2020 H2 Maths Paper 1 Q6 — Worked Solution

Solution

8 marks

Approach

The equation has one real coefficient ( $t$ ) and two non-real ones, so a single complex equation in $z$ becomes, once we substitute the known root $z = k + ki$ , two real equations — real part $\text{=}0$ and imaginary part $\text{=}0$ — which we solve for $k$ and $t$ together.

Let $z = k + ki = k(1 + i)$ . We first compute $z^{2}$ :

z^{2} = k^{2}(1 + i)^{2}

= k^{2}\left( 1 + 2i + i^{2} \right)

= k^{2}(1 + 2i - 1)

= 2k^{2}i

Then $z^{2}(2 + i)$ :

z^{2}(2 + i) = 2k^{2}i(2 + i)

= 4k^{2}i + 2k^{2}i^{2}

= 4k^{2}i - 2k^{2}

And $- 8iz$ :

- 8iz = - 8i \cdot k(1 + i)

= - 8ki - 8ki^{2}

= - 8ki + 8k

Substituting all three into $z^{2}(2 + i) - 8iz + t = 0$ and grouping real and imaginary parts,

\left( 4k^{2}i - 2k^{2} \right) + ( - 8ki + 8k) + t = 0

\Rightarrow \left( - 2k^{2} + 8k + t \right) + \left( 4k^{2} - 8k \right)i = 0

Since $k$ and $t$ are real, both the real part and the imaginary part must vanish. From the imaginary part:

4k^{2} - 8k = 0

\Rightarrow 4k(k - 2) = 0

\Rightarrow k = 0\ \left( \text{rej as k is non}\text{-}\text{zero} \right),\ k = 2

Substituting $k = 2$ into the real part,

- 2(2)^{2} + 8(2) + t = 0

\Rightarrow - 8 + 16 + t = 0

\Rightarrow t = - 8

To check your answer: with

k = 2

z = 2 + 2i

; substituting into

z^{2}(2 + i) - 8iz + t = 0

with

t = - 8

z^{2} = (2 + 2i)^{2} = 8i

, so

8i(2 + i) - 8i(2 + 2i) - 8 = (16i - 8) - (16i - 16) - 8 = 0

. ✓

$\therefore$ $t = - 8$ , $k = 2$ , and one root is $z = 2 + 2i$ .

To find the other root, we use that for a quadratic $az^{2} + bz + c = 0$ , the sum of roots is $- \frac{b}{a}$ . Here $a = 2 + i$ , $b = - 8i$ , $c = t = - 8$ :

z_{1} + z_{2} = - \frac{b}{a} = \frac{8i}{2 + i}

Rationalising by multiplying by $\frac{2 - i}{2 - i}$ ,

\frac{8i}{2 + i}

= \frac{8i(2 - i)}{(2 + i)(2 - i)}

= \frac{16i - 8i^{2}}{5}

= \frac{16i + 8}{5}

= \frac{8}{5} + \frac{16}{5}i

Since $z_{1} = 2 + 2i$ ,

z_{2} = \left( \frac{8}{5} + \frac{16}{5}i \right) - (2 + 2i)

= \left( \frac{8}{5} - \frac{10}{5} \right) + \left( \frac{16}{5} - \frac{10}{5} \right)i

= - \frac{2}{5} + \frac{6}{5}i

To check your answer: solving

z^{2}(2 + i) - 8iz - 8 = 0

directly confirms the two roots are exactly

2 + 2i

and

- \frac{2}{5} + \frac{6}{5}i

. ✓

$\therefore$ the other root is $z = - \frac{2}{5} + \frac{6}{5}i$ .

2020 H2 Maths Paper 1, Question 6

Solution

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