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I have to find where the Cauchy-Reimann equations $\left( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \text{ and } \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}\right)$ are satisfied for the function $$f(x+iy)= \sqrt{\frac{\sqrt{x^2+y^2}+x}{2}}+isign(y)\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}}.$$

I found that $$\frac{\partial u}{\partial x} =\frac{1}{2}\left(\frac{(x^2+y^2)^{1/2}+x}{2}\right)^{-1/2}\frac{1}{2}\left( \frac{1}{2}\left(x^2+y^2\right)^{-1/2}(2x)+1\right),$$

$$\frac{\partial v}{\partial y}= sign(y)\frac{1}{2}\left(\frac{(x^2+y^2)^{1/2}-x}{2}\right)^{-1/2}\frac{1}{2}\left( \frac{1}{2}\left(x^2+y^2\right)^{-1/2}(2y)\right),$$

$$\frac{\partial u}{\partial y} = \frac{1}{2}\left(\frac{(x^2+y^2)^{1/2}+x}{2}\right)^{-1/2}\frac{1}{2}\left( \frac{1}{2}\left(x^2+y^2\right)^{-1/2}(2y)\right),$$

and

$$\frac{\partial v}{\partial x} = sign(y)\frac{1}{2}\left(\frac{(x^2+y^2)^{1/2}-x}{2}\right)^{-1/2}\frac{1}{2}\left( \frac{1}{2}\left(x^2+y^2\right)^{-1/2}(2x)-1\right).$$

Of course, these can be simplified a bit (wolfram alpha will give something more concise), and when I put them in the Cauchy-Riemann equations, some things cancel, but I still am left with something too complicated (for me) to figure out any conditions on $x$ and $y$ that make these hold. I've also tried adding the equations, as well as setting $$\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} + \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} = 0,$$

but couldn't deduce anything from this either.

My instructor hinted to note that $\sqrt{x^2+y^2}\geq |x|,$ but I'm not seeing how to use this fact. This hint made me think that I should look at this function as $$f(z) = \sqrt{\frac{\sqrt{z\bar{z}}+RE(z)}{2}} + isign(IM(z))\sqrt{\frac{\sqrt{z\bar{z}}-RE(z)}{2}}.$$

But this doesn't seem to provide me with anymore insight. I was thinking of going this route because we proved in class that a function is analytic iff you can write the function in terms of only $z.$

I have a hunch that the hint provided will lead to the function not being analytic anywhere, but am not sure if this is true because I don't really have a good sense of what this function does or how it acts on the complex plane.

Thanks in advance for any help / hints you can provide.

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    $\begingroup$ Write $z=x+iy$. What is $(f(z))^2$? I think that could help a little. Also, I think that your instructor's hint was intended to stress that $f$ is given by a pair of real valued functions (because the root is well defined) $\endgroup$ – Brandon Mar 8 '14 at 6:26
  • $\begingroup$ Thanks Brandon. Looking at what you suggested, I found $$f(x+iy)^2 = x + isign(y)\left(\frac{y}{\sqrt{2}}\right).$$ Since this function is not analytic anywhere (because there are no values for which the C-R equations hold), then the original function must not be analytic, correct? $\endgroup$ – Jonathan Mar 8 '14 at 14:41
  • $\begingroup$ You have miscalculated the square, $$f(x+iy)^2 = x + 2i\operatorname{sign}y \sqrt{\frac{y^2}{4}} = x+iy.$$ $\endgroup$ – Daniel Fischer Mar 8 '14 at 14:47
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You can simplify the derivatives a bit more. It is maybe easier to see if we write $r = \sqrt{x^2+y^2}$ and note $\frac{\partial r}{\partial x} = \frac{x}{r},\; \frac{\partial r}{\partial y} = \frac{y}{r}$ when $r > 0$. Then for generic points ($x\neq 0 \neq y$) we find

$$\begin{align} \frac{\partial u}{\partial x} &= \frac{1}{2u}\cdot \frac{1}{2}\left(\frac{\partial r}{\partial x} + \frac{\partial x}{\partial x}\right) = \frac{1}{4u}\left(\frac{x}{r}+1\right) = \frac{x+r}{4r\sqrt{\frac{r+x}{2}}} = \frac{\sqrt{r+x}}{2\sqrt{2}\cdot r},\\ \frac{\partial v}{\partial y} &= \frac{1}{2v}\cdot \frac{1}{2}\left(\frac{\partial r}{\partial y} - \frac{\partial x}{\partial y}\right) = \frac{1}{4v}\frac{y}{r} = \frac{y}{4r\operatorname{sign} y\sqrt{\frac{r-x}{2}}} = \frac{\sqrt{r^2-x^2}}{2\sqrt{2}\cdot r\sqrt{r-x}}, \end{align}$$

from which it is not hard to see that $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$. The cases $x = 0$ or $y =0$, but $r > 0$ need to be handled separately (further distinguishing between positive and negative $x$ resp. $y$; noting also that $v$ isn't continuous everywhere).

The treatment of $\frac{\partial u}{\partial y}$ and $\frac{\partial v}{\partial x}$ is similar.

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