Complex number question?

Given that (2+i) is a root for the equation z^3 + mz + nz +10 = 0.

Find the real values of m and n

3 Answers

Relevance
  • 7 years ago
    Favourite answer

    i) Complex roots always occur in pairs.

    So one root being (2 + i), the other root is (2 - i); and let the 3rd root be t;

    ii) Applying properties of roots of an equation theory,

    Sum of roots = -coefficient of 2nd degree term/leading coefficient

    Sum of product of roots taken in pair = coefficient of 1st degree term/leading coefficient

    Product of all roots = -constant term/leading coefficient.

    ==> t*(2 + i)*(2 - i) = -10

    t*(4 - i²) = t*5 = -10 [Since i² = -1]

    Solving t = -2

    Thus the 3 roots of the equation are: {-2, 2 + i, 2 - i}

    iii) By the properties as explained above:

    a) -2 + 2 + i + 2 - i = -m/1; solving m = -2

    b) -2(2 + i) + (2 + i)(2 - i) + (2 - i)(-2) = n

    Solving n = -3

    Thus (m, n) = (-2, -3)

  • Anonymous
    7 years ago

    I will assume you really meant

    Given that (2+i) is a root for the equation z^3 + mz^2 + nz +10 = 0.

    Find the real values of m and n

    A cubic can have 3 REAL roots or one REAL root and TWO equal but opposite in sign complex roots.

    Let the real root be a, then the two complex roots are (2+i) and (2-i)

    So

    (z-a)(z-[2+i])(z-[2-i]) = z^3 + mz^2 + nz +10 = 0.

    (z-a)(z-2-i)(z-2+i) = z^3 + mz^2 + nz +10

    (looking at LHS Expanding using BOX Method - GOGGLE this simple neat method)

    (z-a)( z^2 –2z -2z –zi +zi –2i +2i –[i^2] +4) (remember i^2 = -1)

    (z-a)(z^2-4z –(-1) +4)

    (z-a)(z^2 –4z+5) = z^3 + mz^2 + nz +10

    So we must discover a. Well it can only be -2 since –(–2) * 5 = +10

    (z+2)(z^2 –4z+5) = z^3 + mz^2 + nz +10

    Again expanding with BOX method and collecting terms

    z^3 –2z^2+z+10 = z^3 + mz^2 + nz +10

    by comparison

    m = -2, n =1

  • 7 years ago

    Sub z=2+i, z^3=2+11i, z^2=3+4i so

    2+11i+m(3+4i)+n(2+i)+10=0

    so (12+3m+2n)+(11+4m+n)i=0

    giving 12+3m+2n=0 and 11+4m+n=0

    and solving simultaneously gives

    m=-2 and n=-3

Still have questions? Get answers by asking now.