3
$\begingroup$

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!

$\endgroup$
3
  • $\begingroup$ Can you show any work that you have done? $\endgroup$
    – Gabe
    Commented Aug 24, 2019 at 15:34
  • $\begingroup$ 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. $\endgroup$
    – Gabe
    Commented Aug 24, 2019 at 15:37
  • $\begingroup$ 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 $a$ $\endgroup$
    – reuns
    Commented Aug 24, 2019 at 15:55

3 Answers 3

1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .