Continuity of Complex Component Functions

Theorem

Suppose we have a function $f : U \to C$ for $U \subseteq C$ given by

f (x + i y) = u (x, y) + i v (x, y)

where $u (x, y), v (x, y) \in R$ . Then $f$ is continuous at a point if and only if both $u$ and $v$ are continuous at the corresponding point.

Proof

Suppose $f$ is as above, and suppose further that it is continuous at a point $z_{0} \in U$ . Then we have

lim_{z \to z_{0}} f (z) = f (z_{0}) .

That is, for any $ϵ > 0$ there exists a $δ$ such that

| z - z_{0} | < δ ⟹ | f (z) - f (z_{0}) | < ϵ .

Writing $z = x + i y$ and $z_{0} = x_{0} + i y_{0}$ for $x, y, x_{0}, y_{0} \in R$ , we have the following equivalences to the premise on the left

\begin{aligned} | z - z_{0} | < δ \\ ⟺ & | x + i y - x_{0} - i y_{0} | < δ \\ ⟺ & | (x - x_{0}) + i (y - y_{0}) | < δ \\ ⟺ & \sqrt{(x - x_{0})^{2} + (y - y_{0})^{2}} < δ \\ ⟺ & | [\begin{array}{c} x \\ y \end{array}] - [\begin{array}{c} x_{0} \\ y_{0} \end{array}] | < δ . \end{aligned}

Furthermore, the consequence on the right implies the following

\begin{aligned} | f (z) - f (z_{0}) | < ϵ \\ ⟹ & | u (x, y) + i v (x, y) - u (x_{0}, y_{0}) - i v (x_{0}, y_{0}) | < ϵ \\ ⟹ & | (u (x, y) - u (x_{0}, y_{0})) + i (v (x, y) - v (x_{0}, y_{0})) | < ϵ \\ ⟹ & \sqrt{(u (x, y) - u (x_{0}, y_{0}))^{2} + (v (x, y) - v (x_{0}, y_{0}))^{2}} < ϵ \\ ⟹ & (u (x, y) - u (x_{0}, y_{0}))^{2} + (v (x, y) - v (x_{0}, y_{0}))^{2} < ϵ^{2} \\ ⟹ & (u (x, y) - u (x_{0}, y_{0}))^{2} < ϵ^{2} and (v (x, y) - v (x_{0}, y_{0}))^{2} < ϵ^{2} \\ ⟹ & | u (x, y) - u (x_{0}, y_{0}) | < ϵ and | v (x, y) - v (x_{0}, y_{0}) | < ϵ . \end{aligned}

As such, we can fix any $ϵ > 0$ , acquire $δ > 0$ as per continuity of $f$ , and then we have that

\begin{aligned} | [\begin{array}{c} x \\ y \end{array}] - [\begin{array}{c} x_{0} \\ y_{0} \end{array}] | < δ & ⟺ | z - z_{0} | < δ \\ ⟹ | f (z) - f (z_{0}) | < ϵ \\ ⟹ | u (x, y) - u (x_{0}, y_{0}) | < ϵ \end{aligned}

and hence we have continuity of $u$ , and equivalently $v$ .

The reverse implication follows trivially from the basic properties of complex limits, namely we can split the limit across the sum and product given the constituent limits exist in this case:

\begin{aligned} lim_{z \to z_{0}} f (z) & = lim_{(x, y) \to (x_{0}, y_{0})} (u (x, y) + i v (x, y)) \\ = lim_{(x, y) \to (x_{0}, y_{0})} u (x, y) + i lim_{(x, y) \to (x_{0}, y_{0})} v (x, y) \\ = u (x_{0}, y_{0}) + i v (x_{0}, y_{0}) \\ = f (z_{0}) . \end{aligned}