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$f(z)$ is defined as $ f(z) = \overline z - 3 re(z) + i $

I need to prove the $f(z)$ is continuous in that point. I only know the epsilon-delta definition of limit.

How you do that? Do you know any method that I can use?

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  • $\begingroup$ What is a definition of $f$ ? $\endgroup$
    – HK Lee
    Commented Oct 23, 2018 at 22:28
  • $\begingroup$ I added the definition $\endgroup$ Commented Oct 23, 2018 at 22:30

2 Answers 2

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Hint: Write your function $f=u+iv$ . Then you can use continuity because $u=u(x,y)$ and $v=v(x,y)$ with $z=x+yi$

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i) $ f(z=x+iy)=x-iy -3x +i = -2x +i(-y+1) \ =(g+ih) $ where $g,\ h$ are a real function on $\mathbb{R}^2$.

ii) $f$ is a continuous at point $z_0$ iff $g,\ h$ are continuous at a point $z_0$

Proof : $\Leftarrow$ For $|z-z_0|<\delta$, assume that $|g(z)-g(z_0) |,\ |h(z)-h(z_0)|<\varepsilon$ so that $$ |f(z)-f(z_0)| = \sqrt{(g(z)-g(z_0))^2+ (h(z)-h(z_0))^2} \leq \sqrt{2}\varepsilon $$

$\Rightarrow $ For $|z-z_0|<\delta$, assume that $|f(z)-f(z_0) | <\varepsilon$ so that $$ \varepsilon > |f(z)-f(z_0)| = \sqrt{(g(z)-g(z_0))^2+ (h(z)-h(z_0))^2} $$

Accordingly, $$|g(z)-g(z_0)|,\ |h(z)-h(z_0)| <\varepsilon $$

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