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

Question

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.

Answer 1

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.