1
$\begingroup$

How to show that $f(z)=\sqrt{|xy|}$ satisfies the Cauchy Riemann equations but isn't differentiable at $z=0$?

My Attempt

$$ f(z)=u+i v,u=\sqrt{|xy|},v=0 $$ $$ v_{x}=0,v_{y}=0,u_{x}=\frac{|y|}{2\sqrt{|x y|}},u_{y}=\frac{|x|}{2\sqrt{|x y|}} $$ $ \lim_{\Delta z\rightarrow 0}\frac{\sqrt{|(x+\Delta x)(y+\Delta y)|}-\sqrt{|(x)(y)|}}{\Delta x+i\Delta y} $

I am not sure how the limit doesn't exist. Also, $u_x,u_y$ seem to become infinite at $z=0$.

  • 0
    Well, there are two tasks here, and each is fairly mechanical. Have you successfully completed either one? Where are you getting stuck? – 2017-01-06
  • 0
    @NickPeterson, I have updated my question. – 2017-01-06
  • 0
    Have you tried to write down the *definition* of $u_x,v_y$ at $(0,0)$? – 2017-01-06
  • 0
    @Jack, They are not equal because $v_{y}=0$ but $u_{x}$ seems to be infinite. $\frac{|0|}{2\sqrt{|0\times 0|}}=\infty$ – 2017-01-06

1 Answers 1

2

You need to calculate by definition the partial derivatives of $u$ and $v$ at $(0,0)$. Note that $$ u_x(0,0)=\lim_{h\to 0}\frac{u(h,0)-u(0,0)}{h}=0. $$ Similarly, you can calculate $v_y,v_x,u_y$ at $(0,0)$ so that you can show the Cauchy-Riemann equation part of the statement.

For differentiability, you want to show that $ \displaystyle\lim_{z\to 0}\dfrac{f(z)}{z} $ does not exist. But direct calculation shows that the limit does not even exist along the long the line $y=x$ since $$ \frac{\sqrt{|x\cdot x|}}{x+ix}=\frac{|x|}{x}\frac{1}{1+i} $$ and the limit $$ \lim_{x\to 0}\frac{|x|}{x}\frac{1}{1+i} $$ does not exist.

  • 0
    You only want to analyze the limit when $x=y=0$, not the general case. – 2017-01-06
  • 0
    It is not that "it has two limits" but that the limit "does not exist". You might want to be careful about the writing. Also, the first equality is incorrect. Please see my edited answer. – 2017-01-06
  • 0
    P.S. A classical textbook example for this sort of exercises is to show that $f(z)=\overline{z}$ is not differentiable at $z=0$. You might want to check that. – 2017-01-06
  • 0
    "its limit is not unique" does not make sense at all because the limit does not exist in the first place. One would never ever say that. – 2017-01-06
  • 0
    Is it correct to set $\Delta y=m\Delta x$ then $ \lim_{\Delta z\rightarrow 0}\frac{\sqrt{|\Delta x\Delta y|}}{\Delta x+i\Delta y}=\lim_{\Delta x\rightarrow 0}\frac{\sqrt{|m\Delta x^2|}}{\Delta x+i m\Delta x}=\lim_{\Delta x\rightarrow 0}\frac{\sqrt{|m|}}{1+i m}=\frac{\sqrt{|m|}}{1+i m} $ So $f^{\prime}(0)$ has different value for each $m$ which means $f^{\prime}(0)$ is not defined. – 2017-01-06