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Edexcel A-level Further Maths revision notes

Complex numbers

Section CP-2
Both years
Both years: this holds AS subject content and content the exam board adds beyond it for the full A-level.
11 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section CP-2

Checked against Edexcel 9FM0 section CP-2. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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CP-2.1

Solve any quadratic equation with real coefficients. Solve cubic or quartic equations with real coefficients.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A quadratic with real coefficients may have two real roots, one repeated real root or a complex-conjugate pair, determined by the sign of b24acb^2-4ac. Use factorisation, completing the square or x=b±b24ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}, with q=iq\sqrt{-q}=i\sqrt q for q>0q>0.
  • For a cubic or quartic, sufficient information will identify a root or factor.
  • Apply the factor theorem or polynomial division, then solve every remaining quadratic factor.
  • A real polynomial cannot have an isolated non-real root: its conjugate is also a root.
  • A complete solution must list all roots, including repeated and non-real roots, rather than stopping after the supplied factor is used.
Worked example

Given that z=1z=1 is a root of z35z2+17z13=0z^3-5z^2+17z-13=0, solve the equation completely.

  1. 1.The factor theorem gives the factor z1z-1.
  2. 2.Polynomial division gives z35z2+17z13=(z1)(z24z+13)z^3-5z^2+17z-13=(z-1)(z^2-4z+13).
  3. 3.z24z+13=(z2)2+9z^2-4z+13=(z-2)^2+9, so (z2)2=9(z-2)^2=-9.
  4. 4.Hence z=1z=1 or z=2±3iz=2\pm3i.

Answer: z=1z=1, z=2+3iz=2+3i or z=23iz=2-3i.

Common mistakes

  • Don't fall into the trap of stopping after finding the supplied real root and leaving the remaining quadratic unsolved.
  • Don't fall into the trap of writing 36=6\sqrt{-36}=-6 instead of 36=6i\sqrt{-36}=6i.
  • Don't fall into the trap of losing a repeated root when reporting the complete solution of a quartic.

Exam tip

For a 'solve completely' question, show the factor division and state every real and non-real root.

Tier 1 · Easy

ORIGINAL

1.

Solve x24x+13=0x^2-4x+13=0, giving the roots in the form a+iba+ib.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Given that x=2x=2 is a root of x32x2+5x10=0x^3-2x^2+5x-10=0, solve the equation completely.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Given that x2+3x+6x^2+3x+6 is a factor of x4+3x2+36x^4+3x^2+36, solve x4+3x2+36=0x^4+3x^2+36=0 completely.

(5)

(Total for Question 1 is 5 marks)

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CP-2.2

Add, subtract, multiply and divide complex numbers in the form x + iy with x and y real. Understand and use the terms 'real part' and 'imaginary part'.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For z=x+iyz=x+iy, where x,yRx,y\in\mathbb R, the real part is Re(z)=x\operatorname{Re}(z)=x and the imaginary part is Im(z)=y\operatorname{Im}(z)=y; the imaginary part is not iyiy. Its modulus is the non-negative length z=x2+y2|z|=\sqrt{x^2+y^2} and an argument is its directed angle from the positive real axis.
  • Addition and subtraction combine corresponding parts.
  • Multiplication uses ordinary expansion followed by i2=1i^2=-1.
  • To divide by c+dic+di, multiply numerator and denominator by its conjugate cdic-di so the denominator becomes c2+d2c^2+d^2, then collect the result into Cartesian form.
  • Equality of complex numbers means equality of their real parts and equality of their imaginary parts, producing two real equations.
Worked example

Find real numbers xx and yy if (2i)(x+iy)=7+4i(2-i)(x+iy)=7+4i.

  1. 1.Expand: (2i)(x+iy)=(2x+y)+i(2yx)(2-i)(x+iy)=(2x+y)+i(2y-x).
  2. 2.Equate real parts to obtain 2x+y=72x+y=7.
  3. 3.Equate imaginary parts to obtain 2yx=42y-x=4.
  4. 4.Solving simultaneously gives x=2x=2 and y=3y=3.

Answer: x=2x=2 and y=3y=3.

Common mistakes

  • Don't fall into the trap of stating Im(x+iy)=iy\operatorname{Im}(x+iy)=iy instead of the real coefficient yy.
  • Don't fall into the trap of using i2=1i^2=1 when expanding a product.
  • Don't fall into the trap of dividing real and imaginary parts separately instead of rationalising with the conjugate.

Exam tip

After any Cartesian calculation, collect the result as x+iyx+iy before reading off or equating its parts.

Tier 1 · Easy

ORIGINAL

1.

Let z=(32i)+(5+7i)z=(3-2i)+(5+7i). Find zz, Re(z)\operatorname{Re}(z) and Im(z)\operatorname{Im}(z).

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Express 4+i2i\dfrac{4+i}{2-i} in the form x+iyx+iy.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

Express (2+3i)(14i)3+i\dfrac{(2+3i)(1-4i)}{3+i} in the form x+iyx+iy.

(5)

(Total for Question 1 is 5 marks)

CP-2.3

Understand and use the complex conjugate. Know that non-real roots of polynomial equations with real coefficients occur in conjugate pairs.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The complex conjugate of z=x+iyz=x+iy is z=xiy\overline z=x-iy, obtained by reflecting the point for zz in the real axis. Important identities are z+z=2Re(z)z+\overline z=2\operatorname{Re}(z) and zz=x2+y2=z2z\overline z=x^2+y^2=|z|^2, which is real.
  • If a polynomial has real coefficients and a+iba+ib is a non-real root, conjugating the whole equation shows that aiba-ib is also a root.
  • The pair produces the real quadratic factor [x(a+ib)][x(aib)]=(xa)2+b2[x-(a+ib)][x-(a-ib)]=(x-a)^2+b^2.
  • This rule depends on the coefficients being real.
  • In polynomial questions, use the conjugate root before dividing or comparing coefficients to find the remaining real factor.
Worked example

A monic cubic with real coefficients has roots 3+2i3+2i, 32i3-2i and 1-1. Find the polynomial in expanded form.

  1. 1.The conjugate pair gives (x32i)(x3+2i)=(x3)2+4(x-3-2i)(x-3+2i)=(x-3)^2+4.
  2. 2.This simplifies to x26x+13x^2-6x+13.
  3. 3.Include the third factor: (x26x+13)(x+1)(x^2-6x+13)(x+1).
  4. 4.Expansion gives x35x2+7x+13x^3-5x^2+7x+13.

Answer: x35x2+7x+13x^3-5x^2+7x+13.

Common mistakes

  • Don't fall into the trap of replacing a+iba+ib by aib-a-ib instead of by its conjugate aiba-ib.
  • Don't fall into the trap of assuming conjugate pairing when the polynomial coefficients are not all real.
  • Don't fall into the trap of expanding the pair as (xa)2b2(x-a)^2-b^2 instead of (xa)2+b2(x-a)^2+b^2.

Exam tip

When a real-coefficient polynomial has one non-real root, write its conjugate and the resulting real quadratic factor immediately.

Tier 1 · Easy

ORIGINAL

1.

For z=54iz=5-4i, write down z\overline z and calculate zzz\overline z.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

The polynomial p(x)=x3x27x+15p(x)=x^3-x^2-7x+15 has real coefficients, and 2+i2+i is a root. Solve p(x)=0p(x)=0 completely.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

A monic quartic polynomial has real coefficients. Two of its roots are 5+i5+i and 1+3i-1+3i. Determine the polynomial in expanded form.

(6)

(Total for Question 1 is 6 marks)

CP-2.4

Use and interpret Argand diagrams.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An Argand diagram represents z=x+iyz=x+iy by the point (x,y)(x,y): the horizontal axis is the real axis and the vertical axis is the imaginary axis.
  • Addition of complex numbers is vector addition, so the point representing z1+z2z_1+z_2 is obtained by completing a parallelogram.
  • The difference z2z1z_2-z_1 is the directed displacement from the point representing z1z_1 to the point representing z2z_2, and z2z1|z_2-z_1| is the corresponding distance.
  • Coordinate geometry can prove gradients, perpendicularity, lengths and areas.
  • A labelled sketch should show exact complex numbers or coordinates, but conclusions such as a right angle or equal length still require algebraic justification.
Complex addition shown as vector addition on an Argand diagram.
Worked example

The point AA is represented by the complex number a=2+ia=2+i, and the point BB is represented by b=5+5ib=5+5i. Find ba|b-a| and the complex number represented by the midpoint of ABAB.

  1. 1.ba=(5+5i)(2+i)=3+4ib-a=(5+5i)-(2+i)=3+4i.
  2. 2.ba=32+42=5|b-a|=\sqrt{3^2+4^2}=5.
  3. 3.The complex number represented by the midpoint is a+b2=7+6i2\dfrac{a+b}{2}=\dfrac{7+6i}{2}.

Answer: AB=5AB=5 and the midpoint is represented by 72+3i\dfrac72+3i.

Common mistakes

  • Don't fall into the trap of plotting x+iyx+iy at (y,x)(y,x) by interchanging the real and imaginary axes.
  • Don't fall into the trap of using z2+z1|z_2+z_1| for the distance between two represented points.
  • Don't fall into the trap of claiming a geometric property from an approximate sketch without an algebraic calculation.

Exam tip

For an Argand-diagram distance, form the difference of the two represented complex numbers first and then take its modulus.

Tier 1 · Easy

ORIGINAL

1.

The complex number z=2+3iz=-2+3i is represented by PP on an Argand diagram. State the coordinates of PP and the quadrant in which it lies.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

The points AA, BB and CC are represented by the complex numbers 1+i1+i, 5+i5+i and 1+4i1+4i, respectively. Determine the exact area of triangle ABCABC.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

The point AA is represented by the complex number 3+i3+i, the point BB is represented by 1+3i-1+3i, and OO is the origin. Prove that triangle OABOAB is right-angled and isosceles, and find its exact area.

(5)

(Total for Question 1 is 5 marks)

CP-2.5

Convert between the Cartesian form and the modulus-argument form of a complex number.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For z=x+iyz=x+iy, the modulus is r=z=x2+y2r=|z|=\sqrt{x^2+y^2} and an argument θ\theta is the directed angle from the positive real axis to the vector representing zz. Thus z=r(cosθ+isinθ)z=r(\cos\theta+i\sin\theta), or z=reiθz=re^{i\theta}.
  • When converting from Cartesian form, calculate rr and use tanθ=y/x\tan\theta=y/x only as a reference angle: the signs of xx and yy determine the correct quadrant.
  • The principal argument lies in (π,π](-\pi,\pi].
  • To return to Cartesian form, use x=rcosθx=r\cos\theta and y=rsinθy=r\sin\theta.
  • Exact special-angle values should remain exact, and radians are used throughout this topic.
Worked example

Express 3+i-\sqrt3+i in modulus-argument form using its principal argument.

  1. 1.r=(3)2+12=2r=\sqrt{(-\sqrt3)^2+1^2}=2.
  2. 2.The point lies in the second quadrant and has reference angle π/6\pi/6.
  3. 3.Its principal argument is ππ/6=5π/6\pi-\pi/6=5\pi/6.

Answer: 2(cos5π6+isin5π6)2\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right), equivalently 2e5πi/62e^{5\pi i/6}.

Common mistakes

  • Don't fall into the trap of using tan1(y/x)\tan^{-1}(y/x) as the final argument without checking the quadrant.
  • Don't fall into the trap of writing the modulus as x2+y2x^2+y^2 instead of x2+y2\sqrt{x^2+y^2}.
  • Don't fall into the trap of giving a degree angle when the complex-number question requires radians.

Exam tip

State the quadrant before choosing an argument, especially when the real part is negative.

Tier 1 · Easy

ORIGINAL

1.

Express 1+3i1+\sqrt3i in the form r(cosθ+isinθ)r(\cos\theta+i\sin\theta), where r>0r>0 and π<θπ-\pi<\theta\leq\pi.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Express 22i-2-2i in modulus-argument form using its principal argument.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

The complex number z=a+ibz=a+ib has modulus 1010, positive real part and principal argument tan1(3/4)-\tan^{-1}(3/4). Determine zz in Cartesian form.

(4)

(Total for Question 1 is 4 marks)

CP-2.6

Multiply and divide complex numbers in modulus-argument form.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • If z1=r1eiθ1z_1=r_1e^{i\theta_1} and z2=r2eiθ2z_2=r_2e^{i\theta_2}, then z1z2=r1r2ei(θ1+θ2)z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)}: moduli multiply and arguments add. For division, z1z2=r1r2ei(θ1θ2)\dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}e^{i(\theta_1-\theta_2)}, provided z20z_2\ne0: moduli divide and arguments subtract.
  • These results also follow from the compound-angle formulae in trigonometric form.
  • Any argument may be adjusted by an integer multiple of 2π2\pi; if a principal argument is requested, place it in (π,π](-\pi,\pi].
  • Geometrically, multiplication combines a scaling by the modulus with a rotation by the argument, which helps check whether the resulting modulus and quadrant are sensible.
  • Final form should retain a positive modulus.
Worked example

Let z1=3e5πi/6z_1=3e^{5\pi i/6} and z2=6e2πi/3z_2=6e^{-2\pi i/3}. Find z1/z2z_1/z_2 with principal argument.

  1. 1.Divide the moduli: 3/6=1/23/6=1/2.
  2. 2.Subtract the arguments: 5π/6(2π/3)=3π/25\pi/6-(-2\pi/3)=3\pi/2.
  3. 3.Subtract 2π2\pi to obtain the principal argument π/2-\pi/2.

Answer: 12eπi/2\dfrac12e^{-\pi i/2}.

Common mistakes

  • Don't fall into the trap of adding the moduli when multiplying complex numbers in modulus-argument form.
  • Don't fall into the trap of subtracting the numerator argument from the denominator argument during division.
  • Don't fall into the trap of leaving an argument such as 3π/23\pi/2 when the principal argument has been requested.

Exam tip

Write separate modulus and argument lines before combining them, then normalise the argument only at the end.

Tier 1 · Easy

ORIGINAL

1.

Express 2eiπ/6×3eiπ/42e^{i\pi/6}\times3e^{i\pi/4} in the form reiθre^{i\theta}.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Express 12e5πi/63eπi/4\dfrac{12e^{5\pi i/6}}{3e^{-\pi i/4}} in the form reiθre^{i\theta} using the principal argument.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

Given that z(2eiπ/7)3=16e2πi/7z(2e^{i\pi/7})^3=16e^{-2\pi i/7}, determine zz in the form reiθre^{i\theta}, where π<θπ-\pi<\theta\leq\pi.

(4)

(Total for Question 1 is 4 marks)

CP-2.7

Construct and interpret simple loci in the Argand diagram such as |z - a| > r and arg(z - a) = theta.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A locus translates complex notation into distance and direction. The equation za=r|z-a|=r is a circle of radius rr centred at the point representing aa; zar|z-a|\le r selects its closed interior and za>r|z-a|>r its exterior without the boundary.
  • The equation za=zb|z-a|=|z-b| is the perpendicular bisector of the segment joining the points representing aa and bb, while zazb|z-a|\le|z-b| selects the side at least as close to the point representing aa.
  • The condition arg(za)=β\arg(z-a)=\beta is a ray from that point at directed angle β\beta, excluding the endpoint.
  • Double argument inequalities select a sector.
  • Sketch boundaries first, then use solid or dashed curves to show inclusion correctly.
A circle locus and an argument-ray locus measured from the same point representing aa.
Worked example

Let z=x+iyz=x+iy. Describe the region z1z(5+2i)|z-1|\le|z-(5+2i)| in Cartesian form.

  1. 1.Square both non-negative distances: (x1)2+y2(x5)2+(y2)2(x-1)^2+y^2\le(x-5)^2+(y-2)^2.
  2. 2.Expand and cancel x2x^2 and y2y^2.
  3. 3.2x+110x+294y-2x+1\le-10x+29-4y, so 2x+y72x+y\le7.
  4. 4.The boundary is the perpendicular bisector, and the inequality selects the side containing the point representing 11.

Answer: The closed half-plane 2x+y72x+y\le7.

Common mistakes

  • Don't centre the circle za=r|z-a|=r at a-a instead of at aa.
  • Don't fall into the trap of including the endpoint z=az=a in an argument locus even though arg0\arg 0 is undefined.
  • Don't fall into the trap of drawing a full line for arg(za)=β\arg(z-a)=\beta instead of the directed ray from aa.

Exam tip

For a locus sketch, label the centre or endpoint, state whether each boundary is included, and shade the side selected by a test point.

Tier 1 · Easy

ORIGINAL

1.

Describe geometrically the locus z(2i)>3|z-(2-i)|>3, stating whether its boundary is included.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Let z=x+iyz=x+iy. Find a Cartesian description of the locus arg(z+1)=π/4\arg(z+1)=\pi/4, including the required restriction on xx.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

A region is defined by z(1i)<4|z-(1-i)|<4 and 0<arg(z(1i))<π/30<\arg(z-(1-i))<\pi/3. Sketch the region and determine its exact area.

(5)

(Total for Question 1 is 5 marks)

CP-2.8

Understand de Moivre's theorem and use it to find multiple angle formulae and sums of series.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • De Moivre's theorem states [cosθ+isinθ]n=cos(nθ)+isin(nθ)[\cos\theta+i\sin\theta]^n=\cos(n\theta)+i\sin(n\theta) for every integer nn.
  • Expanding the left side and equating real or imaginary parts produces multiple-angle formulae for cospθ\cos p\theta and sinqθ\sin q\theta in powers of cosθ\cos\theta and sinθ\sin\theta.
  • The relations z+z1=2cosθz+z^{-1}=2\cos\theta and zz1=2isinθz-z^{-1}=2i\sin\theta, with z=eiθz=e^{i\theta}, also convert powers back into multiple angles and support formulae involving tanrθ\tan r\theta.
  • Finite trigonometric sums are handled by summing a complex geometric series and then taking its real or imaginary part.
  • The chosen part and any non-zero denominator condition must be stated.
Worked example

Use de Moivre's theorem to express cos4θ\cos4\theta in powers of cosθ\cos\theta.

  1. 1.Expand (cosθ+isinθ)4(\cos\theta+i\sin\theta)^4 and equate real parts.
  2. 2.cos4θ=cos4θ6cos2θsin2θ+sin4θ\cos4\theta=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta.
  3. 3.Substitute sin2θ=1cos2θ\sin^2\theta=1-\cos^2\theta throughout.
  4. 4.Collecting powers gives cos4θ=8cos4θ8cos2θ+1\cos4\theta=8\cos^4\theta-8\cos^2\theta+1.

Answer: cos4θ=8cos4θ8cos2θ+1\cos4\theta=8\cos^4\theta-8\cos^2\theta+1.

Common mistakes

  • Don't fall into the trap of equating the complete binomial expansion to only cosnθ\cos n\theta instead of taking its real part.
  • Don't fall into the trap of dropping the powers of ii before replacing i2i^2 by 1-1.
  • Don't fall into the trap of taking the real part of a geometric series when the required sum contains sine terms.

Exam tip

State whether real or imaginary parts are being equated before simplifying a de Moivre expansion.

Tier 1 · Easy

ORIGINAL

1.

Use de Moivre's theorem to write (cosθ+isinθ)5(\cos\theta+i\sin\theta)^5 in modulus-argument form.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1.

Starting from de Moivre's theorem, prove the identity sin3θ=3sinθ4sin3θ\sin3\theta=3\sin\theta-4\sin^3\theta.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

Using a complex geometric series, evaluate exactly S=r=05cos(rπ4)S=\displaystyle\sum_{r=0}^{5}\cos\left(\dfrac{r\pi}{4}\right).

(5)

(Total for Question 1 is 5 marks)

CP-2.9

Know and use the definition e^(i theta) = cos theta + i sin theta and the form z = r e^(i theta).

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Euler's definition eiθ=cosθ+isinθe^{i\theta}=\cos\theta+i\sin\theta identifies a unit complex number with argument θ\theta. Hence z=reiθz=re^{i\theta} has modulus rr and argument θ\theta, and arguments differing by 2πk2\pi k represent the same number.
  • Replacing θ\theta by θ-\theta gives eiθ=cosθisinθe^{-i\theta}=\cos\theta-i\sin\theta. Adding and subtracting these equations yields cosθ=12(eiθ+eiθ)\cos\theta=\dfrac12(e^{i\theta}+e^{-i\theta}) and sinθ=12i(eiθeiθ)\sin\theta=\dfrac1{2i}(e^{i\theta}-e^{-i\theta}).
  • Exponential form turns multiplication into addition of arguments and often simplifies symmetric trigonometric expressions.
  • The trigonometric and exponential forms are interchangeable, so a final answer should use the form requested.
  • It must not be treated as a real exponential: eiθ=1|e^{i\theta}|=1 for every real θ\theta.
Worked example

Use Euler's definition to simplify e3iθe3iθe^{3i\theta}-e^{-3i\theta}.

  1. 1.e3iθ=cos3θ+isin3θe^{3i\theta}=\cos3\theta+i\sin3\theta.
  2. 2.e3iθ=cos3θisin3θe^{-3i\theta}=\cos3\theta-i\sin3\theta.
  3. 3.Subtracting cancels the cosine terms and doubles the imaginary sine term.

Answer: e3iθe3iθ=2isin3θe^{3i\theta}-e^{-3i\theta}=2i\sin3\theta.

Common mistakes

  • Don't fall into the trap of writing eiθ=cosθisinθe^{-i\theta}=-\cos\theta-i\sin\theta instead of cosθisinθ\cos\theta-i\sin\theta.
  • Don't fall into the trap of omitting the factor ii from eiθeiθ=2isinθe^{i\theta}-e^{-i\theta}=2i\sin\theta.
  • Don't fall into the trap of assuming eiθe^{i\theta} has real-exponential growth instead of modulus 11.

Exam tip

Pair exponentials with opposite arguments, then use their sum for cosine or their difference for sine.

Tier 1 · Easy

ORIGINAL

1.

Express 4(cos2π5+isin2π5)4\left(\cos\dfrac{2\pi}{5}+i\sin\dfrac{2\pi}{5}\right) in exponential form.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1.

Using Euler's definition, show that e2iθ+e2iθ=2cos2θe^{2i\theta}+e^{-2i\theta}=2\cos2\theta.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

For π<θ<π-\pi<\theta<\pi, prove that 1+eiθ=2cos(θ/2)eiθ/21+e^{i\theta}=2\cos(\theta/2)e^{i\theta/2}.

(4)

(Total for Question 1 is 4 marks)

CP-2.10

Find the n distinct nth roots of r e^(i theta) for r != 0 and know that they form the vertices of a regular n-gon in the Argand diagram.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • To solve zn=reiθz^n=re^{i\theta} with r>0r>0, allow every argument of the right side: θ+2kπ\theta+2k\pi. Taking nth roots gives z=r1/nei(θ+2kπ)/nz=r^{1/n}e^{i(\theta+2k\pi)/n} for k=0,1,,n1k=0,1,\ldots,n-1.
  • These are the nn distinct roots; larger integer values repeat them.
  • All roots have modulus r1/nr^{1/n} and consecutive arguments differ by 2π/n2\pi/n, so their points are equally spaced on a circle centred at the origin and form a regular nn-gon.
  • The polygon may be rotated, depending on θ\theta, but its size and angular spacing are fixed.
  • Each proposed root can be checked by raising it to the nth power and reducing its argument modulo 2π2\pi.
Ten distinct tenth roots equally spaced on their common modulus circle.
Worked example

Find all solutions of z3=8eiπz^3=8e^{i\pi} in exponential form.

  1. 1.Each root has modulus 81/3=28^{1/3}=2.
  2. 2.The arguments are π+2kπ3\dfrac{\pi+2k\pi}{3} for k=0,1,2k=0,1,2.
  3. 3.These arguments are π/3\pi/3, π\pi and 5π/35\pi/3.

Answer: z=2eiπ/3z=2e^{i\pi/3}, z=2eiπz=2e^{i\pi} or z=2e5πi/3z=2e^{5\pi i/3}.

Common mistakes

  • Don't fall into the trap of using only the principal argument θ\theta and therefore finding only one nth root.
  • Don't fall into the trap of adding 2kπ2k\pi after dividing by nn instead of before dividing.
  • Don't fall into the trap of giving n+1n+1 values by including both endpoints of one complete argument cycle.

Exam tip

Use k=0k=0 to n1n-1 explicitly and check that consecutive root arguments differ by exactly 2π/n2\pi/n.

Tier 1 · Easy

ORIGINAL

1.

Find the three cube roots of 88, giving them in exponential form.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Find the four roots of z4=16eiπ/3z^4=16e^{i\pi/3}, giving their arguments in the interval 0θ<2π0\leq\theta<2\pi.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

The roots of z5=32eiπ/2z^5=32e^{-i\pi/2} are plotted on an Argand diagram. Find all five roots in exponential form and determine the exact area of the regular pentagon they form.

(6)

(Total for Question 1 is 6 marks)

CP-2.11

Use complex roots of unity to solve geometric problems.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The nth roots of unity solve zn=1z^n=1 and are 1,ω,ω2,,ωn11,\omega,\omega^2,\ldots,\omega^{n-1} for ω=e2πi/n\omega=e^{2\pi i/n}. They lie on the unit circle as the vertices of a regular nn-gon.
  • For a non-real root ω1\omega\ne1, factorising ωn1\omega^n-1 gives 1+ω++ωn1=01+\omega+\cdots+\omega^{n-1}=0.
  • Complex methods encode geometry efficiently: the distance between the points representing aa and bb is ab|a-b|, and multiplication by ω\omega rotates every point representing a complex number through 2π/n2\pi/n without changing lengths.
  • Algebraic identities among powers of ω\omega can therefore prove equal sides, angles, parallelism and ratios.
  • Any cancellation involving ω1\omega-1 requires the explicit fact that ω1\omega\ne1.
Worked example

Let ω=e2πi/3\omega=e^{2\pi i/3}. Show that the points represented by 00, 11 and 1+ω1+\omega form an equilateral triangle.

  1. 1.The side from 00 to 11 has length 1=1|1|=1.
  2. 2.The side from 11 to 1+ω1+\omega has length ω=1|\omega|=1.
  3. 3.Since 1+ω+ω2=01+\omega+\omega^2=0, 1+ω=ω21+\omega=-\omega^2, so 1+ω=1|1+\omega|=1.
  4. 4.All three side lengths are 11.

Answer: The three points form an equilateral triangle of side length 11.

Common mistakes

  • Don't fall into the trap of using 1+ω++ωn1=01+\omega+\cdots+\omega^{n-1}=0 for the root ω=1\omega=1.
  • Don't fall into the trap of calculating a geometric distance as aba-b without taking the modulus.
  • Don't fall into the trap of claiming that multiplication by ω\omega changes lengths even though ω=1|\omega|=1.

Exam tip

Translate each geometric claim into a modulus of a difference or a rotation by a unit complex number.

Tier 1 · Easy

ORIGINAL

1.

Let ω=e2πi/3\omega=e^{2\pi i/3}. Find the exact distance between the points represented by 11 and ω\omega.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Let ω=e2πi/3\omega=e^{2\pi i/3}. The points AA, BB and CC are represented by the complex numbers 11, ω\omega and ω2\omega^2, respectively. Prove that triangle ABCABC is equilateral.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

The fifth roots of unity are the vertices of a regular pentagon. Prove that the ratio of a diagonal to a side is 1+52\dfrac{1+\sqrt5}{2}.

(7)

(Total for Question 1 is 7 marks)

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