# Write the quadratic that has real coefficients and complex solutions z1=1+i, z2=1-i.

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### 3 Answers

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.**