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$f$ 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.

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    If $f$ be entire so, is $f'$ entire.? – 2017-01-22
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    How can that help? – 2017-01-22
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    Well, $f'$ will be entire and bounded $|f'|<1$ so it will be constant. – 2017-01-22
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    Here i got $\mathopen|f'^{2}\mathclose| = 1$ ? – 2017-01-22
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    $|zw|=|z|.|w|$. No matter, say $|f'|<2$ – 2017-01-22
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    still not getting? – 2017-01-22
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52234/discussion-between-baymax-and-myglasses). – 2017-01-22

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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 $|a|^2=1$