1. ## Complex Variables

Let $f (z)$ be an entire function such that for all x, y,
$|f (x +iy)| \neq 0$ and $arg(f (x + iy)) = xy$. Find f .

First I thought that working using Euler's notation would make the thing simpler. Let $f (r,\theta)=\rho e^{i\phi}$
The condition are thus $|f(r,\theta)|=\rho \neq 0$
and $arg(f (x + iy)) = \phi = r^2cos(\theta)sin(\theta)$.
Then $f(r,\theta) = \rho e^{ir^2cos(\theta)sin(\theta)}$ where
$\rho \neq 0$. Since this is a composition of entire function it should be an entire function.
I'm not comfortable with complex variables so I ask for some help. The question suggest there might be a single $f (z)$ but I [think I] found many. Does what is above makes any sense?

2. Originally Posted by vincisonfire
Let $f (z)$ be an entire function such that for all x, y,
$|f (x +iy)| \neq 0$ and $arg(f (x + iy)) = xy$. Find f .

First I thought that working using Euler's notation would make the thing simpler. Let $f (r,\theta)=\rho e^{i\phi}$
The condition are thus $|f(r,\theta)|=\rho \neq 0$
and $arg(f (x + iy)) = \phi = r^2cos(\theta)sin(\theta)$.
Then $f(r,\theta) = \rho e^{ir^2cos(\theta)sin(\theta)}$ where
$\rho \neq 0$. Since this is a composition of entire function it should be an entire function.
I'm not comfortable with complex variables so I ask for some help. The question suggest there might be a single $f (z)$ but I [think I] found many. Does what is above makes any sense?

I think you are correct. It says for all $x,y$, $|f(x+iy)| \neq 0$ and $\text{arg}(f(x+iy)) = xy$. So the argument is not constant. Nor is the modulus. Also polynomials of the form $f(x+yi)$ are analytic/entire.