Proof of the form of $f(z)$. $f$ is entire and also if $f = u + iv$, it has given the condition that $u_{x}v_{y} - u_{y}v_{x} = 1$ in $\mathbb{C}$ , Prove that $f$ has the form $az + b$ where $a$ and $b$ are constants with $\mathopen| a \mathclose| = 1$.
How do i claim that $f$ is of that form ?.I tried using C-R equations but that gave me $u_{x}^2 + v_{x}^2 = 1$ or $\mathopen|f'^{2}\mathclose| = 1$ and after this i cannot proceed.
Any help is great.

As you noted, you have that $|f'(z)|² \leq 1$ and by Liouvilles theorem, since f' is also entire, the function has to be constant. Now we can write:
$$
f(z)=f(0)+ \int_{[0,z]}f'(\xi)d\xi=f(0)+ \int_{[0,z]}a \;d\xi=az+b
$$
By the fundamental theorem of complex calculus (or a suting version for curvature integrals).
Now plugging we can rewrite our function as, using $z=x+iy $:
$$
f(x+iy)=(Re(a)x-Im(a)y)+i(Re(a)y+Im(a)x)+b
$$
Differentiatng now yields:
$$
u_x=Re(a) \\
v_x=Im(a)
$$
Plugging that into the equation above, $u_x^2+v_x^2=1 $ gives us $|a|^2=1$
