Question

Please help me solve the following sum on complex numbers (Part 2):

Answer 1

it is given that \[\alpha = -1+3i\] and \[\lambda \alpha^3 + 8\alpha^2 + 34\alpha + \mu = 0\], where lamda and mu are real numbers. show that lamda is 3

Answer 2

Just substitute the value of alpha in the second equation!
What do you get?

Answer 3

@tanvidais13 ?

Answer 4

ohh sorry. yup, i think i've got it :P

Answer 5

but thanks anyways :D

Answer 6

:)

Answer 7

If alpha zeros the poynomial then x-alpha is a factor (factor theorem) Complex roots come in conjugate pairs so (x+1-3i)(x+1+3i) = (x+1)^2+9 = x^2+2x+10 so dividing lamdax^3 +8x^2+34x + mu by x^2+2x+10 must give a zero remainder. During long division you get [34x -10lamdax -16x + 4lamdax] =0 so 18x - 6lamdax = 0 so lamda equal 3