
Joel
  • 16,256