GR 8767# 58

[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!

[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.

[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.

[student27]

and of course Little Picard Theorem is very useful here

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

from this theorem it follows immidiately

[sss]

blitzer6266 and student27, thanks very much!

[PNT]

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