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Is the function $f$ given by

$$f(z) =\left\{ \begin{array}{ll} \frac{(\bar{z})^2}{z}, & \hbox{if }z\neq 0; \\ 0, & \hbox{if }z=0. \end{array} \right.$$ differrentiable at $z=0$?


I start by taking the $$\lim\limits_{z\to 0} \frac{\frac{(\bar{z})^2}{z}-0}{z-0}=\lim\limits_{z\to 0} \frac{(\bar{z})^2}{z^2}$$

Choosing $z=x$, $\bar z=x$, thus $\lim\limits_{x\to 0}\frac{x^2}{x^2}=1$. Also by choosing $z=iy, \bar{z}=-iy$, $\lim\limits_{y\to 0} \frac{-y^2}{-y^2}=1$.

Hence, from the above the function $f$ is differentiable at $z = 0$.

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Looks pretty continuous to me, which can be seen by noting that $|f(z)|=|z|$. However, hint: Cauchy-Riemann equations. –  Henning Makholm Mar 1 '12 at 21:00
    
I hereby retract the Cauchy-Riemann hint -- Didier's argument showed me I had misplaced a sign along the way. –  Henning Makholm Mar 1 '12 at 22:16
    
@Hassan To make a fancy-looking $\lim\limits_{x\to 0}$ use \lim\limits_{x\to 0} or \lim_{x \to 0} for $\lim_{x \to 0}$ –  Pedro Tamaroff Mar 7 '12 at 3:32
    
@PeterT.off: Thanks for the edit. Is my approach correct? –  Hassan Muhammad Mar 7 '12 at 3:48
    
You should be careful to state whether you are considering complex differentiability or real differentiability. –  Willie Wong Mar 9 '12 at 11:59

1 Answer 1

up vote 2 down vote accepted

The function $f$ is $C^\infty$ everywhere except at $0$ and continuous at $0$ since, for example, $|f(x+\mathrm iy)|=\sqrt{x^2+y^2}\to0$ when $x+\mathrm iy\to0$, and $f(0)=0$.

But $f:\mathbb R^2\to\mathbb R^2$ is not differentiable at $0$ because, for $h\to0$, $h$ real, $f(h)=h+o(h)$ and $f(\mathrm ih)=\mathrm ih+o(h)$ but $f((1+\mathrm i)h)=-(1+\mathrm i)h+o(h)\ne (1+\mathrm i)h+o(h)$.

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No (lack of) holomorphy here, but a lack of differential at $(0,0)$. –  Did Mar 1 '12 at 21:08
    
What can you say about the differentiability of $f$ if $f:I\to \mathbb C$ where $I\in \mathbb C$? –  Hassan Muhammad Mar 2 '12 at 5:26
2  
??? $ $ $ $ $ $ –  Did Mar 2 '12 at 6:02

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