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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 $z=0$.

gbd
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  • Well, there are two tasks here, and each is fairly mechanical. Have you successfully completed either one? Where are you getting stuck? – Nick Peterson Jan 06 '17 at 16:35
  • @NickPeterson, I have updated my question. – gbd Jan 06 '17 at 16:53
  • Have you tried to write down the definition of $u_x,v_y$ at $(0,0)$? –  Jan 06 '17 at 17:00
  • @Jack, They are not equal because $v_{y}=0$ but $u_{x}$ seems to be infinite. $\frac{|0|}{2\sqrt{|0\times 0|}}=\infty$ – gbd Jan 06 '17 at 17:02

1 Answers1

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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.

  • You only want to analyze the limit when $x=y=0$, not the general case. –  Jan 06 '17 at 17:10
  • 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. –  Jan 06 '17 at 17:17
  • 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. –  Jan 06 '17 at 17:19
  • "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. –  Jan 06 '17 at 17:30
  • 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. – gbd Jan 06 '17 at 17:42
  • @Jack doesn't $\frac{|x|}{x} = {\pm1}$ since $x \in \mathbb{R}$? It doesn't change the answer, but the limit does exist. It just isn't equal to $0$.? – Peter_Pan Sep 21 '19 at 22:57
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    @Jess: No. $\lim_{x\to0}\frac{|x|}{x}$ does not exist since $\lim_{x\to0-}\frac{|x|}{x}\neq \lim_{x\to0+}\frac{|x|}{x}$. –  Sep 22 '19 at 03:20