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Let $f$ be defined as $f(z)=\frac{z}{1+|z|}$. Is $f$ continuous from $\mathbb{C} \to \mathbb{C}$?


$f$ is clearly well-defined on $\mathbb{C}$, but is not holomorphic (Cauchy-Riemann equations are not satisfied as a result of the '$|z|$' term).

I think we may need to make use of the '$\epsilon$-$\delta$' definition of continuity but I'm really not sure how to apply this to complex-valued functions. Any assistance would be very appreciated.

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up vote 3 down vote accepted

Sure it's continuous, it's composed out of continuous bits. $|z|$ is continuous, so $1+|z|$ is, so $1/(1+|z|)$ is (since the denominator is never zero), and $z$ is, so $z/(1+|z|)$ is.

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Not sure how I didn't realise that! To more formally 'prove' that $|z|$ is continuous, can we make the following argument: Take $z\in\mathbb{C}$ and $\epsilon>0$, then $B_{\epsilon}(f(z))\subset\mathbb{R}$ as $\mathbb{R}$ is open in $\mathbb{C}$. Then for any $0<\delta<\epsilon$, $f(B_{\delta}^{\mathbb{R}}(z))\subset B_{\epsilon}(f(z))$, and so $f$ is continuous for any $z\in\mathbb{C}$, which gives us that $f$ is continuous on $\mathbb{C}$. Is that right? – Mathmo Dec 9 '11 at 11:18
Real mathematicians don't use deltas and epsilons (at any rate, not any more than is absolutely necessary). $|z|=\sqrt{x^2+y^2}$; $x$ is a continuous function of $z$, so is $y$, so $x^2$ and $y^2$ are continuous, so $x^2+y^2$ is, and square root is continuous on the non-negative reals, so $\sqrt{x^2+y^2}$ is continuous. I guess you have to use $\epsilon,\delta$ to prove $x$ is continuous, but that should be easy enough that you can't go wrong. – Gerry Myerson Dec 9 '11 at 11:26
Thanks; you've been a great help! – Mathmo Dec 9 '11 at 11:28

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