0
$\begingroup$

I must show that the Cauchy-Riemann equations are satisfied for $f(z) = \sqrt{xy}$, that $f$ is not differentiable at $z=0$.

How can a complex function be holomorphic / complex-analytic, i.e., $f$ satisfies the C-R equations, but not differentiable at a point $z_0$? I thought holomorphic / complex-analytic meant that $f$ was differentiable everywhere.

$\endgroup$
  • $\begingroup$ Cauchy-Riemann equations are only satisfied at $z=0$. Everywhere else, they fail. So the function is not holomorphic anywhere. So you just have to show it's not diff'ble at the origin. $\endgroup$ – Paul Hurst Mar 8 '15 at 7:21
  • $\begingroup$ Differentiable everywhere would imply it is an entire function. $\endgroup$ – Jamil_V Mar 8 '15 at 7:21
  • $\begingroup$ Holomorphic at a point means it is differentiable in a neighborhood of that point. Not just diff'ble at the point. $\endgroup$ – Paul Hurst Mar 8 '15 at 7:23
  • $\begingroup$ I thought that if a complex function $f$ was holomorphic, it was differentiable everywhere. How can $f = \sqrt{xy}$ satisfy the C-R equations yet...not actually be holomorphic? $\endgroup$ – holoNo Mar 8 '15 at 7:27
  • 1
    $\begingroup$ @holoNo For me that's (i.e., yours) the actual, more-usual, definition. $\endgroup$ – Timbuc Mar 8 '15 at 7:31
1
$\begingroup$

To show C-R equations are satisfied at $z=0$, use $u = \sqrt{xy}$ and $v = 0$ and take the appropriate partials. You should be able to see that they ONLY hold at $z=0$- nowhere else. This means that $f$ NOT differentiable when $z \neq 0$.

You can show it's not differentiable at 0 from the definition. Differentiable means that $$\lim_{h \rightarrow 0} \frac{f(z+h) - f(z)}{h} = 0$$

So for $z=0$, you're looking at $$\lim_{h \rightarrow 0} \frac{f(0+h) - f(0)}{h} = 0$$

Use $h = x+iy$, so $f(0+h) = \sqrt{xy}$, $f(0) = 0$, and you're looking at $$\lim_{(x,y) \rightarrow (0,0)} \frac{\sqrt{xy}}{x+iy} = 0$$

Take this limit along two different paths to show it does NOT exist. Along $x=0$ and along $y=x$ should do it, for example.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.