# Proving continuity of $\operatorname{Re}(z)$ from the definition

Proving, using the definition of a limit that the function is continuous everywhere where $z,a \in C$

$$f(z) = \operatorname{Re}(z)$$

$f:S$ $\subset C \to C$

So I got to $|\operatorname{Re}(z) - \operatorname{Re}(a)| = 0.5|z - a + \bar z - \bar a|$ but I don't know how I can use this expression to show that if $|z - a| < \delta$ then $|\operatorname{Re}(z) - \operatorname{Re}(a)| < \varepsilon$

for any $\varepsilon > 0$

• What you need to do is FIX an arbitrary $\epsilon$. Then FIND a $\delta$ that will work. Hint: Recall that $|x+iy| = \sqrt{x^2 + y^2}$ and so $\min(|x|, |y|) \leq |x+iy| \leq \max(|x|, |y|)$. – 4-ier Aug 22 '18 at 6:17
• Hint for your approach: Remember the triangle inequality. – 4-ier Aug 22 '18 at 6:18
• Thanks, forgot about the triangle inequality – Gordon Ramsey Aug 22 '18 at 6:29

Let $z=(x+iy)$ and $w=(a+ib)$

Note that your function is $$f(x+iy)=x$$and $$f(a+ib)=a$$

$$|f(z)-f(w)|=|f(x+iy)-f(a+ib)|=|x-a|\le \sqrt {(x-a)^2+(y-b)^2}=|z-w|$$

For a given $\epsilon >0$ let $\delta = \epsilon$

If $$|z-w|<\delta$$ then $$|f(z)-f(w)|\le |z-w|<\delta =\epsilon$$

Thus $f$ is continuous.

It is easy to show that $z_n \to z$ implies $\overline{z}_n \to \overline{z}$. Hence, $z_n \to z$ implies $$\operatorname{Re} z_n = z_n + \overline{z}_n \ \to \ z + \overline{z} = \operatorname{Re} z.$$ This shows that $\operatorname{Re}(\cdot)$ is continuous.

Forgot about the Triangle Inequality $|\operatorname{Re}(z) - \operatorname{Re}(a)| = 0.5|z - a + \bar z - \bar a| \leq 0.5(|z - a| + |\bar z - \bar a|) = 0.5(2|z - a|) = |z-a|$ using triangle inequality

then pick $\delta = \varepsilon$

Therefore if $|z - a| < \delta$

$|\operatorname{Re}(z) - \operatorname{Re}(a)| < \delta = \varepsilon$

Thanks the guy in the comments

• Something to keep in mind is that the real and imaginary parts are both smaller than the absolute value. So, $|Re(z-a)| \leq |z-a|$. This is proved in the way that you did it above. – 4-ier Aug 22 '18 at 6:34