19N.2.AHL.TZ0.H_6

pestleMathematicsAAHLPaper 219N· ahl-1-14-complex-roots-of-polynomials-conjugate-roots-de-moivres-powers-&-roots-of-complex-numberssource ↗

Let $P\left( z \right) = a{z^3} - 37{z^2} + 66z - 10$ , where $z \in \mathbb{C}$ and $a \in \mathbb{Z}$ .

One of the roots of $P\left( z \right) = 0$ is $3 + {\text{i}}$ . Find the value of $a$ .

Markscheme / solution

METHOD 1

one other root is $3 - {\text{i}}$ A1

let third root be $\alpha$ (M1)

considering sum or product of roots (M1)

sum of roots $= 6 + \alpha = \frac{{37}}{a}$ A1

product of roots $= 10\alpha = \frac{{10}}{a}$ A1

hence $a = 6$ A1

METHOD 2

one other root is $3 - {\text{i}}$ A1

quadratic factor will be ${z^2} - 6z + 10$ (M1)A1

$P\left( z \right) = a{z^3} - 37{z^2} + 66z - 10 = \left( {{z^2} - 6z + 10} \right)\left( {az - 1} \right)$ M1

comparing coefficients (M1)

hence $a = 6$ A1

METHOD 3

substitute $3 + {\text{i}}$ into $P\left( z \right)$ (M1)

$a\left( {18 + 26{\text{i}}} \right) - 37\left( {8 + 6{\text{i}}} \right) + 66\left( {3 + {\text{i}}} \right) - 10 = 0$ (M1)A1

equating real or imaginary parts or dividing M1

$18a - 296 + 198 - 10 = 0$ or $26a - 222 + 66 = 0$ or $\frac{{10 - 66\left( {3 + {\text{i}}} \right) + 37\left( {8 + 6{\text{i}}} \right)}}{{18 + 26{\text{i}}}}$ A1

hence $a = 6$ A1

[6 marks]

Examiners’ report

[N/A]