2017 H2 Maths Paper 1 Q8 — Worked Solution

Part (a)

3 marks

Approach

This is a quadratic in $z$ with complex coefficients, so we apply the quadratic formula directly (no calculator), then rationalise the denominator to reach cartesian form.

Comparing $z^{2}(1 - i) - 2z + (5 + 5i) = 0$ with $Az^{2} + Bz + D = 0$ , $A = 1 - i$ , $B = - 2$ , $D = 5 + 5i$ .

We first find $(1 - i)(5 + 5i)$ , needed for the discriminant:

(1 - i)(5 + 5i)

= 5 + 5i - 5i - 5i^{2}

= 5 + 5

= 10

By the quadratic formula,

z = \frac{2 \pm \sqrt{4 - 4(1 - i)(5 + 5i)}}{2(1 - i)}

B^{2} - 4AD = 4 - 4(10) = - 36 = 36i^{2}

so $\sqrt{B^{2} - 4AD} = \pm 6i$ , and

z = \frac{2 \pm 6i}{2(1 - i)} = \frac{1 \pm 3i}{1 - i}

Rationalising, multiplying numerator and denominator by $1 + i$ :

z

= \frac{(1 \pm 3i)(1 + i)}{(1 - i)(1 + i)}

= \frac{(1 \pm 3i)(1 + i)}{2}

Taking the $\text{+}$ sign: $(1 + 3i)(1 + i) = 1 + i + 3i + 3i^{2} = - 2 + 4i$ , so $z = \frac{- 2 + 4i}{2} = - 1 + 2i$ .

Taking the $\text{-}$ sign: $(1 - 3i)(1 + i) = 1 + i - 3i - 3i^{2} = 4 - 2i$ , so $z = \frac{4 - 2i}{2} = 2 - i$ .

To check your answer: substituting

z = 2 - i

:

z^{2}(1 - i) = (3 - 4i)(1 - i) = 3 - 3i - 4i + 4i^{2} = - 1 - 7i

, and

- 2z + (5 + 5i) = - 4 + 2i + 5 + 5i = 1 + 7i

; summing gives

- 1 - 7i + 1 + 7i = 0

. ✓

$\therefore$ $z = - 1 + 2i$ or $z = 2 - i$ .

Part (b)(i)

7 marks

Since $\omega = 1 - i$ :

\omega^{2}

= (1 - i)^{2}

= 1 - 2i + i^{2}

= - 2i

\omega^{3}

= \omega^{2} \cdot \omega

= - 2i(1 - i)

= - 2i + 2i^{2}

= - 2 - 2i

\omega^{4}

= \left( \omega^{2} \right)^{2}

= ( - 2i)^{2}

= - 4

$\therefore$ $\omega^{2} = - 2i$ , $\omega^{3} = - 2 - 2i$ , $\omega^{4} = - 4$ .

Approach

Substituting these into the given equation splits it into one real and one imaginary equation in $p$ and $q$ , since $p$ and $q$ are given to be real.

Substituting into $\omega^{4} + p\omega^{3} + 39\omega^{2} + q\omega + 58 = 0$ :

- 4 + p( - 2 - 2i) + 39( - 2i) + q(1 - i) + 58 = 0

(54 - 2p + q) + ( - 2p - 78 - q)i = 0

Comparing real and imaginary parts (since $p,q$ are real):

54 - 2p + q = 0

- 2p - 78 - q = 0

Adding the two equations:

- 4p - 24 = 0

p = - 6

Substituting into $54 - 2p + q = 0$ :

54 + 12 + q = 0

q = - 66

To check your answer:

- 2p - 78 - q = 12 - 78 - ( - 66) = 12 - 78 + 66 = 0

, which matches. ✓

$\therefore$ $p = - 6$ , $q = - 66$ .

Part (b)(ii)

Approach

With $p = - 6$ and $q = - 66$ , the equation $z^{4} + pz^{3} + 39z^{2} + qz + 58 = 0$ has real coefficients and $\omega = 1 - i$ as a root, so its conjugate $\overline{\omega} = 1 + i$ is also a root. The quadratic factor with these two roots is found from their sum and product, and the other quadratic factor follows by division.

Since $z^{4} - 6z^{3} + 39z^{2} - 66z + 58$ has real coefficients and $1 - i$ is a root, $1 + i$ is also a root, so $\left( z - (1 - i) \right)\left( z - (1 + i) \right)$ is a factor.

\left( z - (1 - i) \right)\left( z - (1 + i) \right)

= z^{2} - \left\lbrack (1 - i) + (1 + i) \right\rbrack z + (1 - i)(1 + i)

= z^{2} - 2z + 2

Dividing $z^{4} - 6z^{3} + 39z^{2} - 66z + 58$ by $z^{2} - 2z + 2$ :

z^{4} - 6z^{3} + 39z^{2} - 66z + 58 = \left( z^{2} - 2z + 2 \right)\left( z^{2} + Ez + F \right)

Comparing the coefficient of $z^{3}$ : $E - 2 = - 6 \Rightarrow E = - 4$ . Comparing the constant term: $2F = 58 \Rightarrow F = 29$ .

To check your answer: comparing the coefficient of

z

:

2E - 2F = 2( - 4) - 2(29) = - 8 - 58 = - 66

, which matches. ✓

$\therefore$ $z^{4} - 6z^{3} + 39z^{2} - 66z + 58 = \left( z^{2} - 2z + 2 \right)\left( z^{2} - 4z + 29 \right)$ .

2017 H2 Maths Paper 1, Question 8

Part (a)

Part (b)(i)

Part (b)(ii)

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