Complex Analysis Proofs Let $f$ = $u + iv$ be an entire function satisfying $u(x,y) \geq v(x,y)-1$ for every $(x,y) \in R^2$. Prove that all functions $f, u, v$ are constant. 
Can someone please help me prove this...
 A: With these type of questions, when I see entire and some bound on the function, I immediately try to apply Liouville's theorem. From there, it is just a matter of trying to get a bounded entire function. In this case, the following gives a bounded entire function:
The condition that $u\geq v-1$ can be rewritten to say that $v-u\leq 1$. Now consider
$$ |e^{-f-if}| = |e^{-u-iv-iu+v}| = e^{v-u}\leq e^1 $$
Note that the function $-f-if$ is entire, being the sum of entire functions. Also $e^{-f-if}$ is entire, being the composition of entire functions. The above inequality shows that we have a bounded entire function, Liouville's theorem now implies that 
$$ e^{-f-if} = c$$
Note that, since $e^z$ is never zero, $c\neq 0$. To conclude that $f$ must be constant, differentiate both sides to give
$$ (-f'-if')e^{-f-if} =0$$
hence $(-1-i)f' = 0$ which gives $f'=0$, hence $f$ must be constant.
A: Here's an alternative answer. Consider $$(1-i)f = u +iv - iu +v = u + v + i (v-u).$$
Since $v- u \leq 1$ we get that the imaginary part of $f$ lies entirely in the upper half of the complex plane. In particular $(1-i)f$ is an entire function which misses at least two points so by little Picard must be constant. In other words, $f$ is the constant function.
