Write the quadratic that has real coefficients and complex solutions z1=1+i, z2=1-i.
The required quadratic equation has roots 1 + i and 1 - i.
=> [x - (1 + i)]*[x - (1 - i)] = 0
=> [x - 1 - i][x - 1 + i] = 0
=> x^2 - x + xi - x + 1 - i - xi + i + 1 =0
cancel the common terms
=> x^2 - 2x + 2 = 0
Therefore the required equation is x^2 - 2x + 2 = 0.
To write the quadratic equation with roots z1 = 1+i and z2 = 1-i.
We know that the quadratic equation with roots z1 and z2 is given by the equation:
(x-z1)(x-z2) = 0.
=> x^2-(z1+z2)x+z1z2 = 0.....(1)
z1+z2 = 1+i+1-i = 2.
z1z2 = (1+i)(1-i) = 1^2-i^2 = 1-(-1) = 2.
So we put the value z1+z2 = 2 and z1z2 = 2 in (1):
x^2-(1+1)x+2 = 0.
Therefore the required quadratic equation with roots z1 = 1+i and z2 = 1-i is x^2-2x+2 = 0.
The quadratic equation is: ax^2 + bx + c = 0
We could determine the real coefficients a,b,c, using Viete's relations:
z1 + z2 = -b/a
1 + i + 1 -i = -b/a
We'll combine and eliminate like terms:
2 = -b/a
z1*z2 = c/a
(1+i)(1-i) = c/a
We'll apply the formla of difference of squares:
1^2 - i^2 = c/a
1^2 - (-1) = c/a
1 + 1 = c/a
c/a = 2
The quadratic equation is: x^2 - 2x + 2 = 0.