## 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

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!

blitzer6266
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.

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

### 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.

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

### 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

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

### Re: GR 8767# 58

blitzer6266 and student27, thanks very much!

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

### Re: GR 8767# 58

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