How to find where the complex function $f(z)=|z| \bar z$ is differentiable?

Question

I have tried using Cauchy-Riemann and I dont think it is differentiable anywhere but my solution book says it is on $0$ can someone explain to me why? Thanks for the time.

My calculations so far:

$f(x+iy)=x*\sqrt{x^2+y^2} -i(y*\sqrt{x^2+y^2})$

$u(x,y) = x*\sqrt{x^2+y^2}$

$v(x,y) = -y*\sqrt{x^2+y^2}$

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

$\frac{du}{dy}=\frac{xy}{\sqrt{x^2+y^2}}$

$\frac{dv}{dx}=\frac{-xy}{\sqrt{x^2+y^2}}$

$\frac{dv}{dy}=\frac{-2y^2-x^2}{\sqrt{x^2+y^2}}$

This means that $\frac{du}{dx}=\frac{dv}{dy}$ can never happen and therefore the function is not differentiable in any point rigth?

Answer 1

You have $f(z) = |z|\overline{z}$ and so |${f(z)-f(0) \over z-0}| = |z|$, from which it follows that $f'(0) = 0$.

However, to be analytic at $z=0$, the derivative needs to exist in a neighbourhood of $z=0$, which it doesn't in this case.

Aside: $|u(x,y)| = |x| \sqrt{x^2+y^2} \le \sqrt{x^2+y^2} \sqrt{x^2+y^2} = x^2 + y^2$. Hence $\lim_{(x,y) \to (0,0)} |{ u(x,y) - u(0,0) \over \sqrt{x^2+y^2} } | = 0$. Similarly for $v$.

It would be better to write ${\partial u(x,y) \over \partial x} = \begin{cases} {2 x^2 + y^2 \over \sqrt{x^2+y^2} }, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases}$, etc, etc.

Answer 2

$$f(x+iy)=\sqrt{x^2+y^2}(x-iy)\\ u(x,y)=x\sqrt{x^2+y^2} \\ v(x,y)=-y\sqrt{x^2+y^2}$$

Now, check the CR equations, and beware of the fact that the formulas you get for $u_x,u_y,v_x, v_y$ have a $\sqrt{x^2+y^2}$ in the denominator, thus they don't work at $z=0$. At $z_0$ you can calculate the partial derivatives with the definition, or you can check directly that $f$ is differentiable.