hi there so i'm having a bit of an issue which im sure you the good people of stack exchange will be able to fix quickly.

so heres my problem.

$$\text{let } f:\mathbb{C} \rightarrow \mathbb{C} \text{ with } z \in \mathbb{C}$$

then for any $z \in \mathbb{C}$ $f$ has the representation in the form of

$$f(z) = u(x,y) + iv(x,y)$$

where $u(x,y),v(x,y) \in \mathbb{R}$

i need to understand the similiar forms of the compliments you can get from $f(z)$ namely

$f(\bar{z}),\bar{f(z)} \text{ and } \bar{f(\bar{z})}$ where the bar is the compliment of the respective variable/function.

my thinking was originally this

$$f(\bar{z}) = u(x,y)-i(x,y)$$

but when asked to find $$\bar{f(\bar{z})}$$ i'm stuck. i know from the answer that

$$\bar{f(\bar{z})} = u(x,-y)-iv(x,-y)$$ but dont understand why.

any help would be great thanks for taking the time to read this.


1 Answer 1


If $f(x+yi)=u(x,y)+v(x,yi)$, then$$f\left(\overline{x+yi}\right)=f(x-yi)=u(x,-y)+v(x,-y)i$$and$$\overline{f(x+yi)}=\overline{u(x,y)+v(x,y)i}=u(x,y)-v(x,y)i.$$Can you take it from here?

  • $\begingroup$ Yup. Brilliant, that all makes sense. Thank you for the quick response. $\endgroup$
    – Vaas
    Jan 3, 2018 at 18:49

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