Write a quadratic equation whose root is (1-i)(2+i).

If an equation has a complex root, then the equation
has another root, namely the conjugate of the complex
root.

Because it is a quadratic equation, the number of
roots is 2, and we have already one root, so the second root is the conjugate of the
number (1-i)(2+i).

We'll calculate
(1-i)(2+i).

(1-i)(2+i) = 2 + i - 2i - i^2, where i^2 =
-1

(1-i)(2+i) = 2 - i +
1

(1-i)(2+i) = 3 - i

The first
root is z1 = 3 - i.

The conjugate of the complex number z1
is the second root,

z2 = 3 +
i.

We'll form the quadratic equation using Viete's
relations:

z1 + z2 = 3 - i + 3 +
i

We'll combine like terms:

z1
+ z2 = 6

z1*z2 = (3-i)(3+i) = 9 -
i^2

 z1*z2 = 9+1

z1*z2 =
10

The quadratic equation
is:

z^2 - (z1+z2)*z + (z1*z2) =
0

z^2 - 6z + 10 =
0