$re^{i\omega} \rightarrow re^{2i\phi}$ not holomorphic over $\mathbb{C} \backslash \{0\}$

Question

Try to solve this problem:

Show that function $ f: re^{i\phi} \rightarrow re^{2i\phi}$ not holomorphic over $\mathbb{C} \backslash \{0\}$

My solution: We have $$f(z) =f(x+yi) = f(re^{i\phi}) = re^{2i\phi} = rcos(2\phi) + i sin(2\phi)$$ and $u(x,y) = rcos(2\phi), v(x,y) = sin(2\phi)$ After using Cauchy–Riemann equation we have.

I'm having a hitch here because I don't know what to do next.

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

UPD after comment

But! if i have Q-R equations in polar form: $$ \left( \frac{\partial u}{\partial r}\right) = \frac{1}{r} \left( \frac{\partial v}{\partial \theta}\right) \ \ \ \ \ \text{and} \ \ \ \ \left(\frac{\partial v}{\partial r} \right) = \frac{-1}{r} \left( \frac{\partial u}{\partial \theta}\right) $$ We have: $u(r,\phi) = rcos(2\phi), v(r,\phi) = sin(2\phi)$ and: $$ \left( \frac{\partial u}{\partial r}\right) = cos(2\phi) \ \ 2cos(2\phi)= \frac{1}{r} \left( \frac{\partial v}{\partial \theta}\right) \ \ \text{and} \ \ \left(\frac{\partial v}{\partial r} \right) = sin(2\phi) \ -2sin(2\phi) = \frac{-1}{r} \left( \frac{\partial u}{\partial \theta}\right) $$ we have system:

$$ $$ \begin{cases} cos(2\phi) = 2cos(2\phi) \\ sin(2\phi) = -2sin(2\phi) \end{cases} $$ $$

Answer 1

Observe that $f(z)=\frac {z^{2}} {|z|}$ for all $z \neq 0$. Let us prove by contradiction that this is not holomorphic in $\mathbb C \setminus \{0\}$. If $f$ is holomorphic then so is $\frac 1 {|z|}$ because $\frac 1 {|z|}$ is the product of $f$ and the holomorphic function $\frac 1 {z^{2}}$. But now $|z|$ itself is holomorphic since it is the reciprocal of a holomorphic function with no zeros. Can you finish the proof by showing that $|z|$ is not holomorphic?

No real valued function other than a constant is holomorphic in any domain.