# How to calculate $\lim_{z \rightarrow (3 -2i)} f(z) = \overline z - 3 re(z) + i$?

$$f(z)$$ 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?

• What is a definition of $f$ ? Commented Oct 23, 2018 at 22:28
• I added the definition Commented Oct 23, 2018 at 22:30

Hint: Write your function $$f=u+iv$$ . Then you can use continuity because $$u=u(x,y)$$ and $$v=v(x,y)$$ with $$z=x+yi$$
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, $$|g(z)-g(z_0)|,\ |h(z)-h(z_0)| <\varepsilon$$