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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!

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  • $\begingroup$ Next follow up question: Is it differentiable (holomorphic)? $\endgroup$ – imranfat Feb 12 '18 at 21:41
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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: $f(z)=|z|=f(x,y)=\sqrt{x^{2}+y^{2}}$.

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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.

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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 $$ \big|\|x\|-\|y\|\big| \leq\|x-y\| < \epsilon. $$

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