z1 * z2 = (1 + i)(2 - i) = 2 - i + 2i - i^2 = 2 + 1 + i = 3 + i.

The discriminant is 0, indicating one real root (a repeated root).

The discriminant is 0, hence the roots are real and equal.

For equal roots, the discriminant must be zero: k^2 - 4*1*16 = 0, solving gives k = -8.

Using the sum of roots: p = -(3 + (-2)) = -1.

The equation factors to (x-1)(x-2) = 0.

The discriminant is 0, indicating one distinct real root.

The discriminant is 0, indicating that there is exactly one distinct real root.

Let the roots be r and r+2. Then, r + (r+2) = 6 and r(r+2) = k. Solving gives k = 10.

The real part of z = 4 - 3i is 4.

Using the quadratic formula, x = [4 ± √(16 - 8)] / 4 = [4 ± 2√2] / 4 = 1 ± √2/2. Hence, the roots are not simple fractions.

Using the quadratic formula, the roots are x = [20 ± √(400 - 300)] / 10 = [20 ± 10] / 10 = 3 and 1.

The equation can be factored as (x + 1)^2 = 0, giving a double root at x = -1.

Factoring gives (x-1)(x-2) = 0, so the roots are 1 and 2.

Using Vieta's formulas, the sum of the roots is -(-3)/2 = 3/2.

The sum of the roots is given by -b/a = 4/2 = 2. Setting this equal to 3 gives k = 1.

The sum of the roots is given by -b/a = 12/3 = 4.

The sum of the roots is given by -b/a = 7/1 = 7.

Using the sum of roots formula -b/a, we have 4/2 = 2, thus 2 + 1 = 3, so k = 1.

Using Vieta's formulas, the sum of the roots is -(-12)/3 = 4.

Using Vieta's formulas, the sum of the roots is -(-12)/3 = 4.

Using Vieta's formulas, sum of roots = -b/a = 12/3 = 4, hence k = 8.

The sum of the roots is given by -b/a = 7/1 = 7.

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

sin^(-1)(-1) corresponds to the angle whose sine is -1, which is -π/2.

sin^(-1)(√3/2) corresponds to the angle whose sine is √3/2, which is π/3.

tan^(-1)(√3) corresponds to the angle whose tangent is √3, which is π/3.

a_7 = 2^7 + 3^7 = 128 + 2187 = 2315.

The argument of z = -1 + 0i is arg(z) = π.

The argument of z = -1 - i is θ = tan^(-1)(-1/-1) = -3π/4.