# analytic function that maps the entire complex plane into the real axis

An analytic function that maps the entire complex plane into the real axis must map the imaginary axis onto:

A) the entire real axis

B) a point

C) a ray

D) an open finite interval

E) the empty set

I was thinking that it might be a constant function. Any help would be appreciated!

• Can you show any work that you have done? – Gabe Aug 24 '19 at 15:34
• Considering the complex plane contains the set of all imaginary numbers, they would be mapped onto the real axis as well. A common example is the magnitude function, it maps all complex and imaginary numbers onto the real axis. – Gabe Aug 24 '19 at 15:37
• If $f$ is analytic non-constant then $f(z) = f(a)+C (z-a)^n+O((z-a)^{n+1})$ so it can't be real valued on a complex neighborhood of $a$ – reuns Aug 24 '19 at 15:55

If $$f(\mathbb{C})\subset \mathbb{R}$$, then $$|e^{if(z)}| = 1$$ for all $$z$$. By Liouville's theorem, $$e^{if(z)}$$ and hence $$f(z)$$ is a constant. So $$f(i\mathbb{R})$$ is a single point.

Hint:As a consequence of Open mapping theorem, the dimension of Image space of an analytic $$f$$ can be either $$0$$ or $$2$$. Moreover if it is $$0$$ then $$f$$ is constant.

I think this might answer it?

Let f(x+iy) = u(x,y) where u is a real function. Then it is obvious by C-R equations since it is analytic.

Thanks for all the hints provided!