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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.

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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. – achille hui Feb 4 '13 at 6:32
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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 $f$.

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