2024 H2 Maths Paper 1 Q3 — A Cubic with a Conjugate Pair of Roots

The roots, livethe three roots

Real part of

w

−1.0

Imaginary part of

w

2.0

The real root, 3

w

and the third root The real axis, the mirror

Part (a)

1 mark

The equation $z^{3}+az^{2}+bz+c=0$ has roots $3$ and $w$ , where $w$ is complex. State a condition on $a$ , $b$ and $c$ for the third root to be $w^{*}$ .

Approach

One mark, one sentence. The conjugate root theorem does not say that complex roots always pair up — it says they pair up when the coefficients are real. The condition being asked for is that hypothesis.

$\therefore\ a$ , $b$ and $c$ are all real.

The reasoning behind it: if $a$ , $b$ , $c$ are real, then conjugating the whole equation leaves it unchanged apart from replacing $z$ by $z^{*}$ ,

\left(z^{3}+az^{2}+bz+c\right)^{*}=\left(z^{*}\right)^{3}+a\left(z^{*}\right)^{2}+bz^{*}+c

so if $z=w$ makes the left side zero, $z=w^{*}$ makes the right side zero too. Non-real roots therefore arrive in conjugate pairs.

Where the mark goes

" $a$ , $b$ , $c$ are real" is the answer. " $a$ , $b$ , $c$ are rational", "the coefficients are integers" and "the discriminant is negative" are all either too strong or beside the point. And note the direction of the theorem: real coefficients force the pairing; a conjugate pair appearing does not by itself force the coefficients to be real, though for a cubic with a real third root it does.

Part (b)

3 marks

Given that the condition in part (a) holds and that $w=-1+2i$ , find the values of $a$ , $b$ and $c$ .

Approach

Three roots are now known, so the cubic is known up to a constant — and it is monic, so there is no constant to find. Build it as a product of factors and expand. The trick that keeps it clean is to multiply the two complex factors together first: their product is real, and everything after that is ordinary algebra.

By part (a) the third root is $w^{*}=-1-2i$ , so the three roots are $3$ , $-1+2i$ and $-1-2i$ . Pairing the two complex ones:

\begin{aligned}w+w^{*}&=(-1+2i)+(-1-2i)=-2\\[2pt]w\,w^{*}&=\left|w\right|^{2}=(-1)^{2}+2^{2}=5\end{aligned}

Note — why $w\,w^{*}=\left|w\right|^{2}$ is worth the two seconds

Multiplying $(-1+2i)(-1-2i)$ out is fine, but the sum and product of a conjugate pair are always $2\operatorname{Re}(w)$ and $\left|w\right|^{2}$ — both real, both instant, and neither can go wrong on a sign. It is the same shortcut that turns the quadratic factor below into one line.

A quadratic with roots $w$ and $w^{*}$ is $z^{2}-\left(w+w^{*}\right)z+w\,w^{*}$ , so

z^{3}+az^{2}+bz+c\equiv(z-3)\left(z^{2}+2z+5\right)

Expanding the right-hand side:

\begin{aligned}(z-3)\left(z^{2}+2z+5\right)&=z^{3}+2z^{2}+5z-3z^{2}-6z-15\\[2pt]&=z^{3}+(2-3)z^{2}+(5-6)z-15\\[2pt]&=z^{3}-z^{2}-z-15\end{aligned}

Comparing coefficients with $z^{3}+az^{2}+bz+c$ :

To check your answer:

Substituting the known real root, $z=3$ : $27-9-3-15=0$ ✓. And the coefficients are all real, as part (a) requires — if any had come out complex, the conjugate assumption would have been wrong somewhere. A second check: the sum of the roots is $3+(-1+2i)+(-1-2i)=1$ , which must equal $-a$ , giving $a=-1$ ✓.

$\therefore\ a=-1$ , $b=-1$ , $c=-15$ .

Where the marks go

Substituting $z=-1+2i$ into $z^{3}+az^{2}+bz+c=0$ and comparing real and imaginary parts also works, but it needs $(-1+2i)^{3}$ and then a pair of simultaneous equations — three times the work and three times the chances to slip. Building the cubic from its factors is the intended route.

What this question was really testing

The conjugate root theorem, stated properly. Complex roots pair up because the coefficients are real. The hypothesis is the whole content of part (a).
Roots to polynomial. A monic cubic with known roots is $(z-r_{1})(z-r_{2})(z-r_{3})$ . Nothing needs to be solved; it needs to be multiplied out.
Pairing conjugates first. $(z-w)(z-w^{*})=z^{2}-2\operatorname{Re}(w)\,z+\left|w\right|^{2}$ is real, so the rest of the expansion never touches $i$ .
Checking with a root. Substituting the real root back into the finished cubic costs one line and catches every arithmetic slip in the expansion.

One real root, one complex root, and the condition that supplies the third

Part (a)

Part (b)

What this question was really testing

This is how we mark, and how we teach.