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Suppose that $f:U\subset\mathbb{C}\to\mathbb{C}$, where $U$ is a region in the complex plane, is a holomorphic function.

If $c\in\mathbb{R}$ is a regular value for $\text{Re}(f(z))$ then it follows from implicit function theorem that $\text{Re}(f(z))^{-1}(c)$ is at least locally a differentiable curve in the plane.

Question:

1- If $c$ is a regular value is any connected component of $\text{Re}(f(z))^{-1}(c)$ a global differentiable curve ?

2-If $c$ is not a regular value and $\text{Re}(f(z))^{-1}(c)$ have at least one cluster point is this set locally a curve ?

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What do you mean by "global differentiable curve"? –  Jesse Madnick Nov 28 '10 at 5:43

1 Answer 1

(1) To use your notation, if $c \in \mathbb{R}$ is a regular value, then the level set $Re(f(z))^{-1}(c)$ will be a 1-dimensional embedded submanifold of $U \subset \mathbb{R}^2$. Therefore, every connected component will be a connected 1-manifold.

Now, I'm not sure what exactly you mean by "global differentiable curve."

If you mean "something of the form $f(t) = (x(t), y(t))$," then it follows from the comments in this question that yes, every connected component can be put in that form.

If you mean "something of the form $y = f(x)$ or $x = f(y)$" then the answer is (I think) no. For example, consider $f(z) = \log z$ on $U = R^2 - \{x \geq 0, y = 0\}$, i.e. the plane with the non-negative x-axis deleted. Then $Re(f(z)) = \log(\sqrt{x^2+y^2})$, so the level set $Re(f(z))^{-1}(1)$ is the circle $x^2 + y^2 = e^2$ minus the point $(e,0)$. So, it doesn't seem like you'd be able to represent this curve by a single function $x = f(y)$ or $y = f(x)$.

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Hi Jesse, I realized after I post that this part (1) of the question it is not hard as you said in the first paragraph. Some time after I post the question here I post it also on Mathoverflow and I got from Bruno Martelli the full answer. The idea is to use the local form of holomorphic functions. I will leave the link in case you want to see the solution : –  Leandro Nov 29 '10 at 3:38
    
    
Thanks for the answer Jesse. –  Leandro Nov 29 '10 at 3:38

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