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The question suggests that I proceed by contradiction, and exploit the fact that $ |z| $ is nowhere differentiable (rather than using C-R equations).

I'm struggling to find a way to find a contradiction. Any help would be appreciated. Thanks

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Hint: Quotients, sums and products of holomorphic functions are holomorphic.

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    Well, I did consider that, if $ z \neq 0 $, $ \frac{1+|z|}{z} z - 1 $ must be holomorphic. Contradiction. But what if the point at which $ \frac{z}{1+|z|} $ is holomorphic is at $ z = 0 $?2011-04-16
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    I'm forgetting what being holomorphic at a point actually means. Am I right in saying that if $ \frac{z}{1+|z|} $ is holomorphic at a point, then it must be holomorphic at some non-zero point?2011-04-16
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    @Mathmo Well you'll have to treat the $z = 0$ case separately. But that's not too hard. Just use the definition of the derivative as a limit.2011-04-16
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    @Mathmo: surely (as Leibniz says) at $z=0$ we have $d(z/(1+|z|))=dz$, i.e. the function is "holomorphic" there. But the usual definition of holomorphicity is only for an open set, not for individual points (i.e. holomorphic at a point means that the function has a complex derivative in a neighbourhood of the point)2011-04-16