Let $D$ be an open subset of $\Bbb C$ and let $f:D\rightarrow\Bbb C$ be a function. We say that $f$ is continuous at $z_0 \in D$ if, given $\epsilon > 0$, there exists $\delta >0$ such that for all $z\in D$ $$|z-z_0|<\delta \Rightarrow |f(z)-f(z_0)|<\epsilon.$$ [Stewart & Tall, "Complex Analysis"]
I am asked if the function $f(z)=Re(z)+Im(z)$ is continuous at $z=0$.
The solution provided is as follows:
We have $|Re(z)+Im(z)| \le 2|z|$, whence $|z|<\delta \Rightarrow |f(z)|<2\delta=\epsilon$, so we can take $\delta = \epsilon / 2$.
I'm unsure how the first conclusion ("$|Re(z)+Im(z)| \le 2|z|$") is drawn. Can it be derived from the definitions of $Re(z)$ and $Im(z)$, or is it merely an observation given that $|Re(z)+Im(z)| \sim \sqrt{(Re(z))^2+(Im(z))^2}=|z|$ and hence $2|z|\ge |Re(z)+Im(z)|$?