Is $w=|z|\overline{z}$ differentiable and holomorphic?

Question

Making use of the Cauchy-Riemann equations verify if the following function is differentiable at least in one point and if it is holomorphic $w=|z|\overline{z}$

I tried to solve the problem in the following way:

$|z|\overline{z}=r(r\cos(\phi)-i\sin(\phi))=r^2\cos(\phi)-r^2i\sin(\phi)$ where $r\in\mathbb{R}^+\:,\:\phi\in[-\pi,\pi]$

Taking the $u(r,\phi)=r^2\cos(\phi)$ and $v(r,\phi)=r^2\sin(\phi)$

Computing $\frac{\partial u(r,\phi)}{\partial r}=2r\cos(\phi)\\\frac{\partial v(r,\phi)}{\partial \phi}=r^2\cos(\phi)$

So if we take the equation: $\frac{\partial u(r,\phi)}{\partial r}=\frac{\partial v(r,\phi)}{\partial \phi}\implies r=2$

Now computing: $\frac{\partial u(r,\phi)}{\partial \phi}=-r^2\sin(\phi)\\\frac{\partial v(r,\phi)}{\partial \phi}=2r\sin(\phi)$

Taking the equality: $\frac{\partial u(r,\phi)}{\partial \phi}=-\frac{\partial v(r,\phi)}{\partial r}\implies r=2$

So the equation is differentiable on the circumference $|z|=2$

I guess the function is not holomorphic because it is not differentiable in a neighbourhood of any point in the circumference.

Questions:

Is this right? If not. How should I solve it?

Thanks in advance!

Answer 1

The Cauchy-Riemann equations in polar form are different from their Cartesian counterparts. You can check, for example,

http://math.furman.edu/~dcs/courses/math39/lectures/lecture-14.pdf

Actually, all you need is $\partial w/\partial\bar z=0$, and in your case this condition gives $|z|=0$, that is, your function is differentiable only at the origin.

Of course you can also use the CR equations in polar form and you get $r=0$.

Answer 2

That is not correct. If you want to know about the polar form of the Cauchy-Riemann equations, take a look at this post.

It is easier to study the square of that function. So, let $f(z)=\lvert z\rvert^2\overline z^2$. Then, if $x,y\in\mathbb R$,$$f(x+yi)=(x^2+y^2)(x^2+y^2-2xyi)=(x^2+y^2)^2-2xy(x^2+y^2)i.$$Now, if you solve the Cauchy-Rimeann equations$$\left\{\begin{array}{l}4x(x^2+y^2)=-2x(x^2+3y^2)\\4y(x^2+y^2)=2y(3x^2+y^2),\end{array}\right.$$you'll get a single solution: $(x,y)=(0,0)$. So, $f$ is differentiable at $0$ and only at $0$.

So, $0$ is the only point at which your function can be differentiable. And it actually is! In order to see why, consider the limit$$\lim_{z\to0}\frac{\lvert z\rvert\overline z}z.\tag1$$Since $\left\lvert\frac{\overline z}z\right\rvert=1$, and since $\lim_{z\to0}\lvert z\rvert=0$, the limit $(1)$ is equal to $0$.