Is the Conjugate of a Polynomial the Same as Its Conjugate Field?

[chuy52506]

Let p(x)=a₀+a₁x+a₂x²\in Real Numbers and let z\in Complex Field. Show that p(conjugate of z)=conjugate of p(z)

 

[sponsoredwalk]

I assume that was supposed to be

p(x) = a_o + a₁x + a₂x²

where a_i ∈ ℝ & xⁱ ∈ ℂ.

Just write the expression out in all it's glory, i.e. all the gory algebra, & it should become apparent to you how one side equals the other.

 

[chuy52506]

Would I have the use the quadratic formula to show this? or how would i start?

 

[sponsoredwalk]

If you get stuck with problems like this it's always best to just write out what you know first.

For example:

p(x) = a_o + a₁x + a₂x²

is what you're working with, & this is the regular form of a polynomial of degree 2
when the x's are real. What does a polynomial look like when the x's are complex?
Replace x with z if it's more comfortable.

 

[chuy52506]

would i then replace then x's with a+bi?

 

[sponsoredwalk]

Well if the x's are complex then they can be written in the form x = a + bi so yes you are
correct. Going with the notation of your original post we will say that z = a + bi.
I'm sure you know how to find the conjugate of z so work with the polynomial and see
what happens.

 

[chuy52506]

ok thanks i got it!=]