Is $ f(z)= |z|.\bar z $ analytic?

Question

I want to know whether it is analytic - and if so, to find $ f´(z)$

What I do:

I use the polar form.

$ z= x + iy \\ |z|=r \\ \bar z= (cos (\phi) + isin(\phi))$

then: $ f(z)= r^2(cos (\phi) + isin(\phi)) = r^2cos (\phi) + i r^2sin(\phi)$

Cauchy-Riemann:
$\begin{cases} \\ \frac{\partial u}{\partial r} \ = \frac{1}{ r}\frac{\partial v}{\partial \phi} \\[2ex] \frac{\partial u}{\partial r} \ = \frac{1}{ r}\frac{\partial v}{\partial \phi}\\ \end{cases} \quad \Rightarrow \quad \begin{cases}\\ 2r\cos \phi= -r\cos\phi \\[2ex] -2r\sin \phi=r\sin\phi\\ \end{cases} $

$\text{It is differentiable when $ \ 0\le\phi\le 2\pi \ , \ r=0 \ ?$ }$

$ \text{ Any help would be greatly appreciated. Thanks!}$

Answer 1

Put $z=r.e^{i\theta}$ for polar coordinates then $f(r.e^{i\theta})=r^2(cos\theta-isin\theta)$.

You can see that the real part $u(r,\theta)=r^2.cos\theta$ and imaginary part $v(r,\theta)=-r^2.sin\theta$ satisfy CR-equations

$ru_r=v_\theta, u_\theta=-rv_r$ iff

$3r^2.cos\theta=0$ and $3r^2.sin\theta=0$ i.e. $r=0$

Since CR-equations merely hold at the point $r=0$ (origin) and not in any open neighborhood of it hence $f$ is differentiable at $z=0$ but not analytic.

Answer 2

From the Cauchy-Riemann equations, if $r\neq 0$ then $$ \cos \phi =0 \wedge \sin \phi =0 $$ which is impossible. This excludes any $z\neq 0$.

For $z=0$, write

$$ \frac{f(0+h)-f(0)}{h}=|h|\frac{\bar{h}}{h} \to 0 \text{ when }h\to 0 $$ because $\frac{\bar h}{h}$ has modulus $1$.

Answer 3

Hint If $f(z)$ would be analytic on any disk, then so would be $$g(z)=zf(z)=|z|^3$$

Now, it is trivial to show that this function is nowhere analytic. This follows immediately from CR, or via many other simple methods.

Answer 4

Note that $f(x+iy)=x\sqrt{x^2+y^2}-i y\sqrt{x^2+y^2}=u+iv.$

So $u_x=\frac{2x^2+y^2}{\sqrt{x^2+y^2}}$ and $v_y=-\frac{2y^2+x^2}{\sqrt{x^2+y^2}}.$ From C-R equations: $u_x=v_y$ is possible only for $x=o, y=0$ and this implies f is nowhere analytic.