How to prove $f(z)=|z|$ is continuous?

Question

We were given this hint . (Hint: $||z| − |w|| ≤ |z − w|$ for all $z, w ∈ C.$)

From what I understood in lecture, it's some epsilon delta proof? I don't understand how to prove continuity at all using this method or any other and I'm really lost): please be very explicit with your help if you can because i really suck at this and I've been staring at this question for hours

Answer 1

To prove continuity at a point $z_0$ we need to show that for all $\epsilon > 0$ there exists a $\delta >0$ such that $$ |z-z_0|\le \delta \implies |f(z)-f(z_0)| \le \epsilon.$$

So let $\epsilon>0.$ We can pick $\delta=\epsilon.$ Then if $|z-z_0|\le \delta,$ then $$ |f(z)-f(z_0)| = ||z|-|z_0|| \le |z-z_0| \le \delta =\epsilon.$$

Answer 2

Let $w\in \mathbb C$. To prove: $f$ is continuous at $w$.

Let $\epsilon \gt 0$ be given. We need to produce $\delta \gt 0$ such that $\lvert z-w \rvert \lt \delta \implies \lvert f(z)-f(w)\rvert \lt \epsilon $...

But $f(z)-f(w)=\lvert z\rvert -\lvert w\rvert \implies \lvert f(z)-f(w) \rvert =\lvert \lvert z\rvert-\lvert w\rvert \rvert \le \lvert z-w \rvert \lt \delta $.

$\therefore $ set $\delta =\epsilon $...