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GR 8767# 58

Posted: Sun Oct 07, 2012 9:39 pm
by sss
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
The answer is B

Could anyone help me with it? Thanks a lot!

Re: GR 8767# 58

Posted: Sun Oct 07, 2012 11:59 pm
by blitzer6266
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
by student27
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
by student27
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
by sss
blitzer6266 and student27, thanks very much!

Re: GR 8767# 58

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