# Is this function complex differentiable?

I was solving a problem yesterday and it has bugged me for the whole night, im not sure whether if I got it correct or not.

First I was asked if $f(x+iy)=x^2-y^2 + i\sqrt{|xy|}$ satisfies the C-R equations at $0$. So I found $\frac{\partial u}{\partial x}$ $=$ $\frac{\partial v}{\partial y}$ & $-\frac{\partial u}{\partial y}$ $=$ $\frac{\partial v}{\partial x}$

$u=x^2-y^2$ and $v=\sqrt{|xy|}$

So I found $\frac{\partial u}{\partial x} = 2x$, $-\frac{\partial u}{\partial y}=2y$, $\frac{\partial v}{\partial x}=\frac{1}{2}\frac{\sqrt{y}}{\sqrt{x}}$ and $\frac{\partial v}{\partial y}=\frac{1}{2}\frac{\sqrt{x}}{\sqrt{y}}$.

So obviously it does not satisfy the C-R equations.

1. I was wondering if I have to do anything else because it asks if it satisfies the C-R equations at $0$?

2. The next part of the question asks me if $f$ is differentiable at $0$. And it hints that I should consider $\mathrm{lim}_{r\rightarrow 0} \frac{f(re^{i\theta})}{re^{i\theta}}$. I assume they are meaning $f$ is complex differentiable? (not real differentiable?). How would I determine if f is differentiable at $0$?

Because it says on wikipedia that the sole existence of partial derivatives satisfying the Cauchy-Riemann equations is not enough to ensure complex differentiability at that point. It is necessary to make sure that u and v are real differentiable, which is a stronger condition than the existence of the partial derivatives but it is not necessary to require continuity of these partial derivatives. Then f = u + iv is complex-differentiable at that point if and only if the partial derivatives of u and v satisfy the Cauchy–Riemann equations at that point.

• The question I have is, since $f$ does not satisfy the C-R equations (see my calculations above), is there any need to do anything further (in question 2)? Can I just say its not differentiable at $0$? Why did they ask me to consider that limit?

Thanks alot, Im really stuck and slightly confused with all this..

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Have another look at h(x) = |x| and compute h'. What happens at x= 0? – Tim van Beek Nov 10 '11 at 14:40
First you should ask whether the function is even real differentiable. What is its directional derivative along the line $x = y$? – Willie Wong Nov 10 '11 at 15:07
Since $x^2-y^2$ is the real part of $z^2$, which is entire, thefunction cannot be analytic at (0,0) unless it agrees (in a 'hood of (0,0)) with $i2xy$, the imaginary part of $z^2$ – gary Nov 11 '11 at 0:29

$$f\;'(0) = \lim_{z\to0}\frac{f(z)-f(0)}{z-0} = \lim_{z\to0}\frac{f(z)}{z}.\tag{1}$$ In order that this limit exist, it is necessary that it be equal to the limit as $z$ approaches $0$ along any path that passes through $0$. Thus it must be the same as $$\lim_{r\to0} \frac{f(re^{i\theta})}{re^{i\theta}},$$ which is the limit in (1) as $z$ approaches $0$ along a certain line. The angle between that line and the $x$-axis is $\theta$. Since the function is expressed in terms of real and imaginary parts, write $x+iy=re^{i\theta}=r(\cos\theta+i\sin\theta)$. Then $$f(re^{i\theta})=f(x+iy)= x^2-y^2+i\sqrt{|xy|}=r^2(\cos^2\theta-\sin^2\theta) + i|r|\sqrt{|\cos\theta\sin\theta|}.$$ So $$\frac{f(re^{i\theta})}{re^{i\theta}} = \frac{r(\cos^2\theta-\sin^2\theta) + i\operatorname{sign}(r)\sqrt{|\cos\theta\sin\theta|}}{e^{i\theta}}$$ where $\operatorname{sign}(r)= \pm1$ according as $r$ is positive or negative. So try taking the limit of that as $r\to0$.

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Typos. I've fixed them. I copied and pasted something in which there was a typo, so it occurred twice. – Michael Hardy Nov 11 '11 at 0:17
$x=r\cos\theta$ and $y = r\sin\theta$, so $\sqrt{|r\cos\theta\sin\theta|}= \sqrt{r^2}\sqrt{|\cos\theta\sin\theta|}$. And $\sqrt{r^2}=|r|$. – Michael Hardy Nov 11 '11 at 0:18
You had $|r|$ in the numerator and $r$ in the denominator, and canceled, so that you had $\operatorname{sign}(r)=\pm1$ in the numerator and $1$ in the denominator. I.e., if $r>0$ then $|r|/r = 1$ and if $r<0$ then $|r|/r = -1$. So you have the two one-sided limits: $\lim_{r\to0+}i\sqrt{|\cos\theta\sin\theta|}/e^{i\theta}$ and $\lim_{r\to0-}(-i\sqrt{|\cos\theta\sin\theta|}/e^{i\theta})$. They are unequal except when $\cos\theta\sin\theta=0$. So the limit you seek does not exist except when $\cos\theta\sin\theta=0$, i.e. when $z$ is on the real or imaginary axis. – Michael Hardy Nov 11 '11 at 4:18
Also: Note that when you take $\lim_{r\to\text{something}} A$, if $A$ does not depend on $r$, then the limit is just $A$. – Michael Hardy Nov 11 '11 at 4:21

Since $x^2−y^2$ is the real part of $z^2$, which is entire, the function cannot be analytic at (0,0) unless it agrees (in a 'hood of (0,0)) with $i2xy$, the imaginary part of $z^2$, which it does not, e.g., check along $y=x$

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Thanks, so does that mean f is not differentiable at zero? – John Southall Nov 11 '11 at 0:47
fis not analytic at zero, and, since the partials don't exist there, it cannot be real-differentiable either. – gary Nov 11 '11 at 0:56
John: do you still want a comment? Show that for different directions of approach to (0,0) along lines not on the x-axis, you get different values for the limit, so that the limit does not exist. – gary Nov 11 '11 at 5:43
I wonder why the limit does not exist, could u pls make a comment @gary – Alana Apr 12 at 2:32