Complex Numbers - Worksheets, Questions and Revision

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

FP.CP1 Complex Numbers

EDEXCEL 9FM0 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The complex number z satisfies the equation 2z² - 4z + 5 = 0.

(a)Solve the equation, giving each root in the form a + bi.(4)

(b)For the root with positive imaginary part, find the modulus and the argument, giving the argument in radians to 3 significant figures.(3)

(c)Hence write this root in the form r(cos θ + i sin θ), where r and θ are given to 3 significant figures.(2)

(Total for Question 1 is 9 marks)

A locus C in the Argand diagram is defined by the equation |z - (3 + 4i)| = 5.

(a)Sketch the locus C, marking clearly the points where it intersects the real axis.(3)

(b)Find the greatest and least possible values of |z| for points on C.(3)

(c)The point on C for which arg(z) = π/4 (other than the origin) is z = p + qi. Find the values of p and q.(4)

(Total for Question 2 is 10 marks)

This question is about De Moivre's theorem.

(a)State De Moivre's theorem for a positive integer n.(1)

(b)Hence find, in the form a + bi, the value of (cos(π/6) + i sin(π/6))⁹.(3)

(c)Find the smallest positive integer n for which (cos(π/6) + i sin(π/6))ⁿ is a positive real number.(3)

(Total for Question 3 is 7 marks)

Prove by mathematical induction that, for all positive integers n, (cos θ + i sin θ)ⁿ = cos(n θ) + i sin(n θ).

(a)Show that the result is true for n = 1.(1)

(b)Assuming the result is true for n = k, show that it is true for n = k + 1.(4)

(c)Hence complete the proof by induction, making clear reference to your results in parts (a) and (b).(1)

(Total for Question 4 is 6 marks)

The complex number w = -8i.

(a)Find the three cube roots of w, giving each root in the form r(cos θ + i sin θ), where -π < θ ≤ π.(5)

(b)Show that the sum of the three cube roots is 0.(3)

(c)Plot the three roots on an Argand diagram and state, with a reason, the geometric shape they form.(2)

(Total for Question 5 is 10 marks)

z = 2 - i is one root of the cubic equation z³ - 6z² + 13z - 10 = 0.

(a)Explain why z = 2 + i must also be a root of this equation, and hence find the third root.(5)

(b)State the sum and the product of all three roots of the equation, and verify that these agree with the coefficients of the cubic.(4)

(Total for Question 6 is 9 marks)

Loci C1 and C2 in the Argand diagram are defined by C1: |z - 4| = |z - 2i|, and C2: arg(z - 1) = π/4.

(a)Show that C1 has cartesian equation y = 2x - 3.(4)

(b)State the cartesian equation of C2, including the restriction on x.(2)

(c)Find the point of intersection of C1 and C2, giving your answer in the form a + bi.(3)

(d)Find the exact area of the triangle formed by the point of intersection found in part (c), the point z = 4, and the origin.(3)

(Total for Question 7 is 12 marks)

This question uses De Moivre's theorem to derive a multiple-angle identity.

(a)Use De Moivre's theorem and the binomial expansion of (cos θ + i sin θ)⁵ to show that cos(5 θ) = 16cos⁵(θ) - 20cos³(θ) + 5cos(θ).(5)

(b)Hence solve, for 0 < θ < π, the equation 16cos⁵(θ) - 20cos³(θ) + 5cos(θ) = 0, giving each answer as an exact multiple of π.(4)

(Total for Question 8 is 9 marks)

The complex number w satisfies w⁶ = 64.

(a)Find all six roots of this equation, giving each in the form r(cos θ + i sin θ), with -π < θ ≤ π.(6)

(b)Show that these six roots form the vertices of a regular hexagon in the Argand diagram, and find its exact perimeter.(3)

(c)Find the exact area enclosed by the hexagon.(3)

(Total for Question 9 is 12 marks)

The region R in the Argand diagram is defined by |z - 3| ≤ 3 and 0 ≤ arg(z) ≤ π/4.

(a)Sketch the region R on an Argand diagram.(3)

(b)Show that the circle |z - 3| = 3 has polar equation r = 6cos(θ), for -π/2 ≤ θ ≤ π/2.(3)

(c)Hence find the exact area of the region R.(6)

(Total for Question 10 is 12 marks)

Let z = cos θ + i sin θ.

(a)Show that sin⁴(θ) can be written in the form Acos(4theta) + Bcos(2theta) + C, where A, B and C are constants to be found.(5)

(b)Hence find the exact value of the integral from 0 to π/2 of sin⁴(θ) d θ.(4)

(Total for Question 11 is 9 marks)

A transformation from the z-plane to the w-plane is defined by w = (z - i)/(z + 1), where z ≠ -1. The point z lies on the circle |z| = 1.

(a)Show that, as z varies on the circle |z| = 1, the image point w = u + iv satisfies v = -u.(7)

(b)State the cartesian equation of the locus of w, and explain, with reference to the point z = -1, why this locus is a straight line rather than a circle.(2)

(Total for Question 12 is 9 marks)

Mark scheme · FP.CP1 Complex Numbers

Question 1

(a) M1 correct substitution into the quadratic formula, z = [4 ± √16 - 40] / 4
(a) A1 discriminant simplified to √-24 = 2sqrt(6) i (oe)
(a) M1 simplifies to z = 1 ± (√6/2) i
(a) A1 both roots correct, cao
(a) Answer: z = 1 + (√6/2) i or z = 1 - (√6/2) i
(b) M1 modulus = √1² + (√6/2)²
(b) A1 modulus = √2.5, awrt 1.58
(b) A1 argument = arctan((√6/2) / 1), awrt 0.886 radians, cao
(b) Answer: |z| = √2.5 = awrt 1.58; arg(z) = awrt 0.886 radians
(c) M1 uses the values of r and θ found in part (b)
(c) A1 z = 1.58(cos 0.886 + i sin 0.886), awrt
(c) Answer: z = 1.58 (cos(0.886) + i sin(0.886))

Question 2

(a) M1 sets z = x (real) and forms (x-3)² + 16 = 25
(a) A1 x = 0 and x = 6 found
(a) B1 correct sketch: circle centre 3 + 4i, radius 5, passing through (0,0) and (6,0)
(a) Answer: Circle centre 3 + 4i, radius 5, crossing the real axis at z = 0 and z = 6
(b) M1 finds distance from origin to centre = √3² + 4² = 5
(b) A1 greatest |z| = 5 + 5 = 10
(b) A1 least |z| = 5 - 5 = 0, cao
(b) Answer: Greatest |z| = 10, least |z| = 0
(c) M1 uses arg(z) = π/4 to set y = x, with x > 0
(c) M1 substitutes y = x into (x-3)² + (x-4)² = 25
(c) A1 solves to x = 0 (rejected, this is the origin) or x = 7
(c) A1 z = 7 + 7i, cao
(c) Answer: z = 7 + 7i (p = 7, q = 7)

Question 3

(a) B1 (cos θ + i sin θ)ⁿ = cos(n θ) + i sin(n θ)
(a) Answer: (cos θ + i sin θ)ⁿ = cos(n θ) + i sin(n θ)
(b) M1 uses De Moivre with n θ = 9 x π/6 = 3pi/2
(b) A1 cos(3pi/2) = 0
(b) A1 sin(3pi/2) = -1, giving answer -i, cao
(b) Answer: -i
(c) M1 requires n x π/6 = 2k π for a positive integer k
(c) A1 n = 12k
(c) A1 smallest positive integer n = 12, cao
(c) Answer: n = 12

Question 4

(a) B1 LHS = cos θ + i sin θ = RHS when n = 1, so the result is true for n = 1
(a) Answer: True for n = 1, since both sides equal cos θ + i sin θ
(b) M1 writes (cos θ + i sin θ)^k+1 = (cos θ + i sin θ)^k (cos θ + i sin θ) and substitutes the assumption cos(k θ) + i sin(k θ)
(b) M1 expands the product: [cos(k θ)cos θ - sin(k θ)sin θ] + i[sin(k θ)cos θ + cos(k θ)sin θ]
(b) A1 uses the addition formulae to write the real part as cos(k θ + θ)
(b) A1 writes the imaginary part as sin(k θ + θ), giving cos((k+1)θ) + i sin((k+1)θ), cso
(b) Answer: (cos θ + i sin θ)^k+1 = cos((k+1)θ) + i sin((k+1)θ) (shown)
(c) B1 correct conclusion: since the result is true for n = 1, and true for n = k implies true for n = k+1, the result is true for all positive integers n by mathematical induction
(c) Answer: True for all positive integers n by mathematical induction

Question 5

(a) M1 writes w = 8(cos(-π/2) + i sin(-π/2))
(a) M1 finds the modulus of each root, r = 8^1/3 = 2
(a) M1 uses θ = (-π/2 + 2k π)/3 for k = -1, 0, 1
(a) A1 two of the three roots correct
(a) A1 all three roots correct: 2(cos(-π/6)+i sin(-π/6)), 2(cos(π/2)+i sin(π/2)), 2(cos(-5pi/6)+i sin(-5pi/6))
(a) Answer: 2(cos(-π/6)+i sin(-π/6)), 2(cos(π/2)+i sin(π/2)) [= 2i], 2(cos(-5pi/6)+i sin(-5pi/6))
(b) M1 converts each root to cartesian form: √3 - i, 2i, -√3 - i
(b) A1 all three cartesian forms correct
(b) A1 sums to (√3-√3) + i(-1+2-1) = 0, cso
(b) Answer: 0 (shown)
(c) B1 three points plotted correctly, each at distance 2 from the origin, at arguments -π/6, π/2 and -5pi/6
(c) B1 states an equilateral triangle, since the roots are equally spaced by 2pi/3 around a circle of radius 2
(c) Answer: Equilateral triangle inscribed in a circle of radius 2, centre the origin

Question 6

(a) B1 states that since the coefficients of the cubic are all real, complex roots occur in conjugate pairs, so z = 2 + i is also a root
(a) M1 forms the quadratic factor (z-(2-i))(z-(2+i)) = z² - 4z + 5
(a) M1 divides z³ - 6z² + 13z - 10 by z² - 4z + 5
(a) A1 quotient z - 2
(a) A1 third root z = 2, cao
(a) Answer: z = 2 + i (conjugate root); third root z = 2
(b) B1 sum of roots = (2-i)+(2+i)+2 = 6
(b) B1 product of roots = (2-i)(2+i)(2) = 5 x 2 = 10
(b) M1 compares with -(coefficient of z²)/(coefficient of z³) = 6 and -(constant term)/(coefficient of z³) = 10
(b) A1 both agree, so verified, cso
(b) Answer: Sum = 6, product = 10, both consistent with the coefficients of the cubic

Question 7

(a) M1 writes z = x + iy and squares both sides: (x-4)² + y² = x² + (y-2)²
(a) M1 expands both sides correctly
(a) A1 simplifies to -8x + 16 = -4y + 4
(a) A1 rearranges to y = 2x - 3, cso
(a) Answer: y = 2x - 3 (shown)
(b) B1 y = x - 1
(b) B1 restriction x > 1 (since arg = π/4 requires both x - 1 > 0 and y > 0)
(b) Answer: y = x - 1, for x > 1
(c) M1 solves 2x - 3 = x - 1 simultaneously
(c) A1 x = 2, y = 1
(c) A1 checks x = 2 > 1 is valid; z = 2 + i, cao
(c) Answer: z = 2 + i
(d) M1 uses area = (1/2)|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)| with vertices (2,1), (4,0), (0,0)
(d) A1 correct substitution: (1/2)|2(0-0)+4(0-1)+0(1-0)|
(d) A1 area = 2 square units, cao
(d) Answer: Area = 2 square units

Question 8

(a) M1 writes cos(5theta) + i sin(5theta) = (cos θ + i sin θ)⁵ and expands using the binomial theorem
(a) M1 identifies the real part: cos⁵ θ - 10cos³ θ sin² θ + 5cos θ sin⁴ θ
(a) M1 substitutes sin² θ = 1 - cos² θ throughout
(a) A1 correctly simplifies and collects terms in cos θ
(a) A1 cos(5theta) = 16cos⁵(θ) - 20cos³(θ) + 5cos(θ), cso
(a) Answer: cos(5theta) = 16cos⁵(θ) - 20cos³(θ) + 5cos(θ) (shown)
(b) M1 uses part (a) to rewrite the equation as cos(5theta) = 0
(b) M1 sets 5theta = π/2 + k π
(b) A1 θ = π/10, 3pi/10, π/2, 7pi/10, 9pi/10 (at least 3 correct)
(b) A1 all five values correct within 0 < θ < π, cao
(b) Answer: θ = π/10, 3pi/10, π/2, 7pi/10, 9pi/10

Question 9

(a) M1 writes 64 = 64(cos 0 + i sin 0)
(a) A1 modulus of each root r = 64^1/6 = 2
(a) M1 uses θ = 2k π/6 = k π/3 for six consecutive integers k
(a) A1 arguments 0, π/3, 2pi/3 correct
(a) A1 arguments π, -2pi/3, -π/3 correct
(a) A1 all six roots written correctly in the required form, cao
(a) Answer: 2(cos0+isin0), 2(cos(π/3)+isin(π/3)), 2(cos(2pi/3)+isin(2pi/3)), 2(cos(π)+isin(π)), 2(cos(-2pi/3)+isin(-2pi/3)), 2(cos(-π/3)+isin(-π/3))
(b) B1 all six roots have modulus 2, so all lie on a circle of radius 2 centred at the origin, spaced at equal angles of π/3, forming a regular hexagon
(b) M1 finds the side length using 2 x 2 x sin(π/6)
(b) A1 perimeter = 6 x 2 = 12, cao
(b) Answer: Perimeter = 12
(c) M1 uses area = 6 x (1/2) r² sin(π/3)
(c) A1 substitutes r = 2: 6 x (1/2)(4)sin(π/3)
(c) A1 area = 6sqrt(3), cao
(c) Answer: 6sqrt(3)

Question 10

(a) B1 circle centre 3, radius 3, passing through the origin, drawn correctly
(a) B1 half-lines arg(z) = 0 and arg(z) = π/4 drawn from the origin
(a) B1 region R (bounded by the circle and lying between the two half-lines) correctly identified/shaded
(a) Answer: Region bounded by the circle |z-3|=3 and the half-lines arg(z)=0, arg(z)=π/4, with the origin on the boundary
(b) M1 writes |z-3|=3 as (x-3)²+y²=9 and converts to polar using x = r cos θ, y = r sin θ
(b) M1 expands and uses x²+y² = r² to obtain r² = 6r cos θ
(b) A1 divides by r (r ≠ 0) to get r = 6cos θ, cso
(b) Answer: r = 6cos(θ) (shown)
(c) M1 uses area = integral from 0 to π/4 of (1/2) r² d θ with r = 6cos θ
(c) M1 forms the integral (1/2) x integral of 36cos²(θ) d θ = 18 x integral of cos²(θ) d θ
(c) M1 uses cos²(θ) = (1+cos(2theta))/2 to integrate
(c) A1 correct antiderivative 18[θ/2 + sin(2theta)/4]
(c) M1 substitutes limits θ = π/4 and θ = 0
(c) A1 area = 9pi/4 + 9/2, oe exact form, cao
(c) Answer: Area = 9pi/4 + 9/2 (= (9/4)(π+2))

Question 11

(a) M1 uses 2i sin(θ) = z - 1/z
(a) M1 expands (z - 1/z)⁴ using the binomial theorem
(a) M1 uses z^k + 1/z^k = 2cos(k θ) to group terms
(a) A1 correctly simplifies to (2i sin θ)⁴ = 2cos(4theta) - 8cos(2theta) + 6
(a) A1 divides by 16 to give sin⁴(θ) = (1/8)cos(4theta) - (1/2)cos(2theta) + 3/8, so A=1/8, B=-1/2, C=3/8, cso
(a) Answer: sin⁴(θ) = (1/8)cos(4theta) - (1/2)cos(2theta) + 3/8 (A=1/8, B=-1/2, C=3/8)
(b) M1 integrates term by term: (1/32)sin(4theta) - (1/4)sin(2theta) + (3/8)θ
(b) M1 substitutes limits θ = π/2 and θ = 0
(b) A1 evaluates the sin(4theta) and sin(2theta) terms to 0 at both limits
(b) A1 exact value = 3pi/16, cao
(b) Answer: 3pi/16