A)Find (in rectangular form) the two values of sqrt(-8+6i) Solve the equation z^2 +sqrt(32)iz-6i=0.

a)Find (in rectangular form) the two values of sqrt(-8+6i)
b)Solve the equation z^2 +sqrt(32)iz-6i=0.
Thanks for your help

a) -8 + 6i = -9 + 1 + 6i = -1*9 + 1 + 6i = 9i^2 + 6i + 1 = (3i + 1) ^2 or (-3i - 1)^2.

The two roots are 1 + 3i and -1 - 3i.

b) z^2 + [i * sqrt(32)] z + (-6i) = 0
Use Quadratic Formula:

a = 1, b = [i * sqrt(32)], c = (-6i)

z = { [-i * sqrt(32)] + or - sqrt(-32 + 24i) } / 4

z = { [-i * sqrt(32)] + or - [2 * sqrt(-8 + 6i)] } / 4

z = { [-i * sqrt(32)] + or - [2 * (1 + 3i)] } / 4

z = [-i * 4 * sqrt(2) / 4] + or - (1 + 3i)/2

z = -sqrt(2) * i + 1/2 + 3 * i or -sqrt(2) * i - 1/2 - 3 * i

z = 1/2 + (3 - sqrt(2)) * i or -1/2 + (-3 - sqrt(2)) * i

Asia O`_`O