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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 Cauchyâ€“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_theoremfrom 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

is again another solution