An entire function $f=u+iv$ has the feature that $u_xv_y-u_yv_x=1$ throughout the complex plane. Demostrate that has the form $f(z)=az+b$ An entire function $f=u+iv$ has the feature that $u_xv_y-u_yv_x=1$ throughout the complex plane. Demostrate that has the  form $f(z)=az+b$, where $a$ and $b$ are constants and $|a|=1$.
I am very confused showing this, I have thought about showing that $f$ is an injective function and thus apply the fact that every entire and injective function has this form but I do not know how to do this, could someone help me please? How is it demonstrated that a complex function is injective? The injectivity of a real function is the same as the injectivity of a complex function? Thank you.
 A: Hint: The equation says that $f'$ has constant modulus equal to one. 
A: As has been observed by both Pedro Tamaroff and our OP Nash, the condition
$u_x v_y - u_y v_x = 1 \tag 1$
on the real and imaginary parts of the holomorphic function
$f = u + iv \tag 2$
implies that
$\vert f'(z) \vert^2 = \vert u_x + i v_x \vert^2 = u_x^2 + v_x^2 = u_x v_y + (-u_y) v_x = u_x v_y - u_y v_x = 1, \tag 3$
where we have used the Cauchy-Riemann equations $u_x = v_y$, $u_y = -v_x$ in establishing (3), which then leads to 
$\vert f'(z) \vert = 1 \tag 4$
for every $z \in \Bbb C$, since $f'(z)$, like $f(z)$, is entire and hence defined on all of $\Bbb C$.  (4) shows that $f'(z)$ is bounded, hence by Liouville's theorem, it is constant.  Therefore we may take
$f'(z) = a \in S^1, \tag 5$
where $S^1$ is the unit circle in $\Bbb C$.  It then follows that
$(f(z) - az)' = f'(z) - a = 0 \tag 6$
for all $z \in \Bbb Z$; thus we may write
$(f(z) - az)) - f(0) = (f(z) - az) - (f(0) - a \cdot 0)$
$= \displaystyle \int_{0, \gamma(t)}^z (f(w) - az)'dw = \int_{0, \gamma(t)}^z (f'(w) - a) dw = \int_{0, \gamma(t)}^z 0 \; dw = 0,\tag 7$
where $\gamma(t)$ is any differentiable path in $\Bbb C$ with $\gamma(0) = 0$ and $\gamma(1) = z$; if we now set
$b = f(0), \tag 8$
then (7) yields
$f(z) - az = f(0) = b, \tag 9$
or
$f(z) = az + b, \tag{10}$
as was to be shown.
