# 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

Relevance
• 7 years ago

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