Conjugate of an Integral

A problem with integrals in the form of an example:

"A trigonometric polynomial is a function f:R->C of the form f(x) = Sum{n=1,k} a_n exp(im_n k), x in R, k in N, the a_i in C, m_i in R. Show that <f,g> = lim(T->infinity) (1/2T) Int{-T,T} f(x)g*(x) dx defines an inner product on the linear space of trig polynomials." [g* = conjugate of g]

My problem here is showing that <f,g>=<g,f>* and <f,f> is at least zero and zero iff f=0. Disregard for the moment limits and constants, <g,f>* would then be [Int{-T,T} g(x)f*(x)dx]*, i.e. the "conjugation bar" above the entire integral. How does one calculate this?

The second then; Int f(x)f*(x) dx. What is ff*? I was told that it is |f|^2, which obviously would parallel the same identity for complex numbers, i.e. zz* = (Re z)^2 - (Im z)^2, but how does one reach that conclusion for any arbitrary function?

Also, consider the polynomial 1+(1+i)x for instance (which is not in the space above, but that's not my point here). What is its conjugate; 1+(1-i)x ?

I would very much appreciate help with this, I think the above assignment is really pretty easy otherwise. (I have an exam on this stuff this Friday!) Thank you!

Yes, that's always the way you find a conjugate- replace i by -i.

As for your "zz* = (Re z)^2 - (Im z)^2", that is correct if by "Im z" you are including the "i". That is, if z= a+ bi, Re z= a but some texts use Im z to mean just b, others bi. With the understanding that Re(a+bi)= bi, then, yes, zz*= a^2- (bi)^2= a^2+ b^2.

Oops, I meant (Re z)^2 + (Im z)^2 of course. I've never seen a text define Im(a+bi) as bi.

Anyway, what about the integral?