# Proving a function to be linear: Complex Analysis

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!

• Why do you say that $v'_x \ge 1$ ? In general $a(x)>b(x), \forall x$ does not imply that $a'(x) > b'(x), \forall x$. – PierreCarre Mar 27 at 14:39
• @PierreCarre my bad! Just realised that it need not be true... – Ankit Kumar Mar 27 at 14:49

## 1 Answer

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 $$1/(g(z) + i)$$?

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