2
$\begingroup$

Since we can represent complex functions as $f(z) = u(x,y) + iv(x,y)$, under what conditions do we know that $f(z)$ is continuous? For example, if $u(x,y)$ and $v(x,y)$ are both continuous can we conclude that $f(z)$ is continuous? Does anyone know of a proof of this, or just an intuitive explanation of why or why not this is true? Furthermore, if either $u(x,y)$ or $v(x,y)$ are not continuous does this imply that $f(z)$ is not continuous? This isn't for a particular question I'm just wondering about how arguments involving continuity transfer over to complex valued functions.

$\endgroup$
1
  • $\begingroup$ First, continuous functions are closed under addition and multiplication. This means $u, v$ continuous $\implies f = u + iv$ continuous. Second, continuous functions are closed under function composition. Since taking the real part ( $\Re : x + iy \mapsto x$ ) and imaginary part ( $\Im : x + iy \mapsto y$ ) of a complex number is a continuous function, $f$ continuous $\implies u = \Re \circ f$ and $v = \Im \circ f$ are both continuous. $\endgroup$ – achille hui Feb 4 '13 at 6:32
3
$\begingroup$

The key here is that topologically $\mathbb{C}$ and $\mathbb{R}^2$ are the same thing. Namely, it really behooves us to think of functions $f(x,y)$ which are mappings $\mathbb{R}^2\to\mathbb{R}^2$. Then, it is a common fact (because $\mathbb{R}^2$ is given the so-called product topology) that a map $X\to\mathbb{R}^2$ will be continuous if and only if its two coordinate functions are continuous. Your $u$ and $v$ are precisely the coordinate functions of $f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.