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!
GR 8767# 58

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: GR 8767# 58
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
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
and of course Little Picard Theorem is very useful here
http://en.wikipedia.org/wiki/Picard_theorem
from this theorem it follows immidiately
http://en.wikipedia.org/wiki/Picard_theorem
from this theorem it follows immidiately
Re: GR 8767# 58
blitzer6266 and student27, thanks very much!
Re: GR 8767# 58
Using Liouville's Theorem with $$e^{f(z)i}$$ is again another solution