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I'm moving my first steps in Complex Analysis. I can't tell whether the function $$f(x+iy) = 2|xy|+i(y^2-x^2)$$ is differentiable at $0$ or not.

I tried using the limit of the difference quotient: $$\lim_{z\to 0}\frac{2|xy|+i(y^2-x^2)}{x+iy}$$

The limit is $0$ when $xy>0$ because $x=-iy$ is a root of the numerator, but I can't compute the limit in general...

Kiuhnm
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  • Try using the Cauchy-Riemann Equations – Sheel Stueber Aug 30 '18 at 16:49
  • I'm following a course which says it is, so I was confused. Is it enough to look at the CR equations or do we need a counterexample for the DQ limit? – Kiuhnm Aug 30 '18 at 16:49
  • @SheelStueber $u_x$ doesn't exist (at $0$), but the official solution says one needs to look at the DQ limit... – Kiuhnm Aug 30 '18 at 16:50
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    The function needs to be differentiable in the real sense if you want to use the CR equations. – MSDG Aug 30 '18 at 16:51
  • The CR equations are equivalent to the DQ limit existing – Sheel Stueber Aug 30 '18 at 16:51
  • @SheelStueber The derivative does exist at $z=0$. – Mark Viola Aug 30 '18 at 17:04
  • @SheelStueber No, it's not quite that simple. If the function has continuous partial derivatives, then the C-R equations are equivalent to the existence of the limit of the difference quotient. – David C. Ullrich Aug 30 '18 at 17:32
  • @SheelStueber So in this example we cannot use the CR equations to show $f$ is differentiable at the origin, even though $u_x=u_y=v_x=v_y=0$ at the origin. Because the partials are not even defined in any nieghborhood of the origin, so they're certainly not continuous at the origin. – David C. Ullrich Aug 30 '18 at 17:49
  • @Kiuhnm Actually $u_x$ does exist at $0$, in fact $u_x(0)=0$. There is a more subtle reason we cannot use the C-R equations here - see my other comment and my answer. – David C. Ullrich Aug 30 '18 at 18:02
  • Ah you're correct @DavidC.Ullrich my mistake – Sheel Stueber Aug 30 '18 at 19:36

2 Answers2

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A less miraculous argument somewhat like the other answer: If $z=x+iy$ then it's clear that $|x|\le|z|$ and $|y|\le|z|$, hence $$|f(z)|\le 2|z|^2+|z|^2+|z|^2.$$(So $$\left|\frac{f(z)-f(0)}z\right|\le\frac{4|z|^2}{|z|}=4|z|\to0\quad(z\to0).)$$

This seems preferable because for example it also applies to show the function $g(x+iy)=3|x||y|+x^2$ is differentiable at the origin, without the miraculous simplification.

Regarding the C-R equations, and some erroneous things that have been said in the comments: For this function we have $u_x=u_y=v_x=v_y=0$ at the origin, but the partials are not even defined in a neighborhood of the origin, so they are certainly not continuous at the origin, hence we cannot use the C-R equations to show $f$ is differentiable at the origin.

The theorem is this:

If $u_x,u_y,v_x$ and $v_y$ are continuous at $z$ then the C-R equations at $z$ are equivalent to differentiability at $z$.

Example showing that just knowing the C-R equations at a point does not imply differentiability at that point: Define $$f(x+iy)=\begin{cases}1,&(xy=0), \\0,&(xy\ne0).\end{cases}$$

Then at the origin all four partials exist, and in fact $u_x=u_y=v_x=v_y=0$ at the origin, so the C-R equations are satisfied at the origin. But $f$ is not even continuous at the origin, so it's certainly not differentiable at the origin.

Ok, there is a "pointwise" version of the theorem that could be applied to the example in the question:

Suppose $f:\Bbb C\to\Bbb C$ is Frechet differentiable at $z$ (when regarded as a map from $\Bbb R^2$ to $\Bbb R^2$) and satisfies the C-R equations at $z$. Then $f$ is complex differentiable at $z$.

The function in the question is Frechet differentiable at the origin. But this isn't really of any use, because verifying it's Frechet differentiable at $0$ is almost exactly the same as verifying directly that it's complex-differentiable at $0$.

David C. Ullrich
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HINT:

$$\left|\frac{2|x||y|+i(y^2-x^2)}{x+iy}\right|=\sqrt{\frac{4x^2y^2+(y^2-x^2)^2}{x^2+y^2}}=\sqrt{x^2+y^2}$$

Can you prove that the limit exists?

Mark Viola
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