The modulus of z is given by |z| = √(3^2 + 4^2) = √(9 + 16) = √25 = 5.

The conjugate of z = 4 - 3i is z* = 4 + 3i.

The argument of z = 5 + 12i is θ = arctan(12/5).

z1 * z2 = (2 + 2i)(2 - 2i) = 4 - 4i^2 = 4 + 4 = 8.

z1 + z2 = (2 + 2i) + (3 - i) = 5 + i.

The discriminant is 0, which indicates that the roots are real and equal.

The coefficient of x^3 in (2x - 3)^4 is given by 4C1 * (2)^3 * (-3)^1 = 4 * 8 * (-3) = -96. The coefficient is -108.

The coefficient is C(5,3) * (2)^3 * (-3)^2 = 10 * 8 * 9 = 720.

Using the binomial theorem, the coefficient of x^5 in (3x - 4)^7 is given by 7C5 * (3)^5 * (-4)^2 = 21 * 243 * 16 = 21 * 3888 = 81588.

C(n,3) * b^2 = 60. For n = 5, C(5,3) = 10, which does not satisfy. For n = 6, C(6,3) = 20, which does not satisfy. For n = 7, C(7,3) = 35, which does not satisfy. For n = 8, C(8,3) = 56, which satisfies.

Using the binomial theorem, the coefficient of x^3 in (x + 3)^5 is given by 5C3 * (3)^2 = 10 * 9 = 90.

Using the binomial theorem, the coefficient of x^4 in (x + 3)^6 is given by 6C4 * (3)^2 = 15 * 9 = 135.

The coefficient of x^5 in (x - 1)^8 is C(8,5) * (-1)^3 = 56 * (-1) = -56.

For both roots to be positive, k must be greater than 1 (from Vieta's formulas).

Factoring the equation gives (x - 2)(x - 5) = 0, so the roots are 2 and 5.

The discriminant must be less than zero: (-4)^2 - 4*2*k < 0 leads to k > 8.

This is a perfect square trinomial: (x + 3)(x + 3) = 0.

For both roots to be negative, k must be greater than the square of half the coefficient of x, hence k > 9.

The equation can be factored as (2x - 3)(2x - 3) = 0, hence the roots are 3 and 3.

The sum of the roots is -1 + (-2) = -3, and the product is (-1)(-2) = 2. Thus, k = 2.

If the roots are both 2, then k = 2^2 - 4*2 = 4 - 8 = -4. Thus, k = 4.

Using Vieta's formulas, the sum of the roots is 7 and the product is k = 10.

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

The 4th term is given by C(6,3) * (3x)^3 * (2)^3 = 20 * 27x^3 * 8 = 4320x^3.

The absolute value of -7 is 7.

Using the binomial theorem, the coefficient of x^2 in (2x + 5)^4 is given by 4C2 * (2)^2 * (5)^2 = 6 * 4 * 25 = 600.

The coefficient of x^2 in (2x - 5)^5 is given by 5C2 * (2)^2 * (-5)^3 = 10 * 4 * (-125) = -5000.

The coefficient of x^3 in (3x + 2)^5 is given by 5C3 * (3)^3 * (2)^2 = 10 * 27 * 4 = 1080.

The product of the roots of the equation ax^2 + bx + c = 0 is given by c/a. Here, c = 6 and a = 2, so the product is 6/2 = 3.

Using Vieta's formulas, the product of the roots is c/a = 10/1 = 10.