# Using sequences prove that $f(z)=Re(z)$ and $g(z)=Im(z)$ are continuous.

I've found a few proofs showing the continuity with the Cauchy-Riemann equations but am unsure as to whether they are proved using sequences which the question I'm attempting requires. I may be wrong but I think if I just prove that $$z_j$$ tends towards $$z$$ if and only if Re(zj) tends towards $$Re(z)$$ and $$Im(z_j)$$ tends towards $$Im(z)$$, then this implies that $$Re(z)$$ and $$Im(z)$$ are continuous. If anyone could give the proof that would be much appreciated. Thanks.

• What do you mean by continuity of an equation? – Calvin Khor Feb 21 '19 at 15:04
• I don't see the relation between your question and Cauchy-Riemann equations. – mfl Feb 21 '19 at 15:05
• Your statement '$z_j \rightarrow z$ if and only if $\Re(z_j)\rightarrow \Re(z)$ and $\Im(z_j) \rightarrow \Im(z)$' is correct. Please use $|z_j -z|^2= |\Re(z_j) - \Re(z)|^2+|\Im(z_j) - \Im(z)|^2\leqslant (|\Re(z_j) - \Re(z)|+|\Im(z_j) - \Im(z)|)^2$. – Teebro Prokash Feb 21 '19 at 15:23
• @CalvinKhor continuity of an equation is whether or not it is continuous. – Nick Rilett Feb 21 '19 at 15:31
• @mfl it's related because a lot of the proofs I saw (that don't use sequences) use the CR equations. – Nick Rilett Feb 21 '19 at 15:31

We will use the convergence criterion for continuous maps to prove this. Let $$\{z_n\}_{n=1}^{\infty}$$ be a sequence of complex numbers converging to $$z$$. To prove $$f$$ and $$g$$ to be continuous, it is enough to prove that $$f(z_n) (\text{resp. } g(z_n))$$ converges to $$f(z) (\text{resp. } g(z)).$$
Let $$\epsilon > 0$$ be given. We know that, there exists $$N \in \mathbb{N}$$ such that for all $$n \geq N$$, \begin{align*} |z_n - z| &< \epsilon \\ \implies |Re(z_n - z) + i Im(z_n -z)| &< \epsilon \\ \implies |Re(z_n -z)| < \epsilon &\text{ and } |Im(z_n-z)| < \epsilon \\ \implies |Re(z_n) - Re(z)|< \epsilon &\text{ and } |Im(z_n) -Im(z)| < \epsilon \text{ for all } n \geq N. \end{align*} Thus, $$Re(z_n) \longrightarrow Re(z)$$ and $$Im(z_n) \longrightarrow Im(z)$$. Hence, the functions $$Re(z)$$ and $$Im(z)$$ are continuous.