Linear algebra- set of complex valued functions

1. Apr 30, 2009

gotmilk04

1. The problem statement, all variables and given/known data
Let V be the set of all complex-valued functions f on the real line such that (for all t in R),
f(-t)=f(t) with a bar on top (can't figure out Latex, sorry)
The bar denotes complex conjugation.

Give an example of a function in V which is not real-valued.

2. Relevant equations

3. The attempt at a solution
Not quite sure what this means, just need a place to start really.

2. Apr 30, 2009

Dick

Write f(t)=u(t)+iv(t) where u and v are real valued functions. Equate the real and imaginary parts of both sides of your equation. What are the conditions on u(t) and v(t)?

3. Apr 30, 2009

gotmilk04

I didn't really know there were conditions on u(t) and v(t).
To get an example, can I just make those fuctions whatever real valued function that I want, like t-2, and then plug it in?

4. Apr 30, 2009

Dick

I mean the conditions you derive from f(-t)=conjugate(f(t)). If f(t)=u(t)+iv(t) isn't conjugate(f(t))=u(t)-iv(t)?

5. Apr 30, 2009

gotmilk04

Ohh, yeah, it does. So I have to make f(t) so that when the t's are (-t)'s, u(t)+iv(t) turns into u(t)-iv(t)?

6. Apr 30, 2009

Dick

Yes, so u(-t)+iv(-t)=u(t)-iv(t), right?

7. Apr 30, 2009

gotmilk04

Yep. Now I'm not sure where to go from here.

8. Apr 30, 2009

Dick

Equate real and imaginary parts of both sides. Come on, help me out here.

9. Apr 30, 2009

gotmilk04

I'm sorry, I'm very confused here and I've never learned this before, so it's very frustrating.

When you equate the real and imaginary parts, do you get
u(-t)=u(t) and v(-t)=-v(t)?

10. Apr 30, 2009

Dick

Yes, exactly. Can you find two functions u and v so that u(-t)=u(t) and v(-t)=(-v(t)) and v is not equal to zero? So f is not real valued?

11. Apr 30, 2009

gotmilk04

So if u(t)=t^2+1 and v(t)=t^3-t,
then f(t)=t^2 + 1 + i(t ^3-t)
= it^3 + t^2 - it + 1
Which is not real valued, so that's an example, correct?

12. Apr 30, 2009

Dick

Yes, u(t)=1 and v(t)=t works too. If you want to make it even simpler.

13. Apr 30, 2009

gotmilk04

Ah yes, that is much simpler.
Thanks so much for the help, sorry I was very lost before. I appreciate your patience and guidance!