GR 8767# 58

Forum for the GRE subject test in mathematics.
sss
Posts: 6
Joined: Sun Oct 07, 2012 5:17 pm

GR 8767# 58

Postby sss » 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
The answer is B

Could anyone help me with it? Thanks a lot!

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: GR 8767# 58

Postby blitzer6266 » 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.

student27
Posts: 3
Joined: Tue Oct 02, 2012 1:28 pm

Re: GR 8767# 58

Postby student27 » 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.

student27
Posts: 3
Joined: Tue Oct 02, 2012 1:28 pm

Re: GR 8767# 58

Postby student27 » 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

sss
Posts: 6
Joined: Sun Oct 07, 2012 5:17 pm

Re: GR 8767# 58

Postby sss » Tue Oct 09, 2012 10:54 pm

blitzer6266 and student27, thanks very much!

PNT
Posts: 37
Joined: Fri Mar 11, 2011 9:01 pm

Re: GR 8767# 58

Postby PNT » Thu Oct 11, 2012 7:33 am

Using Liouville's Theorem with e^{f(z)i} is again another solution




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