Is $f(z)= |z|$ continuous on the complex plane? So I understand that the absolute value of $z=a+b\mathbf i=\sqrt{a^2+b^2}$, I just don't know if its enough to just say that this is continuous so $f$ is continuous or if I have to go through an epsilon-delta proof. A brief explanation of the structure of the proof would be greatly appreciated. 
Thank you for any help!

By triangle inequality, 
$$||z_1|-|z_2|| \leq |z_1-z_2|$$
As $z_1-z_2\to 0$, $f(z_1)-f(z_2) \to 0$.
Hence yes, it is continuous.
