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.

  • $\begingroup$ 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

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.


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