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How do complex numbers extend the real numbers, and how do you represent, manipulate and find roots of them?

Arithmetic of complex numbers, the Argand diagram, modulus-argument form, de Moivre's theorem, nth roots, complex roots of polynomials and loci.

A focused answer to the Edexcel A-Level Further Mathematics complex numbers content, covering arithmetic, the Argand diagram, modulus-argument and exponential form, de Moivre's theorem, nth roots, roots of unity, complex roots of polynomials and loci.

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  1. What this dot point is asking
  2. Arithmetic and the conjugate
  3. The Argand diagram and modulus-argument form
  4. De Moivre's theorem and nth roots
  5. Roots of polynomials
  6. Loci on the Argand diagram
  7. Examples in context
  8. Try this

What this dot point is asking

Edexcel wants you to work fluently with complex numbers in Cartesian, modulus-argument and exponential form, plot them on an Argand diagram, use de Moivre's theorem to find powers and roots, solve polynomial equations with complex roots, and sketch loci defined by modulus and argument conditions. Complex numbers appear in every Core Pure paper, both as standalone questions and as a tool inside differential-equation and matrix problems.

Arithmetic and the conjugate

For z=x+iyz = x + iy the real part is Re(z)=x\operatorname{Re}(z) = x and the imaginary part is Im(z)=y\operatorname{Im}(z) = y. Addition is component-wise: (a+bi)+(c+di)=(a+c)+(b+d)i(a + bi) + (c + di) = (a + c) + (b + d)i. Multiplication uses i2=1i^2 = -1:

(a+bi)(c+di)=(acbd)+(ad+bc)i.(a + bi)(c + di) = (ac - bd) + (ad + bc)i.

The complex conjugate of z=x+iyz = x + iy is z=xiyz^* = x - iy (Edexcel also writes zˉ\bar{z}). Multiplying a number by its conjugate clears the imaginary part:

zz=(x+iy)(xiy)=x2+y2=z2,zz^* = (x + iy)(x - iy) = x^2 + y^2 = |z|^2,

which is always real and non-negative. This is the key to division: to divide, multiply numerator and denominator by the conjugate of the denominator so the denominator becomes real.

The Argand diagram and modulus-argument form

On an Argand diagram the horizontal axis is the real part and the vertical axis is the imaginary part, so z=x+iyz = x + iy is the point (x,y)(x, y). The modulus is the distance from the origin,

r=z=x2+y2,r = |z| = \sqrt{x^2 + y^2},

and the argument θ=argz\theta = \arg z is the angle measured anticlockwise from the positive real axis, taken in the principal range π<θπ-\pi < \theta \le \pi. Always sketch the point first so you choose the correct quadrant for θ\theta, because arctanyx\arctan\frac{y}{x} alone cannot tell the second quadrant from the fourth.

De Moivre's theorem and nth roots

For any integer nn,

(cosθ+isinθ)n=cosnθ+isinnθ.(\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta.

Equivalently, in exponential form, (reiθ)n=rneinθ(re^{i\theta})^n = r^n e^{in\theta}. This single result is the engine for three things: raising a complex number to a power, finding the nn distinct nnth roots of a complex number, and deriving multiple-angle identities such as cos3θ=4cos3θ3cosθ\cos 3\theta = 4\cos^3\theta - 3\cos\theta by expanding (cosθ+isinθ)3(\cos\theta + i\sin\theta)^3 and comparing real parts.

To find nnth roots, write the number with a general argument by adding 2πk2\pi k, then divide the argument by nn and let kk run over nn consecutive integers.

Roots of polynomials

For a polynomial with real coefficients, complex roots occur in conjugate pairs. If 2+3i2 + 3i is a root then so is 23i2 - 3i. The sum and product of a conjugate pair are both real, so they generate a real quadratic factor.

Loci on the Argand diagram

Conditions on zz describe curves in the plane. The three standard loci are:

  • za=r|z - a| = r is a circle of radius rr centred at the point aa.
  • za=zb|z - a| = |z - b| is the perpendicular bisector of the segment joining aa and bb.
  • arg(za)=θ\arg(z - a) = \theta is a half-line (ray) starting at aa and pointing at angle θ\theta, with the start point itself excluded.

For exam work, translate the algebra into geometry, sketch it, and then read off distances. For a maximum or minimum of z|z| on a circle, use the centre-to-origin distance plus or minus the radius.

Examples in context

Complex numbers thread through the whole Core Pure course. De Moivre's theorem is the standard route to summing series such as coskθ\sum \cos k\theta by treating them as the real part of a geometric series in eiθe^{i\theta}. The exponential form z=reiθz = re^{i\theta} reappears when you solve second-order differential equations with complex auxiliary roots, where the solution e(α±iβ)xe^{(\alpha \pm i\beta)x} becomes eαx(Acosβx+Bsinβx)e^{\alpha x}(A\cos\beta x + B\sin\beta x) via Euler's formula. In matrices, the eigenvalues of a rotation matrix are complex conjugates cosθ±isinθ\cos\theta \pm i\sin\theta, tying the topic back to modulus-argument form. Loci questions, meanwhile, blend complex algebra with coordinate geometry, and frequently ask for the greatest or least value of z|z| or argz\arg z subject to a constraint.

Try this

Q1. Express z=1+iz = 1 + i in modulus-argument form. [3 marks]

  • Cue. r=2r = \sqrt{2}, θ=π4\theta = \frac{\pi}{4}, so z=2(cosπ4+isinπ4)z = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right).

Q2. Given that 32i3 - 2i is a root of x2+bx+c=0x^2 + bx + c = 0 with real bb and cc, find bb and cc. [3 marks]

  • Cue. The other root is 3+2i3 + 2i; sum of roots =6=b= 6 = -b so b=6b = -6, product =13=c= 13 = c.

Q3. Find the four fourth roots of 16-16, giving them in the form reiθre^{i\theta}. [5 marks]

  • Cue. 16=16ei(π+2πk)-16 = 16e^{i(\pi + 2\pi k)}, so z=2ei(π+2πk)/4z = 2e^{i(\pi + 2\pi k)/4} for k=0,1,2,3k = 0, 1, 2, 3, giving arguments π4,3π4,3π4,π4\frac{\pi}{4}, \frac{3\pi}{4}, -\frac{3\pi}{4}, -\frac{\pi}{4}.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20196 marksThe complex number z=22iz = 2 - 2i. Express zz in the form r(cosθ+isinθ)r(\cos\theta + i\sin\theta), where r>0r > 0 and π<θπ-\pi < \theta \le \pi. Hence find z4z^4, giving your answer in the form a+bia + bi.
Show worked answer →

Find the modulus and argument first, then apply de Moivre.

Modulus: r=z=22+(2)2=8=22r = |z| = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2} (M1 A1).

Argument: zz lies in the fourth quadrant, so θ=arctan22=π4\theta = -\arctan\frac{2}{2} = -\frac{\pi}{4} (M1 A1). Hence z=22(cos(π4)+isin(π4))z = 2\sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right).

By de Moivre, z4=(22)4(cos(π)+isin(π))z^4 = (2\sqrt{2})^4\left(\cos\left(-\pi\right) + i\sin\left(-\pi\right)\right). Now (22)4=(22)2×(22)2=8×8=64(2\sqrt{2})^4 = (2\sqrt{2})^2 \times (2\sqrt{2})^2 = 8 \times 8 = 64 (M1), and cos(π)=1\cos(-\pi) = -1, sin(π)=0\sin(-\pi) = 0, so z4=64(1+0i)=64z^4 = 64(-1 + 0i) = -64 (A1).

Edexcel 20217 marksSolve the equation z3=8iz^3 = 8i, giving the three roots in the form reiθr e^{i\theta} where r>0r > 0 and π<θπ-\pi < \theta \le \pi. Show the roots on a single Argand diagram.
Show worked answer →

Write the right-hand side in exponential form, then take cube roots by adding 2πk2\pi k.

8i8i has modulus 88 and argument π2\frac{\pi}{2}, so 8i=8ei(π/2+2πk)8i = 8 e^{i(\pi/2 + 2\pi k)} for integer kk (M1).

Then z=81/3ei(π/2+2πk)/3=2ei(π/6+2πk/3)z = 8^{1/3} e^{i(\pi/2 + 2\pi k)/3} = 2 e^{i(\pi/6 + 2\pi k/3)} (M1 A1 for modulus 22, A1 for the π6\frac{\pi}{6} base argument).

Taking k=0,1,1k = 0, 1, -1 keeps the arguments in range: z=2eiπ/6z = 2e^{i\pi/6}, z=2ei5π/6z = 2e^{i5\pi/6}, z=2eiπ/2z = 2e^{-i\pi/2} (A1 A1). On the Argand diagram the three points lie on a circle of radius 22, equally spaced 2π3\frac{2\pi}{3} apart (B1 for a correct sketch with equal spacing).

Edexcel 20235 marksThe point PP represents the complex number zz on an Argand diagram. Given that z34i=5|z - 3 - 4i| = 5, find the Cartesian equation of the locus of PP and determine the maximum value of z|z|.
Show worked answer →

Interpret the modulus condition as a circle, then use geometry for the maximum.

z(3+4i)=5|z - (3 + 4i)| = 5 is a circle centre (3,4)(3, 4) radius 55 (M1 A1), with Cartesian equation (x3)2+(y4)2=25(x - 3)^2 + (y - 4)^2 = 25 (A1).

z|z| is the distance from the origin to PP. The centre is at distance 32+42=5\sqrt{3^2 + 4^2} = 5 from the origin (M1). The maximum of z|z| is the distance to the centre plus the radius: 5+5=105 + 5 = 10 (A1). (Note the circle passes through the origin, so the minimum is 00.)

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