Applications and Problem Solving

This final lesson brings together all the complex number techniques covered in this course. We focus on exam-style problem solving, combining multiple ideas within a single question — exactly as you will encounter in the AQA Further Mathematics exam.

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Strategy for Complex Number Problems

1. Read the question carefully — identify what form is needed (Cartesian, polar, exponential).
2. Choose the right representation — Cartesian for addition/subtraction, polar for multiplication/powers/roots.
3. Show all working — examiners award method marks.
4. Check your answer — substitute back, verify modulus/argument, check conjugate pairs.

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Type 1: Polynomial Equations

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

Solution:

Since coefficients are real, $z = -1 - i$z=−1−i is also a root.

Quadratic factor: $(z - (-1 + i))(z - (-1 - i)) = (z + 1)^2 + 1 = z^2 + 2z + 2$(z−(−1+i))(z−(−1−i))=(z+1)2+1=z2+2z+2.

Divide $z^4 + 2z^3 + 6z^2 + 8z + 8$z4+2z3+6z2+8z+8 by $z^2 + 2z + 2$z2+2z+2:

$z^4 + 2z^3 + 6z^2 + 8z + 8 = (z^2 + 2z + 2)(z^2 + 4)$z4+2z3+6z2+8z+8=(z2+2z+2)(z2+4)

Solving $z^2 + 4 = 0$z2+4=0: $z = \pm 2i$z=±2i.

All four roots: $-1 + i, -1 - i, 2i, -2i$−1+i,−1−i,2i,−2i.

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Type 2: Combining Modulus, Argument, and Algebra

Problem 2: Given $z = \frac{(1 + \sqrt{3}\, i)^3}{(1 - i)^2}$z=(1−i)2(1+3​i)3​, find $|z|$∣z∣ and $\arg(z)$arg(z), and express $z$z in Cartesian form.

Solution:

Numerator: $1 + \sqrt{3}\, i$1+3​i has $r = 2$r=2, $\theta = \frac{\pi}{3}$θ=3π​.

$(1 + \sqrt{3}\, i)^3 = 8\left(\cos\pi + i\sin\pi\right) = -8$(1+3​i)3=8(cosπ+isinπ)=−8

Denominator: $1 - i$1−i has $r = \sqrt{2}$r=2​, $\theta = -\frac{\pi}{4}$θ=−4π​.

$(1 - i)^2 = 2\left(\cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2})\right) = -2i$(1−i)2=2(cos(−2π​)+isin(−2π​))=−2i

$z = \frac{-8}{-2i} = \frac{8}{2i} = \frac{4}{i} = \frac{4}{i} \cdot \frac{-i}{-i} = \frac{-4i}{1} = -4i$z=−2i−8​=2i8​=i4​=i4​⋅−i−i​=1−4i​=−4i

$|z| = 4$∣z∣=4, $\arg(z) = -\frac{\pi}{2}$arg(z)=−2π​.

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Type 3: Loci and Regions

Problem 3: On a single Argand diagram, sketch the loci:

(a) $|z - 2 - 2i| = 2$∣z−2−2i∣=2

(b) $\arg(z - 2 - 2i) = \frac{\pi}{4}$arg(z−2−2i)=4π​

Find the point(s) where these loci intersect.

Solution:

(a) Circle centred at $(2, 2)$(2,2) with radius 2.

(b) Half-line from $(2, 2)$(2,2) at angle $\frac{\pi}{4}$4π​ (i.e., along the direction making 45° with the positive real axis).

The half-line in direction $\frac{\pi}{4}$4π​ from $(2, 2)$(2,2) has parametric form $z = (2 + t) + (2 + t)i$z=(2+t)+(2+t)i for $t > 0$t>0.

Substituting into the circle equation:

$|(2 + t) + (2 + t)i - 2 - 2i| = |t + ti| = |t||1 + i| = t\sqrt{2} = 2$∣(2+t)+(2+t)i−2−2i∣=∣t+ti∣=∣t∣∣1+i∣=t2​=2

So $t = \frac{2}{\sqrt{2}} = \sqrt{2}$t=2​2​=2​.

The intersection point is $z = (2 + \sqrt{2}) + (2 + \sqrt{2})i$z=(2+2​)+(2+2​)i.

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Type 4: Proving Geometric Properties

Problem 4: Points $A$A, $B$B, $C$C on an Argand diagram correspond to complex numbers $z_1$z1​, $z_2$z2​, $z_3$z3​ respectively. Show that triangle $ABC$ABC is equilateral if and only if:

$z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$z12​+z22​+z32​=z1​z2​+z2​z3​+z3​z1​

Solution:

Triangle $ABC$ABC is equilateral if and only if $z_3$z3​ can be obtained from $z_1$z1​ and $z_2$z2​ by a rotation of $\pm 60°$±60° about $z_1$z1​.

This is equivalent to: $\frac{z_3 - z_1}{z_2 - z_1} = \omega$z2​−z1​z3​−z1​​=ω where $\omega$ω is a primitive cube root of unity (since $|\omega| = 1$∣ω∣=1 and $\arg(\omega) = \pm \frac{2\pi}{3}$arg(ω)=±32π​, but by symmetry we can use the condition cyclically).

The condition is equivalent to $z_1^2 + z_2^2 + z_3^2 - z_1 z_2 - z_2 z_3 - z_3 z_1 = 0$z12​+z22​+z32​−z1​z2​−z2​z3​−z3​z1​=0, which can be verified by factoring:

$z_1^2 + z_2^2 + z_3^2 - z_1 z_2 - z_2 z_3 - z_3 z_1 = \frac{1}{2}[(z_1 - z_2)^2 + (z_2 - z_3)^2 + (z_3 - z_1)^2]$z12​+z22​+z32​−z1​z2​−z2​z3​−z3​z1​=21​[(z1​−z2​)2+(z2​−z3​)2+(z3​−z1​)2]

Wait — this is zero only when all three are equal. Instead, the correct characterisation uses the rotation argument: the triangle is equilateral iff $(z_1 - z_2)\omega = z_1 - z_3$(z1​−z2​)ω=z1​−z3​ or $(z_1 - z_2)\omega^2 = z_1 - z_3$(z1​−z2​)ω2=z1​−z3​, where $\omega = e^{2\pi i/3}$ω=e2πi/3. Since $1 + \omega + \omega^2 = 0$1+ω+ω2=0, this leads to:

$z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$z12​+z22​+z32​=z1​z2​+z2​z3​+z3​z1​

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Type 5: De Moivre and Trigonometric Results

Problem 5: Use De Moivre's Theorem to show that:

$\cos(5\theta) = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$cos(5θ)=16cos5θ−20cos3θ+5cosθ

Solution:

By De Moivre: $(\cos\theta + i\sin\theta)^5 = \cos(5\theta) + i\sin(5\theta)$(cosθ+isinθ)5=cos(5θ)+isin(5θ).

Expanding the left side using the binomial theorem:

$\binom{5}{0}c^5 + \binom{5}{1}c^4(is) + \binom{5}{2}c^3(is)^2 + \binom{5}{3}c^2(is)^3 + \binom{5}{4}c(is)^4 + \binom{5}{5}(is)^5$(05​)c5+(15​)c4(is)+(25​)c3(is)2+(35​)c2(is)3+(45​)c(is)4+(55​)(is)5

where $c = \cos\theta$c=cosθ, $s = \sin\theta$s=sinθ.