Core Pure: Complex Numbers Depth - Worksheets, Questions and Revision

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A-Level · Further Pure Mathematics 1 and 2 (Complex Numbers Depth)

FP.CP12 Core Pure: Complex Numbers Depth

EDEXCEL 9FM0 · Calculator allowed · about 165 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Let z = 3 - 3*√3*i.

(a)Find the modulus of z.(1)

(b)Find the argument of z, giving your answer in radians in the range -π < θ ≤ π.(2)

(Total for Question 1 is 3 marks)

Let z = 1 + i.

(a)Find the modulus and argument of z.(2)

(b)Hence use de Moivre's theorem to find z⁸, giving your answer as a real integer. You must show your method.(2)

(Total for Question 2 is 4 marks)

The complex number z satisfies |z + 2 - 3*i| = 4.

(a)Show that the locus of z can be written as the Cartesian equation (x + 2)² + (y - 3)² = 16, where z = x + i*y.(2)

(b)State the centre and radius of this circle.(2)

(c)Determine, with a reason, whether the point representing z = 2 + i lies inside, on, or outside this circle.(1)

(Total for Question 3 is 5 marks)

Let z = √2*(cos(π/6) + i*sin(π/6)).

(a)Use de Moivre's theorem to find z⁶, giving your answer in the form a + b*i, where a and b are integers.(4)

(b)Find the smallest positive integer n for which zⁿ is a positive real number.(3)

(Total for Question 4 is 7 marks)

Let z1 = -1 + i*√3 and z2 = 2 - 2*i.

(a)Express z1 in the form r*(cos(θ) + i*sin(θ)), where r > 0 and -π < θ ≤ π.(3)

(b)Express z2 in the form r*(cos(θ) + i*sin(θ)), where r > 0 and -π < θ ≤ π.(3)

(c)Using your answers to parts (a) and (b), find z1*z2 and z1/z2, giving each in the form r*(cos(θ) + i*sin(θ)) where r > 0 and -π < θ ≤ π.(4)

(d)Hence show that z1/z2 = -(√3+1)/4 + i*(√3-1)/4, using exact trigonometric values for cos(11*π/12) and sin(11*π/12).(4)

(Total for Question 5 is 14 marks)

The complex number z satisfies |z - (6 + 8*i)| = 5.

(a)Sketch the locus of z on an Argand diagram, indicating clearly the centre and radius of the circle.(3)

(b)Find the greatest and least possible values of |z|.(4)

(c)Find the range of possible values of arg(z), giving your answer in radians to 3 significant figures.(5)

(Total for Question 6 is 12 marks)

The complex number z satisfies simultaneously arg(z - (1 + i)) = π/3 and |z - 6| = |z - 2*i|.

(a)Show that the locus |z - 6| = |z - 2*i| can be written as y = 3*x - 8, where z = x + i*y.(3)

(b)Show that the locus arg(z - (1 + i)) = π/3 can be written as y = √3*x + 1 - √3, stating the restriction on x.(3)

(c)Hence find, in the form p + q*i where p and q are surds, the single complex number z satisfying both loci simultaneously.(4)

(Total for Question 7 is 10 marks)

This question uses de Moivre's theorem to derive and apply a multiple-angle identity in cos(θ).

(a)Using de Moivre's theorem and the binomial expansion of (cos(θ) + i*sin(θ))⁵, show that cos(5*θ) = 16*cos⁵(θ) - 20*cos³(θ) + 5*cos(θ).(7)

(b)Hence solve, for 0 ≤ θ < π, the equation 16*cos⁵(θ) - 20*cos³(θ) + 5*cos(θ) = 0, giving each solution as an exact multiple of π.(5)

(Total for Question 8 is 12 marks)

Let z = cos(θ) + i*sin(θ).

(a)Show that z - 1/z = 2*i*sin(θ).(2)

(b)Hence show that 32*i*sin⁵(θ) = z⁵ - 5*z³ + 10*z - 10*z^-1 + 5*z^-3 - z^-5.(3)

(c)Hence show that sin⁵(θ) = (1/16)*(sin(5*θ) - 5*sin(3*θ) + 10*sin(θ)).(4)

(d)Hence find the exact value of the integral from θ = 0 to θ = π/2 of sin⁵(θ) with respect to θ.(4)

(Total for Question 9 is 13 marks)

Let w = -8 - 8*√3*i.

(a)Express w in the form r*(cos(θ) + i*sin(θ)), where r > 0 and -π < θ ≤ π.(3)

(b)Hence use de Moivre's theorem to find, in the form x + i*y, the four roots of the equation z⁴ = w.(6)

(c)Show that the sum of these four roots is 0, and explain briefly why this must be the case.(3)

(Total for Question 10 is 12 marks)

The complex number z satisfies |z - 9| = 2*|z|.

(a)Show that the locus of z is a circle with Cartesian equation (x + 3)² + y² = 36, and state its centre and radius.(6)

(b)Find the greatest and least possible values of |z| for points on this locus.(4)

(c)Given that w is the point on the locus with the greatest possible imaginary part, find w exactly, in the form p + q*i.(4)

(Total for Question 11 is 14 marks)

The complex number u = cos(2*π/5) + i*sin(2*π/5) is a fifth root of unity.

(a)Write down, in a similar form, the other four fifth roots of unity (i.e. the five roots of z⁵ = 1), for k = 0, 1, 2, 3, 4.(2)

(b)Show that 1 + u + u² + u³ + u⁴ = 0.(3)

(c)The five roots of z⁵ = 1 are represented on an Argand diagram by points forming a regular pentagon inscribed in a circle of radius 1 centred at the origin. Show that the area of this pentagon is (5/2)*sin(2*π/5).(5)

(d)Hence find the area of the pentagon, giving your answer to 3 significant figures.(2)

(Total for Question 12 is 12 marks)

A transformation from the z-plane to the w-plane is defined by w = 1/z, where z is not equal to 0. The point z lies on the circle |z - 3| = 3 (excluding the origin).

(a)Show that, in the w-plane, the image of this circle is the line with equation 6*Re(w) = 1.(7)

(b)State, with a reason, whether every point on the line 6*Re(w) = 1 is the image of a point on the circle |z-3|=3 (excluding the origin).(2)

(c)Find the point z on the circle such that the image w = 1/z also satisfies Im(w) = 1/6.(4)

(Total for Question 13 is 13 marks)

Mark scheme · FP.CP12 Core Pure: Complex Numbers Depth

Question 1

(a) B1 |z| = √3² + (3sqrt(3))² = 6, cao
(a) Answer: |z| = 6
(b) M1 attempts arctan(3sqrt(3)/3) = arctan(√3) = π/3 with correct fourth-quadrant sign consideration (real part positive, imaginary part negative)
(b) A1 arg(z) = -π/3, cao
(b) Answer: arg(z) = -π/3

Question 2

(a) B1 |z| = √1²+1² = √2
(a) B1 arg(z) = π/4, cao
(a) Answer: |z| = √2, arg(z) = π/4
(b) M1 applies de Moivre's theorem: z⁸ = (√2)⁸*(cos(8*π/4) + i*sin(8*π/4)), simplifying (√2)⁸ = 16 and 8*π/4 = 2*π
(b) A1 z⁸ = 16, cao
(b) Answer: z⁸ = 16

Question 3

(a) M1 writes |(x+2) + i(y-3)| = 4 and squares both sides
(a) A1 (x+2)² + (y-3)² = 16, cso
(a) Answer: (x+2)² + (y-3)² = 16
(b) B1 centre = (-2, 3)
(b) B1 radius = 4
(b) Answer: Centre (-2, 3), radius 4
(c) B1 correct conclusion (outside) with correct reason: (2+2)²+(1-3)² = 20 > 16
(c) Answer: Outside the circle, since (2+2)² + (1-3)² = 20 > 16

Question 4

(a) M1 z⁶ = (√2)⁶*(cos(6*π/6) + i*sin(6*π/6))
(a) A1 (√2)⁶ = 8
(a) M1 simplifies the angle 6*π/6 = π, giving cos(π) = -1 and sin(π) = 0
(a) A1 z⁶ = -8, cao
(a) Answer: z⁶ = -8
(b) M1 zⁿ = (√2)ⁿ*(cos(n*π/6) + i*sin(n*π/6)), and requires n*π/6 to be a multiple of 2*π
(b) M1 n/6 = 2k for integer k, i.e. n = 12k
(b) A1 smallest positive integer n = 12, cao
(b) Answer: n = 12

Question 5

(a) M1 |z1| = √(-1)² + (√3)² = √4 = 2
(a) M1 reference angle arctan(√3/1) = π/3, second quadrant, so θ = π - π/3 = 2*π/3
(a) A1 z1 = 2*(cos(2*π/3) + i*sin(2*π/3)), cao
(a) Answer: z1 = 2*(cos(2*π/3) + i*sin(2*π/3))
(b) M1 |z2| = √2² + (-2)² = √8 = 2*√2
(b) M1 reference angle arctan(2/2) = π/4, fourth quadrant, so θ = -π/4
(b) A1 z2 = 2*√2*(cos(-π/4) + i*sin(-π/4)), cao
(b) Answer: z2 = 2*√2*(cos(-π/4) + i*sin(-π/4))
(c) M1 |z1*z2| = 2*2*√2 = 4*√2 and arg(z1*z2) = 2*π/3 - π/4 = 5*π/12
(c) A1 z1*z2 = 4*√2*(cos(5*π/12) + i*sin(5*π/12))
(c) M1 |z1/z2| = 2/(2*√2) = √2/2 and arg(z1/z2) = 2*π/3 - (-π/4) = 11*π/12
(c) A1 z1/z2 = (√2/2)*(cos(11*π/12) + i*sin(11*π/12)), cao
(c) Answer: z1*z2 = 4*√2*(cos(5*π/12) + i*sin(5*π/12)); z1/z2 = (√2/2)*(cos(11*π/12) + i*sin(11*π/12))
(d) M1 uses 11*π/12 = π - π/12 to write cos(11*π/12) = -cos(π/12) and sin(11*π/12) = sin(π/12)
(d) A1 cos(π/12) = (√6+√2)/4 and sin(π/12) = (√6-√2)/4, found using the addition formulae with π/12 = π/3 - π/4
(d) M1 substitutes into (√2/2)*(cos(11*π/12)+i*sin(11*π/12)) and simplifies the surds, using √2*√6 = 2*√3
(d) A1 z1/z2 = -(√3+1)/4 + i*(√3-1)/4, cso
(d) Answer: z1/z2 = -(√3+1)/4 + i*(√3-1)/4

Question 6

(a) B1 circle drawn, correctly positioned not passing through or enclosing the origin
(a) B1 centre marked and labelled (6, 8)
(a) B1 radius 5 indicated (e.g. circle shown touching x=1, x=11, y=3, y=13)
(a) Answer: Circle, centre (6, 8), radius 5
(b) M1 finds the distance from the origin to the centre: √6²+8² = 10
(b) B1 correct reasoning that both extreme values of |z| occur on the line joining the origin to the centre, extended to meet the circle
(b) A1 greatest |z| = 10 + 5 = 15
(b) A1 least |z| = 10 - 5 = 5, cao
(b) Answer: Greatest |z| = 15, least |z| = 5
(c) M1 finds arg(centre) = arctan(8/6)
(c) M1 finds the half-angle subtended at the origin by the circle: arcsin(radius/|OC|) = arcsin(5/10) = arcsin(0.5) = π/6
(c) A1 arctan(4/3) = 0.927 (awrt)
(c) M1 range = arctan(4/3) - π/6 to arctan(4/3) + π/6
(c) A1 0.404 ≤ arg(z) ≤ 1.45 (awrt, 3 s.f.)
(c) Answer: 0.404 ≤ arg(z) ≤ 1.45 (radians, 3 s.f.)

Question 7

(a) M1 (x-6)² + y² = x² + (y-2)²
(a) M1 expands and simplifies to -12x + 36 = -4y + 4
(a) A1 y = 3x - 8, cso
(a) Answer: y = 3x - 8
(b) M1 tan(π/3) = (y-1)/(x-1), with x > 1 (and y > 1) since arg = π/3 requires both components positive
(b) A1 y - 1 = √3*(x - 1)
(b) A1 y = √3*x + 1 - √3, for x > 1, cso
(b) Answer: y = √3*x + 1 - √3, for x > 1
(c) M1 sets 3x - 8 = √3*x + 1 - √3
(c) M1 solves for x: x*(3 - √3) = 9 - √3
(c) A1 rationalises to x = 4 + √3
(c) A1 y = 3x - 8 = 4 + 3*√3, giving z = (4 + √3) + i*(4 + 3*√3), cao
(c) Answer: z = (4 + √3) + i*(4 + 3*√3)

Question 8

(a) M1 writes cos(5*θ) + i*sin(5*θ) = (cos(θ) + i*sin(θ))⁵ by de Moivre's theorem
(a) M1 attempts the binomial expansion of the right-hand side
(a) A1 correct unsimplified expansion: cos⁵(θ) + 5i*cos⁴(θ)sin(θ) - 10cos³(θ)sin²(θ) - 10i*cos²(θ)sin³(θ) + 5cos(θ)sin⁴(θ) + i*sin⁵(θ)
(a) M1 equates real parts: cos(5*θ) = cos⁵(θ) - 10cos³(θ)sin²(θ) + 5cos(θ)sin⁴(θ)
(a) dM1 substitutes sin²(θ) = 1 - cos²(θ) throughout (dependent on the previous method mark)
(a) A1 correctly expands and collects terms in cos(θ)
(a) A1 cos(5*θ) = 16*cos⁵(θ) - 20*cos³(θ) + 5*cos(θ), cso
(a) Answer: cos(5*θ) = 16*cos⁵(θ) - 20*cos³(θ) + 5*cos(θ)
(b) M1 uses part (a) to rewrite the equation as cos(5*θ) = 0
(b) M1 sets the general solution 5*θ = π/2 + k*π
(b) M1 θ = π/10 + k*π/5
(b) A1 at least three correct values within the range
(b) A1 all five values correct with no extras: θ = π/10, 3*π/10, π/2, 7*π/10, 9*π/10
(b) Answer: θ = π/10, 3*π/10, π/2, 7*π/10, 9*π/10

Question 9

(a) B1 1/z = cos(θ) - i*sin(θ), using de Moivre's theorem with n = -1 (or since |z|=1, 1/z equals the conjugate of z)
(a) B1 z - 1/z = (cos(θ)+i*sin(θ)) - (cos(θ)-i*sin(θ)) = 2*i*sin(θ), cso
(a) Answer: z - 1/z = 2*i*sin(θ)
(b) M1 raises both sides of part (a) to the power 5: (2*i*sin(θ))⁵ = (z - 1/z)⁵, and attempts the binomial expansion of the right-hand side
(b) A1 correct expansion z⁵ - 5*z³ + 10*z - 10*z^-1 + 5*z^-3 - z^-5
(b) A1 (2*i)⁵ = 32*i (since i⁵ = i), giving 32*i*sin⁵(θ) as required, cso
(b) Answer: 32*i*sin⁵(θ) = z⁵ - 5*z³ + 10*z - 10*z^-1 + 5*z^-3 - z^-5
(c) M1 groups terms as (z⁵ - z^-5) - 5*(z³ - z^-3) + 10*(z - z^-1)
(c) M1 uses zⁿ - z^-n = 2*i*sin(n*θ) for n = 5, 3, 1
(c) A1 32*i*sin⁵(θ) = 2*i*sin(5*θ) - 10*i*sin(3*θ) + 20*i*sin(θ)
(c) A1 divides both sides by 2*i to obtain sin⁵(θ) = (1/16)*(sin(5*θ) - 5*sin(3*θ) + 10*sin(θ)), cso
(c) Answer: sin⁵(θ) = (1/16)*(sin(5*θ) - 5*sin(3*θ) + 10*sin(θ))
(d) M1 integrates term by term using part (c): integral = (1/16)*[-cos(5*θ)/5 + (5/3)*cos(3*θ) - 10*cos(θ)]
(d) A1 correct integration (follow through on sign errors)
(d) M1 substitutes limits θ = π/2 and θ = 0, using cos(5*π/2) = 0, cos(3*π/2) = 0, cos(π/2) = 0 and cos(0) = 1
(d) A1 exact value = 8/15, cao
(d) Answer: Integral from 0 to π/2 of sin⁵(θ) d(θ) = 8/15

Question 10

(a) M1 |w| = √8² + (8*√3)² = √64+192 = √256 = 16
(a) A1 reference angle arctan((8*√3)/8) = arctan(√3) = π/3, third quadrant (both parts negative), so θ = -(π - π/3) = -2*π/3
(a) A1 w = 16*(cos(-2*π/3) + i*sin(-2*π/3)), cao
(a) Answer: w = 16*(cos(-2*π/3) + i*sin(-2*π/3))
(b) M1 uses z_k = r^1/4*(cos((θ+2*k*π)/4) + i*sin((θ+2*k*π)/4)) for k = 0, 1, 2, 3
(b) M1 finds r^1/4 = 16^1/4 = 2 and the angle formula (-2*π/3 + 2*k*π)/4 = -π/6 + k*π/2
(b) A1 the four arguments are -π/6, π/3, 5*π/6 and -2*π/3, all reduced to the principal range
(b) A1 first two roots in x+iy form: √3 - i and 1 + i*√3
(b) A1 remaining two roots in x+iy form: -√3 + i and -1 - i*√3
(b) A1 all four roots correct and fully simplified, cao
(b) Answer: z = √3 - i, or z = 1 + i*√3, or z = -√3 + i, or z = -1 - i*√3
(c) M1 adds real parts (√3+1-√3-1 = 0) and imaginary parts (-1+√3+1-√3 = 0)
(c) A1 sum = 0, cso
(c) B1 explains that z⁴ - w = 0 is a quartic in z with zero coefficient of z³, and the sum of the roots of a quartic equals minus the coefficient of z³ divided by the coefficient of z⁴, which is 0 here
(c) Answer: Sum of the four roots = 0

Question 11

(a) M1 squares both sides: (x-9)² + y² = 4*(x²+y²)
(a) M1 expands both sides: x²-18x+81+y² = 4x²+4y²
(a) M1 rearranges to x² + y² + 6x - 27 = 0
(a) A1 completes the square: (x+3)² + y² = 36, cso
(a) B1 centre (-3, 0)
(a) B1 radius 6
(a) Answer: (x+3)² + y² = 36; centre (-3, 0); radius 6
(b) M1 distance from the origin to the centre = √(-3)²+0² = 3, noting this is less than the radius 6, so the origin lies inside the circle
(b) A1 least |z| = 6 - 3 = 3
(b) A1 greatest |z| = 6 + 3 = 9
(b) B1 identifies the corresponding points z = 3 (least) and z = -9 (greatest), both on the real axis
(b) Answer: Least |z| = 3 (at z = 3); greatest |z| = 9 (at z = -9)
(c) M1 recognises that the point of greatest imaginary part on a circle of centre (a,b) and radius r is (a, b+r)
(c) B1 correct reasoning: the topmost point of the circle lies directly above the centre
(c) A1 coordinates (-3, 6)
(c) A1 w = -3 + 6*i, cao
(c) Answer: w = -3 + 6*i

Question 12

(a) B1 roots given by cos(2*k*π/5) + i*sin(2*k*π/5) for k = 0, 1, 2, 3, 4
(a) B1 correctly identifies k=0 gives z=1 and the others are u, u², u³, u⁴
(a) Answer: The five roots are 1, u, u², u³, u⁴, i.e. cos(2*k*π/5) + i*sin(2*k*π/5) for k = 0, 1, 2, 3, 4
(b) M1 recognises the sum as a geometric series with sum (u⁵ - 1)/(u - 1), valid since u is not equal to 1
(b) A1 u⁵ = 1 (as u is a fifth root of unity), so the numerator is 0
(b) A1 hence the sum is 0, cso
(b) Answer: 1 + u + u² + u³ + u⁴ = 0
(c) M1 splits the pentagon into 5 congruent isosceles triangles from the centre, each with two sides of length 1 and included angle 2*π/5
(c) M1 area of one triangle = (1/2)*1*1*sin(2*π/5), using Area = (1/2)*a*b*sin(C)
(c) A1 correct triangle area (1/2)*sin(2*π/5)
(c) M1 total pentagon area = 5*(1/2)*sin(2*π/5)
(c) A1 Area = (5/2)*sin(2*π/5), cso
(c) Answer: Area = (5/2)*sin(2*π/5)
(d) M1 evaluates sin(2*π/5), approximately 0.951056
(d) A1 area = awrt 2.38 (3 s.f.)
(d) Answer: Area is approximately 2.38 (3 s.f.)