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### GR 8767# 58

Posted: Sun Oct 07, 2012 9:39 pm
If f(z) is an analytic function that maps the entire finite complex plane into the real axis, then the imaginary axis must be mapped onto
(A) the entire real axis
(B) a point
(C) a ray
(D) an open finite interval
(E) the empty set

Could anyone help me with it? Thanks a lot!

### Re: GR 8767# 58

Posted: Sun Oct 07, 2012 11:59 pm
nonconstant analytic functions are open maps (this is an important theorem in complex analysis). Since the real line is not open in the complex plane, this function must be constant.

### Re: GR 8767# 58

Posted: Mon Oct 08, 2012 9:58 am
I think that you can also use Cauchy–Riemann here. If f(x+iy)=u+iv, and it is given that v=0, then since du/dx=dv/dy and du/dy=-dv/dx you get that u==const and hence the image of f is just a point.

### Re: GR 8767# 58

Posted: Mon Oct 08, 2012 10:01 am
and of course Little Picard Theorem is very useful here

http://en.wikipedia.org/wiki/Picard_theorem

from this theorem it follows immidiately

### Re: GR 8767# 58

Posted: Tue Oct 09, 2012 10:54 pm
blitzer6266 and student27, thanks very much!

### Re: GR 8767# 58

Posted: Thu Oct 11, 2012 7:33 am
Using Liouville's Theorem with $e^{f(z)i}$ is again another solution