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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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    BTW, your edited approach is incorrect: just because a function has partial derivatives in two distinct directions does **not** mean it is (real) differentiable.2012-03-12

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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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    ??? $ $ $ $ $ $2012-03-02