# Check if the complex function is differentiable

1. Jan 26, 2014

### Fabio010

The question is to check where the following complex function is differentiable.

$$w=z \left| z\right|$$

$$w=\sqrt{x^2+y^2} (x+i y)$$

$$u = x\sqrt{x^2+y^2}$$
$$v = y\sqrt{x^2+y^2}$$
Using the Cauchy Riemann equations

$$\frac{\partial }{\partial x}u=\frac{\partial }{\partial y}v$$
$$\frac{\partial }{\partial y}u=-\frac{\partial }{\partial x}v$$

my results:

$$\frac{x^2}{\sqrt{x^2+y^2}}=\frac{y^2}{\sqrt{x^2+y^2}}$$
$$\frac{x y}{\sqrt{x^2+y^2}}=0$$

solutions says that it's differentiable at (0,0). But doesn't it blow at (0,0)?

2. Jan 26, 2014

### DrewD

If you just plug in $y=0$ and $x=0$ you will get an indeterminate form which is meaningless. If you evaluate the limits, I think that you get all expressions equal to $0$, but double check that.

3. Jan 26, 2014

### FactChecker

Division by zero is not allowed in complex analysis, so your final equations are not defined at x=y=0. They are not equal.

4. Jan 27, 2014

### DrewD

That is true, but this function is differentiable at $z=0$. If you evaluate the two limits along the real and imaginary axes (with $h\in\mathbb{R}$)

$\displaystyle\lim_{h\rightarrow0}\displaystyle\frac{(0+0i+h)\left|(0+0i+h)\right|}{h}=0$

$\displaystyle\lim_{h\rightarrow0}\displaystyle\frac{(0+0i+ih)\left|(0+0i+ih)\right|}{ih}=0$

So the function is differentiable at $0$. I don't remember enough from my complex analysis course (which had a number of students who had not taken real analysis, so it was a bit less rigorous than some courses) to reconcile this. My recollection is that the limits of the Cauchy-Riemann equations could be evaluated, but a quick look online showed that my recollection was incorrect. Perhaps, since the partial derivatives are undefined at 0 the Cauchy-Riemann equations are not applicable?

5. Jan 27, 2014

### Dick

I would use the definition of the derivative as a difference quotient to show it's differentiable at z=0.

6. Jan 27, 2014

### DrewD

Yes, but $w(0)=0$, so I left it out.

7. Jan 27, 2014

### Dick

Sure, I'm just saying there is no need work along any particular axes. The complex derivative f'(0) is defined by the limit |h|->0 of (f(0+h)-f(0))/h where h is complex. You can also conclude that that is zero.