The task is as follows:
$$ \text{Check, for which } z\in\mathbb{C} \text{ function } f(z) = |z|^2 +2z \text{ satisfies the Cauchy-Riemann equations.} $$ I figured I need to transform $f(z)$ to $f(x,y) = u(x,y) + i \cdot v(x,y)$.
I started by transforming $f(z)$ to $f(x+iy)$ as such: $$ f(x+iy) = \sqrt{x^2+y^2} + 2(x + iy) = (\sqrt{x^2+y^2}+2x)+(2y)\cdot i $$ Therefore: $$ \begin{cases} u(x,y)=\sqrt{x^2+y^2} + 2x\\ v(x,y) = 2y \end{cases} $$
Unfortunately, when checking I started to check the Cauchy-Riemann equations I got the solution "$f(z)$ does not satisfy the Cauchy-Riemann equations for any $z\in \mathbb{C}$" which is wrong.
Here's how I got this erroneous solution: $$ \frac{\partial u}{\partial x} = \frac{-x}{\sqrt{x^2+y^2}}+2 \\ \frac{\partial v}{\partial y} = 2 \\ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \Leftrightarrow x=0 \wedge y\neq0 $$ $$$$ $$ \frac{\partial u}{\partial y} = \frac{-y}{\sqrt{x^2+y^2}} \\ -\frac{\partial v}{\partial x} = 0 \\ \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} \Leftrightarrow y=0 \wedge x\neq0 $$
And just like that I got the wrong answer. Where did I go wrong? How do I do this task correctly?
You calculated $|z|^2$ wrong. It should be $x^2+y^2$.
Since $h(z)=2z$ is analytic, you can focus on $g(z)=|z|^2=x^2+y^2$ where $z=x+iy$.
Now $u(x,y)=x^2+y^2$ and $v(x,y)=0$ for $g=u+iv$. You can go on and check the C-R equations.