3
$\begingroup$

If $f(z)=\sqrt{|xy|}$, how do I find the partial derivatives of $u$ and $v$? How can we show it is not analytic at $(0,0)$?

$\endgroup$
8
  • 2
    $\begingroup$ $\sqrt{\lvert xy\rvert}$ isn't even real-differentiable in $0$. $\endgroup$ Feb 16, 2014 at 14:49
  • $\begingroup$ @Daniel, can you help me prove that mathematically? $\endgroup$
    – user128964
    Feb 16, 2014 at 14:51
  • 2
    $\begingroup$ The partial derivatives are both $0$. But on the line $x = y$, you have a kink, like the real absolute value function. $\endgroup$ Feb 16, 2014 at 14:55
  • 1
    $\begingroup$ @JonathanY. Complex differentiability is real differentiability + CR. Partial differentiability + CR, as is the case here, is not enough: $$\frac{f(t(1+i))}{t(1+i)} = \frac{\sqrt{2t^2}}{t(1+i)} = \pm \frac{1-i}{\sqrt{2}}\neq \lim_{t\searrow 0} \frac{f(t)}{t} = 0$$ for real $t$, so $f$ is not complex differentiable in $0$. $\endgroup$ Feb 16, 2014 at 16:02
  • $\begingroup$ @JonathanY. I didn't mean to belittle it. However, to show $f$ is not analytic at $0$, it is sufficient to show that it isn't even complex differentiable in $0$. And to show that, it is sufficient to show that it isn't even real differentiable in $0$. Each condition is weaker than the preceding, so if even the weaker condition isn't satisfied, the stronger is a fortiori not satisfied. With what word should that be expressed if "even" carries wrong connotations? $\endgroup$ Feb 16, 2014 at 16:32

1 Answer 1

6
$\begingroup$

In the real case $\lim_{h\to 0} \frac {|f(h,0)|}{|h|}=0=\frac {|f(0,h)|}{|h|}$ hence if it is differentiable at $0$ the derivative would be $0$ (Since if $f$ where differentiable the derivative would be the partial derivatives in a matrix) but $\lim_{(h,h)\to 0}\frac{|f(h,h)-f(0,0)-0|}{|(h,h)|}=\frac 1 {\sqrt{2}}$ which mean the limit $\lim_{(h,k)\to 0}\frac{|f(h,k)-f(0,0)-0|}{|(h,k)|}$ is not $0$ which shows $f$ is not real differentiable at $0$.

$\endgroup$
1
  • $\begingroup$ @user128964 if you like user10444's answer, consider accepting it. If you'd like to see something more, don't be shy about asking for it ;) $\endgroup$ Feb 16, 2014 at 16:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .