Finding complex solution to quadratic equation

[adichy]

Homework Statement

A (complex) solution to the quadratic equation of the form
z^2 + α z + β = 0, is z = 1 + i.
(where α and β are real numerical constants),

Write down the other solution and hence determine the values of α and β.

Homework Equations

The Attempt at a Solution

this question had me staring at it for a while since it has 2 unknown. I am thinking that i have to use some kind of simultaneous equation method but i don't have 2 equations.
How do i get rid of one of the unknowns or is there another way to solving this

i still got it wrong not sure what it is i did wong but here's what i did

z=1+i , z=1-i

a=α and B=β
therefore

(1+i)^2 +a(1+i)+B=(1-i)^2 +a(1-i)+B
(1+i)^2 - (1-i)^2= a(1-i) - a(1+i)
4i+2i^2=-2ai
2i(2+i)=-2ai
a=-2-i

question asks for a real solution my 1s still imaginary :|

 

[adichy]

i jus found out that another solution to this quadratic is its conjugate...if that helps ^^

 

[eumyang]

If z = c is a root of the (complex) quadratic equation
$$z^2 + \alpha z + \beta = 0$$
then (z - c) must be a factor of the quadratic.

You have the two solutions,
z = 1 + i and z = 1 - i. So that means that
$$z^2 + \alpha z + \beta = [z - (1 + i)][z - (1 - i)]$$
. Expand the right-hand side and equate the corresponding coefficients to find α and β.

 

[Sethric]

*edit* Sniped and with tex, eumyang is correct

 

[adichy]

so i expanded the RHS and got

a=2
B= 1-i^2=2

is that correct?

 

[Amok]

i jus found out that another solution to this quadratic is its conjugate...if that helps ^^

Tip: this is always the case. Can anyone come up with a more precise statement of this theorem?

 

[HallsofIvy]

If z = c is a root of the (complex) quadratic equation
$$z^2 + \alpha z + \beta = 0$$
then (z - c) must be a factor of the quadratic.

You have the two solutions,
z = 1 + i and z = 1 - i. So that means that
$$z^2 + \alpha z + \beta = [z - (1 + i)][z - (1 - i)]$$
. Expand the right-hand side and equate the corresponding coefficients to find α and β.

Exactly right. And notice that this can be written $[(z-1)- i][(z-1)+ i]$
the product a "sum and difference" which makes it especially easy to evaluate:
$(a- b)(a+ b)= a^2- b^2$

 

[eumyang]

so i expanded the RHS and got

a=2
B= 1-i^2=2

is that correct?

b is, but not a. I think you have a mistake in your algebra somewhere.

Exactly right. And notice that this can be written $[(z-1)- i][(z-1)+ i]$
the product a "sum and difference" which makes it especially easy to evaluate, so one could use this:
$(a- b)(a+ b)= a^2- b^2$

Nice one. The precalculus book I use in my class actually gives a formula for this type of expansion:
$$[x - (a + bi)][x - (a - bi)] = x^2 - 2ax + (a^2 + b^2)$$