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Let $$f(z)=u(x,y)+iv(x,y)$$$$v(x,y)\geq x\text{ } \forall x,y$$be a complex valued function, where $z=x+iy \text{ and }u(x,y), v(x,y)\text{ are real valued functions.}$ If $f$ is an entire function, prove that it is a polynomial of degree $1$.

My try:

I tried to use CR equations, to obtain no result. All I got was $$v_x'\geq 1$$ which then implies $$v_x''\geq 0\text{ and }v_y''\leq 0$$I've no idea how to proceed with it after this.

I also tried assuming it to be a polynomial(since it is an entire function, we can anyway use taylor series to show the same too)$$f(z)=a_0+a_1z+a_2z^2+\cdots $$I then tried different values for $z$ and used $v\geq x$, but things just got more and more complicated with that...

I'd appreciate any help in the same. Thanks!

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    $\begingroup$ Why do you say that $v'_x \ge 1$ ? In general $a(x)>b(x), \forall x$ does not imply that $ a'(x) > b'(x), \forall x$. $\endgroup$ – PierreCarre Mar 27 at 14:39
  • $\begingroup$ @PierreCarre my bad! Just realised that it need not be true... $\endgroup$ – Ankit Kumar Mar 27 at 14:49
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Let $g(z) = f(z) - i z$. Then your inequality says $\text{Im}(g(z)) \ge 0$ for all $z$. What can you say about $1/(g(z) + i)$?

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  • $\begingroup$ Can you please give a little more hint... $\endgroup$ – Ankit Kumar Mar 27 at 15:09
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    $\begingroup$ $|g(z)+i| \ge 1$ for all $z$. Think Liouville. $\endgroup$ – copper.hat Mar 27 at 15:50
  • $\begingroup$ I got it! We can use contour integration now! Thanks a lot! $\endgroup$ – Ankit Kumar Mar 27 at 19:10
  • $\begingroup$ @copper.hat Ya, thanks $\endgroup$ – Ankit Kumar Mar 27 at 19:11
  • $\begingroup$ It's got nothing to do with contour integration. $\endgroup$ – Robert Israel Mar 27 at 19:54

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