Find all entire holomorphic on $ \mathbb{C} $ which satisfies the following condition

Question

I want to find all entire holomorphic on $ \mathbb{C} $ functions $ f=u+iv $ such that $u(z) =|z|^2 $.

Here's what I have tried;

Assume $ f\left(z\right)=|z|^{2}+iv\left(z\right)=\left(x^{2}+y^{2}\right)+iv\left(x,y\right) $. I'll try to identify $ f $.

Since $ f $ is holomorpgic, Cauchy-Riemman equation should hold. So:

$$ \begin{cases} u_{x}=v_{y}\\ u_{y}=-v_{x} \end{cases}\Rightarrow\begin{cases} 2x=v_{y}\\ -2y=v_{x} \end{cases} $$

Thus integrate both sides of the equation we get $$ \begin{cases} 2xy+C_{1}=v\left(x,y\right)\\ -2xy+C_{2}=v\left(x,y\right) \end{cases} $$

And thus if we'll sum the equations and divide by $ 2 $ we get $ v\left(x,y\right)=\frac{C_{1}+C_{2}}{2} $. That is, $v $ is a constant function, so

$$ \left\{ f:f=|z|^{2}+iv,\thinspace\thinspace f\thinspace\thinspace is\thinspace\thinspace\thinspace entire\right\} =\left\{ f:f=|z|^{2}+ic,\thinspace\thinspace\thinspace where\thinspace\thinspace\thinspace c\in\mathbb{R}\right\} $$

I kind of treated the functions as one variable functions when I integrated in order to find $ v(x,y) $ so im not sure with my result.

My questions are:

1. Does my way correct?
2. Is there a simpler way?

Thank in advance, it is highly appreciated.

Answer 1

Actually, there is no such function. The function $(x,y)\mapsto x^2+y^2$ is not harmonic, but the real and the imaginary parts of a holomorphic functions are always harmonic functions.

Concerning your approach, from $v_y=2x$ and $v_x=-2y$, you should deduce that $v=2xy+C_1(x)$ and $v=-2xy+C_2(y)$, for some functions $C_1$ and $C_2$. But$$2xy+C_1(x)=-2xy+C_2(y)\iff4xy=C_2(y)-C_1(x).$$And there are no two such functions $C_1$ and $C_2$.