0
$\begingroup$

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 $z = 0$.

  • 0
    Looks pretty continuous to me, which can be seen by noting that $|f(z)|=|z|$. However, **hint**: Cauchy-Riemann equations.2012-03-01
  • 0
    I hereby retract the Cauchy-Riemann hint -- Didier's argument showed me I had misplaced a sign along the way.2012-03-01
  • 0
    @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}$2012-03-07
  • 0
    @PeterT.off: Thanks for the edit. Is my approach correct?2012-03-07
  • 0
    You should be careful to state whether you are considering [complex differentiability](http://mathworld.wolfram.com/ComplexDifferentiable.html) or real differentiability.2012-03-09
  • 0
    @WillieWong: What do you think I should do?2012-03-11
  • 0
    Edit the question to say whether by "differentiable" you mean it is "complex differentiable" (and hence solves the Cauchy-Riemann equations and is holomorphic), or you mean it is "real differentiable" as a function from the Euclidean plane to itself.2012-03-12
  • 0
    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

1 Answers 1

2

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 $f((1+\mathrm i)h)=-(1+\mathrm i)h+o(h)\ne (1+\mathrm i)h+o(h)$.

  • 0
    No (lack of) holomorphy here, but a lack of differential at $(0,0)$.2012-03-01
  • 0
    What can you say about the differentiability of $f$ if $f:I\to \mathbb C$ where $I\in \mathbb C$?2012-03-02
  • 2
    ??? $ $ $ $ $ $2012-03-02