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I'm currently taking my first course in Complex analysis and both the lectures and textbook have very minimal examples and it has been extremely frustrating.

Prove that f is continuous on $\mathbb{C}$ when: $$f(z) = \bar{z}$$

Would I be able to use the Cauchy_Riemann Theorem to prove that this is holomorphic and in turn say that it is continuous?

Let $u(x,y) = x$ and $v(x,y) = -iy$.

Therefore, $u_x = 1 , v_y = -i$. Showing that $u_x \neq v_y$.

This shows that the function is not continuous.

Thank you for any guidance.

P.S If anyone has any recommendations on books on undergraduate Complex analysis with a lot of examples please let me know.

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2 Answers

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A complex-valued function is continuous if and only if both, its real part and its imaginary part are continuous.

$f$ sends $$x+iy\mapsto x-iy$$

$f(x+iy)=u(x,y)+iv(x,y)$, where $u,v$ are real valued functions.

$u(x,y)=x$ is continuous and $v(x,y)=-y$ is also continuous so $f$ is continuous.

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We have $|f(z)-f(z_0)|=|\overline{z-z_0}|=|z-z_0| \to 0$ when $z\to 0$. ($|\overline{Z}|=|Z|$).

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