anonymous
  • anonymous
Please help me solve the following sum on complex numbers (Part 2):
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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
AkashdeepDeb
  • AkashdeepDeb
Just substitute the value of alpha in the second equation! What do you get?
AkashdeepDeb
  • AkashdeepDeb
@tanvidais13 ?

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anonymous
  • anonymous
ohh sorry. yup, i think i've got it :P
anonymous
  • anonymous
but thanks anyways :D
AkashdeepDeb
  • AkashdeepDeb
:)
AkashdeepDeb
  • AkashdeepDeb
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anonymous
  • anonymous
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

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