Identify the quotient in the form a+bi 5-8i/3+2i

Answer 1

First multiply the numerator and denominator by the complex conjugate of the denominator.

This uses the fact that (x-yi)(x+yi) = x²+y²

(5-8i)(3-2i)/((3+2i)(3-2i)) = (15-10i-24i-16)/(9+4) =(-1-34i)/13

= (-1/13) + (-34/13)i

Here, a = -1/13 and b = -34/13

Answer 2

To simplify this, we need to remove the "i" part from the bottom. Multiply by (3-2i) to top and bottom.

(3 - 2i)(5 - 8i)/[(3 - 2i)(3 + 2i)]

Top part: (3 - 2i)(5 - 8i) = (3)(5) + (3)(-8i) + (-2i)(5) + (-2i)(-8i) =15 - 24i - 10i - 16 = (-1 - 34i)

(remember that i² = -1)

Bottom part: (3-2i)(3+2i) = (3)(3) + (3)(2i) + (3)(-2i) + (-2i)(2i) = 9 + 6i - 6i + 4 = 9 + 4 = 13

So we get (-1-34i)/13 = (-1/13) - (34i)/(13), a = -1/13 and b = -34/13

Answer 3

We multiply and divide by 3 - 2i: (5 - 8i)/(3 + 2i) = (5 - 8i)(3 - 2i)/(3 + 2i)(3 - 2i) = (15 - 10i - 24i + 16i²)/(9 + 4) =

(- 1 - 34i)/13 = - 1/13 - 34/13i. Keep in mind: i² = - 1