# 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!