# Is $f(z)= |z|$ continuous on the complex plane?

So I understand that the absolute value of $z=a+b\mathbf i$ is $(a^2+b^2)^{1/2}$, 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!

• Next follow up question: Is it differentiable (holomorphic)? – imranfat Feb 12 '18 at 21:41

It is standard to use $\epsilon$-argument to deduce that $h(x,y)=x^{2}+y^{2}$ is continuous on the plane, now use the fact that composition with continuous map is still continuous to conclude that $f$ is continuous: $\xi(u)=\sqrt{u}$ and $f=\xi\circ h$.

The continuity of a complex-valued function is the same issue with the continuity of the corresponding two variables function: $f(z)=|z|=f(x,y)=\sqrt{x^{2}+y^{2}}$.

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.

Every norm is uniformly continuous with respect to itself. Let $(X,\|.\|)$ be a normed space. Let us recall what uniform continuity means. $$\forall \epsilon>0 \;\exists \delta >0 \;\text{such that}\; \big|\|x\| - \|y\|\big|<\epsilon \;\text{if}\; \|x-y\|<\delta$$ We can choose $\delta = \epsilon$ since the reverse triangle inequality yields $$\big|\|x\|-\|y\|\big| \leq\|x-y\| < \epsilon.$$