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Let $f(z)$ de defined by

$f(z)=2z/(z+1)$ for $z\not=0$

and $f(z)=1$ for $z=0$. So I want to assume the function is continuous at $z=0$ and then come to a contradiction. So i have $|2z/(z+1)-1|<\epsilon$ which makes:

$|z-1|/|z+1|<\epsilon$ and i do not really know what to do here. Any help will be appreciated.

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    Doesn't it feel like $f(0)$ should be what it would be without the restriction that $z \neq 0$? I mean, the function *would* only have a problem at $z=-1$... – 2017-02-12
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    @pjs36 f(0)=0.. if it would be without the restriction. It would be a different function. I mean clearly the function is not continuous, i just want to prove it. – 2017-02-12
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    $\lim_{z \to 0} \dfrac{2z}{z+1} = 0$. – 2017-02-12
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    I guess the wording of the post led me to believe you were under the impression it *was* continuous at $z=0$. – 2017-02-12
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    @pjs36 My bad, i see what you are saying. I wanted to somehow show a contradiction. – 2017-02-12

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By the triangle inequality $|1+z|\geq |1|-|z|=1-|z|.$

$$\text {So }\quad |z|<1/3\implies |1+z|\geq 1-1/3\implies |1+z|^{-1}\leq 3/2\implies $$ $$\implies |z(1+z)^{-1}|=|z|\cdot |(1+z)^{-1}|=$$ $$=|z|\cdot |1+z|^{-1}\leq |z|\cdot 3/2\leq (1/3)(3/2)=1/2.$$

$$\text {Therefore }\quad0<|z|<1/3\implies |f(0)-f(z)|\geq |f(0)|-|f(z)|=$$ $$=1-|f(z)|\geq 1-1/2=1/2.$$

So when $\epsilon=1/2$ there does NOT exist any $\delta >0$ such that $|z|<\delta \implies |f(0)-f(z)|<\epsilon.$

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By a simple change of variables $w := z+1$ we have $f(w) = 2(w-1)/w = 2 - 2/w$, so your function has no hope of being continuous at this point - it has a pole there.

To finish your argument, make the same change of variables, and then note that as $w \to 0$, you can make paths that blow up any way you want in the complex plane, so in particular the modulus blows up.

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    Not that it makes a difference to your argument, but it should be that $f(w) = 2 - 2/w$ – 2017-02-12