2016 H2 Maths Paper 1 Q7 — Worked Solution

Part (a)

5 marks

Substituting $w = - 1 + 5i$ into $w^{2} + ( - 1 - 8i)w + ( - 17 + 7i)$ ,

w^{2} = ( - 1 + 5i)^{2} = 1 - 10i + 25i^{2} = - 24 - 10i

( - 1 - 8i)w = ( - 1 - 8i)( - 1 + 5i) = 1 - 5i + 8i - 40i^{2} = 41 + 3i

Adding these to the constant term,

w^{2} + ( - 1 - 8i)w + ( - 17 + 7i) = ( - 24 - 10i) + (41 + 3i) + ( - 17 + 7i) = ( - 24 + 41 - 17) + ( - 10 + 3 + 7)i = 0.

Since this equals $0$ , $- 1 + 5i$ satisfies the equation, so it is a root. $\ \$ (shown)

Approach

Since all the coefficients of this quadratic are complex, the roots are not a conjugate pair; instead we use the fact that the sum of the roots of $w^{2} + pw + q = 0$ is $- p$ , which needs only the coefficient of $w$ , not the constant term.

For $w^{2} + ( - 1 - 8i)w + ( - 17 + 7i) = 0$ , the sum of the two roots is

- ( - 1 - 8i) = 1 + 8i

Since one root is $- 1 + 5i$ , the second root is

(1 + 8i) - ( - 1 + 5i)

= 1 + 8i + 1 - 5i

= 2 + 3i

To check your answer: the product of the roots should equal

- 17 + 7i

. Indeed

( - 1 + 5i)(2 + 3i) = - 2 - 3i + 10i + 15i^{2} = - 17 + 7i

. ✓

$\therefore$ the second root is $p + iq = 2 + 3i$ .

Part (b)

5 marks

Approach

All the coefficients of $z^{3} - 5z^{2} + 16z + k$ are real, so non-real roots occur in conjugate pairs. If $1 + ai$ is a root, so is $1 - ai$ , and the third root is real (it cannot pair with anything else). Rather than dividing out the known factor, it is faster here to use the sum and sum-of-products relations between the roots and coefficients directly.

Let the third root be $r \in \mathbb{R}$ . Since $1 + ai$ and $1 - ai$ are both roots,

Sum of roots $\text{=}5$ :

(1 + ai) + (1 - ai) + r = 5

2 + r = 5

r = 3

Sum of products of pairs of roots $\text{=}16$ :

(1 + ai)(1 - ai) + (1 + ai)(r) + (1 - ai)(r) = 16

\left( 1 + a^{2} \right) + r\left\lbrack (1 + ai) + (1 - ai) \right\rbrack = 16

\left( 1 + a^{2} \right) + 2r = 16

Substituting $r = 3$ ,

1 + a^{2} + 6 = 16

a^{2} = 9

a = 3\ \left( \text{rej}\ a = - 3\ \text{as}\ a > 0 \right)

Product of roots $\text{=} - k$ :

(1 + ai)(1 - ai)(r) = - k

\left( 1 + a^{2} \right)r = - k

Substituting $a = 3$ , $r = 3$ ,

(1 + 9)(3) = - k

30 = - k

k = - 30

To check your answer: with

k = - 30

, the cubic

z^{3} - 5z^{2} + 16z - 30

has roots

3,\ 1 + 3i,\ 1 - 3i

(by GC / by direct expansion), matching

r = 3

and

a = 3

. ✓

$\therefore$ $a = 3$ and $k = - 30$ .

2016 H2 Maths Paper 1, Question 7

Part (a)

Part (b)

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