# Proving that the “real part” function is continuous

I want to use the definition of a limit, $|f(z) - w_0| < \varepsilon$ whenever $0 < |z - z_0| < \delta$

to prove $$\lim_{z \to z_0} \mathop{\rm Re}(z) = \mathop{\rm Re}(z_0)$$

By intuition this is obvious but I dont know how to show it using the defn. of a limit. This is question 1(a) from the book Complex Variables and Applications.

Here's the basic manipulation I have made going by an example in the book, I dont know where to go from here... $$|\mathop{\rm Re}(z)-\mathop{\rm Re}(z_0)| = |x - x_0| = |x| - |x_0| = x - x_0$$

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We have $|z_0-z|^2=(\Re(z_0-z))^2+(\Im(z_0-z))^2\geq (\Re(z_0-z))^2$ so $|z_0-z|\geq |z_0-z|$. Now, check that $\delta=\varepsilon$ in the definition of the limit works.
If you have a complex number $z = a + ib$ where $a$ and $b$ are real, then $$|z| = \sqrt{a^2 + b^2} \ge \sqrt{a^2} = |a|.$$ From this we see that for any complex number $z$, $$|z| \ge |\Re(z)| \ge \Re(z).$$ Now apply this to $z - z_0$ and your limit follows right away.