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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.

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

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  • $\begingroup$ Perfect thank you $\endgroup$ – Nick Rilett Feb 21 '19 at 15:25

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