Applications and Problem Solving

This final lesson is where the separate techniques of the strand — arithmetic, conjugates, Argand geometry, modulus–argument form, polar multiplication, loci, De Moivre and roots of unity — stop being topics and become a toolkit. Real AQA Further Maths questions rarely test one idea in isolation; they hand you a multi-part problem that demands you choose the right representation, switch fluently between Cartesian and polar, and assemble two or three results into a single argument. The skill here is strategy: recognising which tool a sub-question is reaching for, executing it cleanly, and checking the answer. This lesson drills exactly that across the six most common exam archetypes.

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Where this sits in AQA 7367

This is compulsory pure synthesis for Papers 1 & 2. By design it is the most AO3-heavy lesson of the strand (solve problems in mathematics, often in several steps and switching technique), while still drawing on AO1 routine execution and AO2 reasoning and proof. Every archetype below recombines earlier lessons: polynomial roots use the conjugate root theorem and division; the modulus–argument problems use polar multiplication and De Moivre; the loci problems use the four standard loci; the geometric proofs use roots of unity and rotation. Treat this as your exam-rehearsal lesson.

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Core strategy: choose the representation, then execute and check

A reliable four-step habit underlies every problem:

1. Read for the required form — Cartesian ($a + bi$a+bi), modulus–argument, or exponential. The wording ("in the form $a + bi$a+bi", "find $|z|$∣z∣ and $\arg z$argz") tells you where to finish.
2. Choose the representation that suits the operation. Use Cartesian for addition, subtraction and equating real/imaginary parts; use polar for multiplication, division, powers and roots. Mixed problems often need a switch mid-solution.
3. Show every step — examiners award method marks for the structure (conjugate written, polar form stated, parts equated), not just the final number.
4. Check — substitute back, verify a modulus or argument, confirm conjugate pairs, or test that a root really satisfies the equation. A 30-second check routinely rescues a dropped accuracy mark.

If the sub-question involves…	Reach for…
a polynomial with a given complex root	conjugate root theorem + real quadratic factor + division
powers, products or quotients	modulus–argument form + De Moivre
"sketch the locus" / "find where they meet"	the four standard loci + simultaneous equations
a proof about a triangle/polygon	rotation by multiplication + roots of unity
$\cos n\theta$cosnθ or $\cos^n\theta$cosnθ	De Moivre + binomial / $z + 1/z$z+1/z
sums/products of roots	symmetric functions (Vieta)

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Worked examples (with mark scheme)

Example 1 — a quartic from one complex root (Type: polynomial equations)

The quartic $z^4 + 2z^3 + 6z^2 + 8z + 8 = 0$z4+2z3+6z2+8z+8=0 has a root $z = -1 + i$z=−1+i. Find all four roots.

Real coefficients $\Rightarrow$⇒ the conjugate $z = -1 - i$z=−1−i is also a root. (M1 conjugate root.) These give the real quadratic factor

$(z - (-1 + i))(z - (-1 - i)) = (z + 1)^2 - (i)^2 = (z+1)^2 + 1 = z^2 + 2z + 2. \quad (\text{M1 form quadratic factor})$(z−(−1+i))(z−(−1−i))=(z+1)2−(i)2=(z+1)2+1=z2+2z+2.(M1 form quadratic factor)

Divide the quartic by $z^2 + 2z + 2$z2+2z+2:

$z^4 + 2z^3 + 6z^2 + 8z + 8 = (z^2 + 2z + 2)(z^2 + 4). \quad (\text{M1 polynomial division})$z4+2z3+6z2+8z+8=(z2+2z+2)(z2+4).(M1 polynomial division)

Solving $z^2 + 4 = 0$z2+4=0 gives $z = \pm 2i$z=±2i. So all four roots are $-1 + i,\ -1 - i,\ 2i,\ -2i$−1+i, −1−i, 2i, −2i. (A1 all four. Check the factorisation: $(z^2+2z+2)(z^2+4) = z^4 + 4z^2 + 2z^3 + 8z + 2z^2 + 8 = z^4 + 2z^3 + 6z^2 + 8z + 8$(z2+2z+2)(z2+4)=z4+4z2+2z3+8z+2z2+8=z4+2z3+6z2+8z+8 ✓.)

Example 2 — combining modulus, argument and algebra

Given $z = \dfrac{(1 + \sqrt 3\, i)^3}{(1 - i)^2}$z=(1−i)2(1+3​i)3​, find $|z|$∣z∣ and $\arg z$argz, and express $z$z in the form $a + bi$a+bi.

Work the numerator and denominator in polar form. Numerator: $1 + \sqrt 3\, i = 2\,\operatorname{cis}\frac{\pi}{3}$1+3​i=2cis3π​, so by De Moivre $(1 + \sqrt 3\, i)^3 = 2^3\,\operatorname{cis}\pi = -8$(1+3​i)3=23cisπ=−8. Denominator: $1 - i = \sqrt 2\,\operatorname{cis}\!\left(-\frac{\pi}{4}\right)$1−i=2​cis(−4π​), so $(1 - i)^2 = 2\,\operatorname{cis}\!\left(-\frac{\pi}{2}\right) = -2i$(1−i)2=2cis(−2π​)=−2i. (M1 each part to polar and powered.)

$z = \frac{-8}{-2i} = \frac{4}{i} = \frac{4}{i}\cdot\frac{-i}{-i} = \frac{-4i}{1} = -4i. \quad (\text{M1 divide / realise; A1})$z=−2i−8​=i4​=i4​⋅−i−i​=1−4i​=−4i.(M1 divide / realise; A1)

So $z = -4i$z=−4i, giving $|z| = 4$∣z∣=4 and $\arg z = -\frac{\pi}{2}$argz=−2π​. (M1 polar forms; M1 the division; A1 for $-4i$−4i with modulus and argument. Cross-check via polar: $|z| = \frac{8}{2} = 4$∣z∣=28​=4, $\arg z = \pi - \left(-\frac{\pi}{2}\right) = \frac{3\pi}{2} \equiv -\frac{\pi}{2}$argz=π−(−2π​)=23π​≡−2π​ ✓.)

Example 3 — two loci and their intersection

On one Argand diagram, sketch $C: |z - 2 - 2i| = 2$C:∣z−2−2i∣=2 and $L: \arg(z - 2 - 2i) = \frac{\pi}{4}$L:arg(z−2−2i)=4π​, and find their point of intersection.

$C$C is a circle, centre $(2, 2)$(2,2), radius $2$2. $L$L is the half-line from $(2, 2)$(2,2) (excluded) in the direction $\frac{\pi}{4}$4π​. (M1 both loci.) On $L$L, points have the form $z = (2 + t) + (2 + t)i$z=(2+t)+(2+t)i for $t > 0$t>0 (equal steps in the real and imaginary directions). Substitute into $C$C:

$|z - 2 - 2i| = |t + ti| = |t|\,|1 + i| = t\sqrt 2 = 2 \;\Longrightarrow\; t = \frac{2}{\sqrt 2} = \sqrt 2. \quad (\text{M1 substitute; A1 } t)$∣z−2−2i∣=∣t+ti∣=∣t∣∣1+i∣=t2​=2⟹t=2​2​=2​.(M1 substitute; A1 t)

So the intersection is $z = (2 + \sqrt 2) + (2 + \sqrt 2)i$z=(2+2​)+(2+2​)i. (M1 loci; M1 parametrise and substitute; A1 intersection. There is just one point because the ray starts at the centre and leaves the disc once.)

Example 4 — De Moivre to a trigonometric identity, then an application

Use De Moivre's theorem to show $\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$cos5θ=16cos5θ−20cos3θ+5cosθ.

By De Moivre, $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta$(cosθ+isinθ)5=cos5θ+isin5θ. Expand the left side ($c = \cos\theta, s = \sin\theta$c=cosθ,s=sinθ):

$(c + is)^5 = c^5 + 5ic^4 s - 10c^3 s^2 - 10ic^2 s^3 + 5cs^4 + is^5. \quad (\text{M1 binomial; M1 powers of } i)$(c+is)5=c5+5ic4s−10c3s2−10ic2s3+5cs4+is5.(M1 binomial; M1 powers of i)

Take the real part and substitute $s^2 = 1 - c^2$s2=1−c2:

$\begin{aligned} \cos 5\theta &= c^5 - 10c^3 s^2 + 5cs^4 = c^5 - 10c^3(1 - c^2) + 5c(1 - c^2)^2 \\ &= c^5 - 10c^3 + 10c^5 + 5c - 10c^3 + 5c^5 = 16c^5 - 20c^3 + 5c. \end{aligned} \quad (\text{A1})$cos5θ​=c5−10c3s2+5cs4=c5−10c3(1−c2)+5c(1−c2)2=c5−10c3+10c5+5c−10c3+5c5=16c5−20c3+5c.​(A1)

Hence $\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$cos5θ=16cos5θ−20cos3θ+5cosθ. (M1 expansion; M1 simplify powers of $i$i and take real part; A1 the identity after $s^2 = 1 - c^2$s2=1−c2.)

Example 5 — square root of a complex number (equate parts)

Find the square roots of $3 + 4i$3+4i, i.e. solve $w^2 = 3 + 4i$w2=3+4i.

Let $w = a + bi$w=a+bi with $a, b$a,b real. Then $w^2 = a^2 - b^2 + 2abi$w2=a2−b2+2abi, so equating parts with $3 + 4i$3+4i:

$a^2 - b^2 = 3, \qquad 2ab = 4 \;\Rightarrow\; b = \frac{2}{a}. \quad (\text{M1 set up two real equations})$a2−b2=3,2ab=4⇒b=a2​.(M1 set up two real equations)

Substitute $b = 2/a$b=2/a into the first: $a^2 - \dfrac{4}{a^2} = 3$a2−a24​=3, so $a^4 - 3a^2 - 4 = 0$a4−3a2−4=0, i.e. $(a^2 - 4)(a^2 + 1) = 0$(a2−4)(a2+1)=0. (M1 form and solve the quadratic in $a^2$a2.) Since $a$a is real, $a^2 = 4$a2=4, giving $a = \pm 2$a=±2; then $b = 2/a$b=2/a gives $b = \pm 1$b=±1 with the same sign. So the square roots are $w = 2 + i$w=2+i and $w = -2 - i$w=−2−i, i.e. $\pm(2 + i)$±(2+i). (A1 both roots.) (M1 equate parts; M1 the quadratic in $a^2$a2 rejecting $a^2 = -1$a2=−1; A1 for $\pm(2 + i)$±(2+i). Check: $(2 + i)^2 = 4 + 4i + i^2 = 3 + 4i$(2+i)2=4+4i+i2=3+4i ✓.)

Example 6 — a modulus relation (use $|w|^2 = w\bar w$∣w∣2=wwˉ)

Show that $|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2)$∣z1​+z2​∣2+∣z1​−z2​∣2=2(∣z1​∣2+∣z2​∣2) for all complex $z_1, z_2$z1​,z2​ (the parallelogram law).

Use $|w|^2 = w\bar w$∣w∣2=wwˉ and $\overline{z_1 \pm z_2} = \bar z_1 \pm \bar z_2$z1​±z2​​=zˉ1​±zˉ2​:

$|z_1 + z_2|^2 = (z_1 + z_2)(\bar z_1 + \bar z_2) = |z_1|^2 + z_1\bar z_2 + \bar z_1 z_2 + |z_2|^2, \quad (\text{M1 expand the first})$∣z1​+z2​∣2=(z1​+z2​)(zˉ1​+zˉ2​)=∣z1​∣2+z1​zˉ2​+zˉ1​z2​+∣z2​∣2,(M1 expand the first) $|z_1 - z_2|^2 = (z_1 - z_2)(\bar z_1 - \bar z_2) = |z_1|^2 - z_1\bar z_2 - \bar z_1 z_2 + |z_2|^2. \quad (\text{M1 expand the second})$∣z1​−z2​∣2=(z1​−z2​)(zˉ1​−zˉ2​)=∣z1​∣2−z1​zˉ2​−zˉ1​z2​+∣z2​∣2.(M1 expand the second)

Adding, the cross terms $\pm(z_1\bar z_2 + \bar z_1 z_2)$±(z1​zˉ2​+zˉ1​z2​) cancel, leaving $2|z_1|^2 + 2|z_2|^2$2∣z1​∣2+2∣z2​∣2, as required. (M1 each expansion via $w\bar w$wwˉ; A1 the cancellation and conclusion. Geometrically: the sum of the squares of a parallelogram's diagonals equals the sum of the squares of its four sides.)

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Specimen-style exam question

(specimen-style — not from any past paper)

The cubic $z^3 + pz + q = 0$z3+pz+q=0, with $p, q$p,q real, has roots $2$2, $-1 + i\sqrt 3$−1+i3​ and $-1 - i\sqrt 3$−1−i3​. (a) Find $p$p and $q$q. (b) Hence describe the three roots geometrically.

(a) Use the symmetric functions of the roots for $z^3 + 0\cdot z^2 + pz + q = 0$z3+0⋅z2+pz+q=0. Sum of roots $= 2 + (-1 + i\sqrt 3) + (-1 - i\sqrt 3) = 0$=2+(−1+i3​)+(−1−i3​)=0, consistent with the absent $z^2$z2 term ✓. Sum of products in pairs:

$2(-1 + i\sqrt 3) + 2(-1 - i\sqrt 3) + (-1 + i\sqrt 3)(-1 - i\sqrt 3) = -2 + 2i\sqrt 3 - 2 - 2i\sqrt 3 + (1 + 3) = 0,$2(−1+i3​)+2(−1−i3​)+(−1+i3​)(−1−i3​)=−2+2i3​−2−2i3​+(1+3)=0,

and this sum equals $p$p, so $p = 0$p=0. Product of roots $= 2\big((-1)^2 + (\sqrt 3)^2\big) = 2(1 + 3) = 8$=2((−1)2+(3​)2)=2(1+3)=8; for a monic cubic $\alpha\beta\gamma = -q$αβγ=−q, so $8 = -q$8=−q, giving $q = -8$q=−8.

So the cubic is $z^3 - 8 = 0$z3−8=0, i.e. $p = 0,\ q = -8$p=0, q=−8.

(b) $z^3 = 8$z3=8 has the three cube roots $2,\ 2\omega,\ 2\omega^2$2, 2ω, 2ω2 with $\omega = e^{2\pi i/3}$ω=e2πi/3; here $2\omega = -1 + i\sqrt 3$2ω=−1+i3​ and $2\omega^2 = -1 - i\sqrt 3$2ω2=−1−i3​. Geometrically the roots are the vertices of an equilateral triangle on the circle $|z| = 2$∣z∣=2, equally spaced by $\frac{2\pi}{3}$32π​. (The "Hence" rewards recognising $z^3 - 8 = 0$z3−8=0 as a roots-of-unity problem rather than re-deriving from scratch.)

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Synoptic links

• Conjugate root theorem & polynomial division (lesson 3) — the backbone of every "given a root, find the rest" problem.
• Roots and coefficients / Vieta (further-algebra strand) — sums and products of roots, used in the specimen question.
• De Moivre + binomial / $z + 1/z$z+1/z (lesson 8) — the route to $\cos n\theta$cosnθ and $\cos^n\theta$cosnθ.
• Loci (lesson 7) — circles, half-lines and bisectors, plus simultaneous solution for intersections.
• Roots of unity (lesson 9) — equilateral-triangle and regular-polygon geometry.
• Rotation by multiplication (lesson 6) — the engine of the geometric-proof archetype.

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Mark-scheme literacy