# Proof of Complex Number Conjugates

1. Jan 30, 2010

### typeinnocent

1. The problem statement, all variables and given/known data
Consider: y = f(x) = ax^2+bx+c, where a, b and c are real constants. Prove that y*=f(x*)

2. Relevant equations
conjugate of a complex number x=a+jb and x*=a-jb

3. The attempt at a solution
You can show the answer of f(x*) by substituting (a-jb) for each x in f(x), but I was confused on how you do the conjugate of an entire function. Would you try to find the roots of y and f(x*) each and then take the conjugate roots of y, and if they are the same then y*= f(x*)?

2. Jan 30, 2010

### Hurkyl

Staff Emeritus
In everything you've written, conjugation is only being applied to complex values, so I'm not sure why you're asking this.

Actually, in everything you've written, the only expressions that have been conjugated are the variables themselves. But even in an expression like
f(x)*
complex conjugation is being applied to the value of f(x), not to f.

As an aside, you have an error in your grammar (which may betray a misundersanding of logic): you probably meant
(Definition of the) conjugate of a complex number: If x=a+jb then x*=a-jb​
which means something very different from what you wrote.

You'll probably never encounter it so it probably doesn't matter, but the conjugate of the actual function f is f itself. The definition of conjugation should insist that
f*(x*) = f(x)*
and if we invoke the theorem you are trying to prove along with substituting x* for x, we get:
f*(x) = f(x*)* = f(x)​
so the functions f* and f are equal. (Because they are "pointwise equal" -- they have the same value on every input)

Last edited: Jan 30, 2010
3. Jan 30, 2010

### typeinnocent

I'm sorry, but I'm really confused by your explanation. You said that for f(x)*, "complex conjugation is being applied to the value of f(x)." Do you mean then, that the complex conjugation is being applied to the roots of f(x), since those give the values of f(x)?

Perhaps I'm reading my question wrong, so let me try to phrase it the way I'm understanding it - maybe then you can help me clarify my thinking. I also didn't realize until now that we're supposed to let x = $$\alpha$$+j$$\beta$$, so that may also be a component of my misunderstanding.

I solved f(x) for x=$$\alpha$$+j$$\beta$$, which ended up equaling (a*$$\alpha$$^2+b*$$\alpha$$+c) + j(2*a*$$\alpha$$*$$\beta$$-a*$$\beta$$^2+b*$$\beta$$). When I solved f(x) for x = $$\alpha$$-j$$\beta$$ (the complex conjugate), I got (a*$$\alpha$$^2+b*$$\alpha$$+c) - j(2*a*$$\alpha$$*$$\beta$$+a*$$\beta$$^2+b*$$\beta$$). My guess now would be to take the complex conjugate of the solution of f(x) to see if the two solutions will be equal, but that doesn't happen. The complex conjugate of f(x) is (a*$$\alpha$$^2+b*$$\alpha$$+c) - j(-2*a*$$\alpha$$*$$\beta$$+a*$$\beta$$^2-b*$$\beta$$).

So now perhaps you can tell me if I'm still not understanding the question, or if I'm headed down the wrong path in trying to get the solution. I really appreciate the help!

4. Jan 30, 2010

### Hurkyl

Staff Emeritus
Your method is indeed a reasonable one. But you've made an arithmetic error.

P.S. once you get this worked out, you might try doing another proof that uses the algebraic properties of *, rather than splitting everything into real and imaginary parts.

I really have no idea what you mean by that....