# Linear algebra- set of complex valued functions

• gotmilk04
In summary, the function f in the domain V is real valued if and only if f(-t)=f(t) for all t in R, and is not real-valued if and only if f(t) is not conjugate(f(-t)).f

## Homework Statement

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.

## The Attempt at a Solution

Not quite sure what this means, just need a place to start really.

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)?

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?

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)?

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)?

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

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

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

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)?

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?

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?

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

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