1

I'm having trouble in this homework question. Let's suppose $U=\mathbb{C}-\{z\in\mathbb{C}:Re(z)\leq 0\}$ and $n\in\mathbb{Z}^*_+$. I need to find all holomorphic functions $f$ that satisfies the functional equation $z=(f(z))^n$, for all $z\in U$.

How should I approach this problem? The domain suggests the use of logarithm, but I'm not sure how.

I tried working with $f(z)=|f(z)|(\cos(\theta(z))+i\sin(\theta(z)))$ where $\theta$ is its principal argument, that is, $\theta(z)\in (-\frac{\pi}{2},\frac{\pi}{2})$, which led me to these $n$ functions $$ f_k(z)=\sqrt[n]{|z|}\left(\cos\left(\frac{\theta(z)+2k\pi}{n}\right)+i\sin\left(\frac{\theta(z)+2k\pi}{n}\right)\right) $$ , $k=0,1,...,n-1$, and each one of these satisfies $(f_k)^n(z)=z$. But if this is correct why aren't there functions defined on all $z$?

OhMyGod
  • 511
  • The title does not correspond entirely to the question. – Marra Sep 11 '13 at 01:59
  • 1
    Just fixed it ;) – OhMyGod Sep 11 '13 at 01:59
  • Firstly, you've got a typo with the $k$'s in the fractions. Secondly, the problem originates from the manner in which you go about defining $\theta(z)$. – Jonathan Y. Sep 11 '13 at 02:00
  • Also there's something wrong on the arguments of cosine and sine of each $f_k$. – Marra Sep 11 '13 at 02:00
  • @JonathanY. could you be more specific? – OhMyGod Sep 11 '13 at 02:01
  • I think it is correct now. – OhMyGod Sep 11 '13 at 02:03
  • Looks good to me now (EDIT: of course, we'd need to specify why $k$ is fixed for each selection of $f$). Regarding the issues with finding such an entire function, can you continuously define $\theta(z)$ on the entire plane (or punctured at the origin, if that's an issue)? – Jonathan Y. Sep 11 '13 at 02:04
  • @JonathanY. I'm so sorry, I forgot to say and edited again; $z$ must belong in $\mathbb{C}-{z\in\mathbb{R}:Re(z)\leq z}$, that is, the complex plane minus the non-positive real axis. – OhMyGod Sep 11 '13 at 02:09
  • I still don't figure out why can't z be a negative real number (non-zero). – OhMyGod Sep 11 '13 at 02:10
  • That's quite alright. I was answering your question regarding the reason we can't define an entire $f$ which satisfies this relation. EDIT: let me repeat: your definition assumes $\theta(z)$ is well-defined and continuous on $f$'s domain (why?) – Jonathan Y. Sep 11 '13 at 02:11
  • And why is that? – OhMyGod Sep 11 '13 at 02:12
  • Well, show that $f_k$ is holomorphic. In the process, note where you require that $z\mapsto(\theta(z)+2\pi ik)$ is a continuous function. – Jonathan Y. Sep 11 '13 at 02:18
  • Oohh I think I know. Because I must do $\theta(z)=\arctan\left(\frac{v(z)}{u(z)}\right)+2m\pi$, for some $m\in\mathbb{Z}$, where $z=u(z)+ iv(z)$, and on the non-positive real axis $u$ equals zero? – OhMyGod Sep 11 '13 at 02:23
  • Actually, I really recommend not to think of $\theta(z)$ that way (one reason is that it appears $\theta$ doesn't distinguish between $(-u,v),(u,-v)$--while you'd really like it to), but essentially that's true. – Jonathan Y. Sep 11 '13 at 02:32
  • In this case it should work because the real part is always positive. – OhMyGod Sep 11 '13 at 02:36

0 Answers0