Free worked solutions · 2026 Edition

Chapter 6: Complex numbers worked solutions

Worked solutions to the practice questions in Chapter 6 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[7 marks]

(a) Using i^2 = -1, find i^3, i^4, i^5 and i^6.

[2]

(b) Find (i) i^{43}; (ii) i^{100}; (iii) \dfrac{1}{i}.

[3]

(c) Find the value of i + i^2 + i^3 + \ldots + i^{22}.

[2]

Show solutionHide solution

(a) i^3 = i^2 \times i = -i, \ i^4 = (i^2)^2 = 1

i^5 = i^4 \times i = i, \ i^6 = i^4 \times i^2 = -1

(b)(i) 43 = 4 \times 10 + 3, so i^{43} = \left(i^4\right)^{10} i^3 = -i

(ii) i^{100} = \left(i^4\right)^{25} = 1

(iii) \dfrac{1}{i} = \dfrac{i}{i^2} = -i

(c) i + i^2 + i^3 + i^4 = i - 1 - i + 1 = 0, so every block of four consecutive powers sums to 0 and i + \ldots + i^{20} = 0

sum = i^{21} + i^{22} = i - 1 = -1 + i

Note: accept the geometric series \dfrac{i\left(1 - i^{22}\right)}{1 - i} = \dfrac{2i}{1 - i} = -1 + i.

Final answer

(a) i^3=-i, i^4=1, i^5=i, i^6=-1
(b)(i) -i
(ii) 1
(iii) -i
(c) -1+i

Question 2

[8 marks]

Write each number in the form a + bi, where a, b \in \mathbb{R}.

(a) \dfrac{10}{3 - i}

[2]

(b) \dfrac{4 + 7i}{2 + i}

[2]

(c) \dfrac{3 - 5i}{i}

[2]

(d) \dfrac{1 + 8i}{4 - 3i}

[2]

Show solutionHide solution

(a) \dfrac{10(3 + i)}{(3 - i)(3 + i)} = \dfrac{10(3 + i)}{10}

= 3 + i

(b) \dfrac{(4 + 7i)(2 - i)}{(2 + i)(2 - i)} = \dfrac{8 - 4i + 14i - 7i^2}{5}

= \dfrac{15 + 10i}{5} = 3 + 2i

(c) \dfrac{(3 - 5i)(-i)}{i(-i)} = \dfrac{-3i + 5i^2}{1}

= -5 - 3i

(d) \dfrac{(1 + 8i)(4 + 3i)}{(4 - 3i)(4 + 3i)} = \dfrac{4 + 3i + 32i + 24i^2}{25}

= \dfrac{-20 + 35i}{25} = -\dfrac{4}{5} + \dfrac{7}{5}i

Note: the M1 is for multiplying numerator and denominator by the conjugate of the denominator (in (c), by -i or i).

Final answer

(a) 3+i
(b) 3+2i
(c) -5-3i
(d) -\tfrac45+\tfrac75 i

Question 3

[8 marks]

Write each complex number in the form r\,\mathrm{cis}\,\theta, where r > 0 and -\pi < \theta \le \pi. Give exact values.

(a) 5\sqrt{3} + 5i

[2]

(b) -4 + 4i

[2]

(c) -3 - 3\sqrt{3}\,i

[2]

(d) \sqrt{2} - \sqrt{6}\,i

[2]

Show solutionHide solution

(a) r = \sqrt{75 + 25} = 10

first quadrant, \tan\theta = \dfrac{5}{5\sqrt{3}} = \dfrac{1}{\sqrt{3}}: \ 10\,\mathrm{cis}\,\dfrac{\pi}{6}

(b) r = \sqrt{16 + 16} = 4\sqrt{2}

second quadrant, reference angle \dfrac{\pi}{4}: \ 4\sqrt{2}\,\mathrm{cis}\,\dfrac{3\pi}{4}

(c) r = \sqrt{9 + 27} = 6

third quadrant, reference angle \dfrac{\pi}{3}: \ 6\,\mathrm{cis}\left(-\dfrac{2\pi}{3}\right)

(d) r = \sqrt{2 + 6} = 2\sqrt{2}

fourth quadrant, \tan(\text{reference angle}) = \dfrac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}: \ 2\sqrt{2}\,\mathrm{cis}\left(-\dfrac{\pi}{3}\right)

Note: the range -\pi < \theta \le \pi is stated, so an argument outside it (e.g. \frac{4\pi}{3} or \frac{5\pi}{3} instead of -\frac{2\pi}{3} or -\frac{\pi}{3}) does not earn the final A1. In (b) and (c), \tan^{-1}\frac{b}{a} alone gives the wrong quadrant: award A0 for -\frac{\pi}{4} or \frac{\pi}{3}.

Final answer

(a) 10\operatorname{cis}\frac{\pi}{6}
(b) 4\sqrt2\operatorname{cis}\frac{3\pi}{4}
(c) 6\operatorname{cis}\!\left(-\frac{2\pi}{3}\right)
(d) 2\sqrt2\operatorname{cis}\!\left(-\frac{\pi}{3}\right)

Question 4

[8 marks]

Write each complex number in Cartesian form a + bi, giving exact values.

(a) 8e^{i\frac{3\pi}{4}}

[2]

(b) 2\sqrt{2}\,e^{-i\frac{3\pi}{4}}

[2]

(c) 8\,\mathrm{cis}\left(-\dfrac{\pi}{6}\right)

[2]

(d) 10e^{i\frac{2\pi}{3}}

[2]

Show solutionHide solution

(a) 8\left(\cos\dfrac{3\pi}{4} + i\sin\dfrac{3\pi}{4}\right) = 8\left(-\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i\right)

= -4\sqrt{2} + 4\sqrt{2}\,i

(b) 2\sqrt{2}\left(-\dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{2}}{2}i\right)

= -2 - 2i

(c) 8\left(\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i\right)

= 4\sqrt{3} - 4i

(d) 10\left(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right)

= -5 + 5\sqrt{3}\,i

Final answer

(a) -4\sqrt2+4\sqrt2\,i
(b) -2-2i
(c) 4\sqrt3-4i
(d) -5+5\sqrt3\,i

Question 5

[8 marks]

Use De Moivre's theorem to write each power in the form a + bi, where a, b \in \mathbb{R}, giving exact values.

(a) \left(2\,\mathrm{cis}\,\dfrac{\pi}{8}\right)^4

[2]

(b) \left(\sqrt{3}\,\mathrm{cis}\left(-\dfrac{\pi}{9}\right)\right)^6

[2]

(c) \left(\dfrac{1}{2}e^{i\frac{3\pi}{10}}\right)^5

[2]

(d) \left(\sqrt{2}\,e^{-i\frac{5\pi}{6}}\right)^6

[2]

Show solutionHide solution

(a) 2^4\,\mathrm{cis}\,\dfrac{4\pi}{8} = 16\,\mathrm{cis}\,\dfrac{\pi}{2}

= 16i

(b) \left(\sqrt{3}\right)^6\mathrm{cis}\left(-\dfrac{6\pi}{9}\right) = 27\,\mathrm{cis}\left(-\dfrac{2\pi}{3}\right)

= -\dfrac{27}{2} - \dfrac{27\sqrt{3}}{2}i

(c) \dfrac{1}{32}e^{i\frac{15\pi}{10}} = \dfrac{1}{32}e^{i\frac{3\pi}{2}}

= -\dfrac{1}{32}i

(d) \left(\sqrt{2}\right)^6e^{-i\frac{30\pi}{6}} = 8e^{-5\pi i}

= 8e^{i\pi} = -8

Note: M1 for raising the modulus to the power and multiplying the argument by the power.

Final answer

(a) 16i
(b) -\frac{27}{2}-\frac{27\sqrt3}{2}i
(c) -\frac{1}{32}i
(d) -8

Question 6

[7 marks]

Solve each equation for z \in \mathbb{C}, giving the roots in the form a + bi, where a, b \in \mathbb{R}.

(a) z^2 + 4z + 29 = 0

[2]

(b) z^2 - 10z + 34 = 0

[2]

(c) 2z^2 - 2z + 5 = 0

[2]

(d) What do you notice about the two roots in each of parts (a), (b) and (c)?

[1]

Show solutionHide solution

(a) z = \dfrac{-4 \pm \sqrt{16 - 116}}{2} = \dfrac{-4 \pm \sqrt{-100}}{2}

z = -2 \pm 5i

(b) z = \dfrac{10 \pm \sqrt{100 - 136}}{2} = \dfrac{10 \pm 6i}{2}

z = 5 \pm 3i

(c) z = \dfrac{2 \pm \sqrt{4 - 40}}{4} = \dfrac{2 \pm 6i}{4}

z = \dfrac{1}{2} \pm \dfrac{3}{2}i

(d) in each case the discriminant is negative and the two roots are complex conjugates of each other (w and w^{*})

Note: completing the square is equally valid, e.g. (a) (z + 2)^2 = -25, {z + 2 = \pm 5i}.

Final answer

(a) -2\pm5i
(b) 5\pm3i
(c) \frac12\pm\frac32 i
(d) each pair is a conjugate pair

Question 7

[6 marks]

Each number below is a root of a quadratic equation with real coefficients. In each part, write down the other root and find a quadratic equation az^2 + bz + c = 0, with a, b, c \in \mathbb{Z} and a > 0, that has these two roots.

(a) 5i

[2]

(b) 4 - 5i

[2]

(c) \dfrac{3 + \sqrt{5}\,i}{2}

[2]

Show solutionHide solution

(a) the coefficients are real, so the other root is the conjugate -5i

(z - 5i)(z + 5i) = z^2 + 25, so z^2 + 25 = 0

(b) other root 4 + 5i; sum of roots = 8, product = 4^2 + 5^2 = 41

z^2 - 8z + 41 = 0

(c) other root \dfrac{3 - \sqrt{5}\,i}{2}; sum = 3, product = \dfrac{9 + 5}{4} = \dfrac{7}{2}

z^2 - 3z + \dfrac{7}{2} = 0, i.e. 2z^2 - 6z + 7 = 0

Note: accept any positive integer multiple of each equation; in (b) and (c) award the M1 for (z - w)(z - w^{*}) expanded.

Final answer

(a) -5i; z^2+25=0
(b) 4+5i; z^2-8z+41=0
(c) \frac{3-\sqrt5 i}{2}; 2z^2-6z+7=0

Question 8

[6 marks]

The equation z^3 + az^2 + bz - 146 = 0, where a, b \in \mathbb{R}, has a root 3 + 8i.

(a) Write down a second root of the equation.

[1]

(b) By considering the product of the roots, find the third root.

[3]

(c) Find the value of a and the value of b.

[2]

Show solutionHide solution

(a) 3 - 8i (the coefficients are real)

(b) (3 + 8i)(3 - 8i) = 9 + 64 = 73

product of the roots = \dfrac{(-1)^3 a_0}{a_3} = \dfrac{-(-146)}{1} = 146, so 73\gamma = 146

\gamma = 2

(c) sum of the roots = (3 + 8i) + (3 - 8i) + 2 = 8 = -a, so a = -8

z = 2 is a root: 8 + 4a + 2b - 146 = 0 \Rightarrow 8 - 32 + 2b - 146 = 0 \Rightarrow b = 85

Note: accept expanding (z^2 - 6z + 73)(z - 2) = z^3 - 8z^2 + 85z - 146 for (c); FT from (b).

Final answer

(a) 3-8i
(b) 2
(c) a=-8, b=85

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 9

[7 marks]

Let z = 1 + 2i.

(a) Show that z^2 = -3 + 4i.

[1]

(b) Hence find z^4.

[2]

(c) Find \dfrac{1}{z} + \dfrac{1}{z^2}, giving your answer in the form a + bi.

[4]

Show solutionHide solution

(a) (1+2i)^2 = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i.

[1]

(b) z^4 = (z^2)^2 = (-3+4i)^2 = 9 - 24i + 16i^2

[1]

= -7 - 24i.

[1]

(c) \dfrac{1}{z} = \dfrac{1-2i}{(1+2i)(1-2i)} = \dfrac{1-2i}{5}.

[2]

\dfrac{1}{z^2} = \dfrac{1}{-3+4i} = \dfrac{-3-4i}{9+16} = \dfrac{-3-4i}{25}.

[1]

Sum: \dfrac{5(1-2i) + (-3-4i)}{25} = \dfrac{2 - 14i}{25} = \dfrac{2}{25} - \dfrac{14}{25}i.

[1]

Final answer

(b) -7-24i
(c) \frac{2}{25}-\frac{14}{25}i

Question 10

[4 marks]

The number \dfrac{p + 3i}{2 - i}, where p \in \mathbb{R}, is purely imaginary.

(a) Find the value of p.

[3]

(b) Hence write the number in the form bi.

[1]

Show solutionHide solution

(a) Realise the denominator: \dfrac{(p+3i)(2+i)}{(2-i)(2+i)} = \dfrac{2p + pi + 6i + 3i^2}{5}.

[1]

= \dfrac{(2p - 3) + (p + 6)i}{5}.

[1]

Purely imaginary means the real part is zero: 2p - 3 = 0, so p = \dfrac{3}{2}.

[1]

(b) With p = \tfrac32 the number is \dfrac{(\tfrac32 + 6)i}{5} = \dfrac{3}{2}i.

[1]

Final answer

(a) p=\frac32
(b) \frac32 i

Question 11

[6 marks]

(a) Find \left(\sqrt{3} - i\right)^7 in the form a + bi, where a, b \in \mathbb{R}.

[4]

(b) Find the smallest positive integer n for which \left(\sqrt{3} - i\right)^n is a negative real number, and write down the value of \left(\sqrt{3} - i\right)^n for this n.

[2]

Show solutionHide solution

(a) \sqrt{3} - i has modulus 2 and argument -\dfrac{\pi}{6}, so \sqrt{3} - i = 2e^{-i\pi/6}.

[1]

\left(\sqrt{3} - i\right)^7 = 2^7 e^{-7\pi i/6} = 128e^{-7\pi i/6}.

[2]

Since -\dfrac{7\pi}{6} + 2\pi = \dfrac{5\pi}{6}: 128\left(\cos\dfrac{5\pi}{6} + i\sin\dfrac{5\pi}{6}\right) = 128\left(-\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i\right) = -64\sqrt{3} + 64i.

[1]

(b) The argument of \left(\sqrt{3} - i\right)^n is -\dfrac{n\pi}{6}. The number is a negative real number when -\dfrac{n\pi}{6} = \pi + 2\pi m for an integer m, i.e.; n = -6 - 12m.

[1]

The smallest positive value is n = 6 (take m = -1), and \left(\sqrt{3} - i\right)^6 = 2^6 e^{-i\pi} = -64.

[1]

Final answer

(a) -64\sqrt3+64i
(b) n=6; -64

Question 12

[5 marks]

Solve the equation (z - 2)^3 = 27i, giving your answers in the form a + bi.

[5]

Show solutionHide solution

Let u = z - 2, so u^3 = 27i.

[1]

27i = 27e^{i\pi/2}, so u_k = 3e^{i(\pi/2 + 2\pi k)/3} for k = 0, 1, 2.

[1]

u = 3e^{i\pi/6} = \dfrac{3\sqrt{3}}{2} + \dfrac{3}{2}i, \quad u = 3e^{5\pi i/6} = -\dfrac{3\sqrt{3}}{2} + \dfrac{3}{2}i, \quad u = 3e^{-i\pi/2} = -3i.

[1]

Now z = u + 2.

[1]

z = 2 + \dfrac{3\sqrt{3}}{2} + \dfrac{3}{2}i, \quad z = 2 - \dfrac{3\sqrt{3}}{2} + \dfrac{3}{2}i, \quad z = 2 - 3i.

[1]

Final answer

2+\frac{3\sqrt3}{2}+\frac32 i,
2-\frac{3\sqrt3}{2}+\frac32 i,
2-3i

Question 13

[5 marks]

The complex number z satisfies z + 2iz^* = 5 + 7i. Find z.

[5]

Show solutionHide solution

The unknown appears with its conjugate, so put z = x + yi with x, y \in \mathbb{R}.

[1]

2iz^* = 2i(x - yi) = 2y + 2xi, so z + 2iz^* = (x + 2y) + (y + 2x)i.

[1]

Equate real and imaginary parts: x + 2y = 5 and 2x + y = 7.

[1]

Solve: from the first, x = 5 - 2y; then 10 - 4y + y = 7, so y = 1 and x = 3.

[1]

z = 3 + i.

[1]

Final answer

z=3+i

Question 14

[8 marks]

(a) Find the two square roots of 21 - 20i, giving each in the form a + bi with a, b \in \mathbb{Z}.

[5]

(b) Hence solve the equation z^2 - 2z - 20 + 20i = 0.

[3]

Show solutionHide solution

(a) Let (a+bi)^2 = 21 - 20i. The real parts give a^2 - b^2 = 21 and the imaginary parts give 2ab = -20.

The modulus of 21 - 20i is \sqrt{21^2 + 20^2} = 29, so a^2 + b^2 = 29.

M1A1

Adding and subtracting: a^2 = 25 and b^2 = 4.

Since ab = -10 < 0, the signs differ: the square roots are 5 - 2i and -5 + 2i.

(b) Complete the square: (z-1)^2 = 1 + 20 - 20i = 21 - 20i.

So z - 1 = \pm(5 - 2i) by part (a).

z = 6 - 2i or z = -4 + 2i.

Final answer

(a) \pm(5-2i)
(b) z=6-2i or z=-4+2i

Question 15

[8 marks]

The cubic equation z^3 + az^2 + bz + c = 0, where a, b, c \in \mathbb{R}, has the root 2 + i. The sum of the squares of the three roots of the equation is 22.

(a) Write down the other non-real root.

[1]

(b) Find the possible values of the third root.

[3]

(c) Hence find the possible values of a, b and c.

[4]

Show solutionHide solution

(a) The coefficients are real, so the other non-real root is 2 - i.

[1]

(b) Let the third root be r (it is real, because non-real roots come in pairs). (2 + i)^2 = 3 + 4i and (2 - i)^2 = 3 - 4i.

[1]

Sum of squares: (3 + 4i) + (3 - 4i) + r^2 = 22, so 6 + r^2 = 22 and r^2 = 16.

[1]

r = 4 or r = -4.

[1]

(c) The cubic is (z - (2+i))(z - (2-i))(z - r) = (z^2 - 4z + 5)(z - r).

[1]

= z^3 - (4 + r)z^2 + (5 + 4r)z - 5r.

[1]

For r = 4: z^3 - 8z^2 + 21z - 20, so a = -8,\ b = 21,\ c = -20.

[1]

For r = -4: z^3 + 0z^2 - 11z + 20, so a = 0,\ b = -11,\ c = 20.

[1]

Final answer

(a) 2-i
(b) 4 or -4
(c) a=-8, b=21, c=-20; or a=0, b=-11, c=20

Question 16

[7 marks]

The points O, A and B represent the complex numbers 0, a = 2 + 2i and b respectively. The point B is the image of A under a rotation of \dfrac{\pi}{3} anticlockwise about O.

(a) Find b in the form x + yi, where x, y \in \mathbb{R}.

[3]

(b) Show that \mathrm{AB} = \mathrm{OA}, and hence state the type of triangle OAB.

[2]

(c) Find the exact area of triangle OAB.

[2]

Show solutionHide solution

(a) A rotation of \dfrac{\pi}{3} anticlockwise about O is multiplication by e^{i\pi/3} = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i, so b = (2 + 2i)\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right).

[1]

= 1 + \sqrt{3}\,i + i + \sqrt{3}\,i^2.

[1]

b = (1 - \sqrt{3}) + (1 + \sqrt{3})i.

[1]

(b) \mathrm{AB} = |a - b| = \left|a\left(1 - e^{i\pi/3}\right)\right| = |a|\left|\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i\right| = |a| \times 1 = \mathrm{OA}.

[1]

Also \mathrm{OB} = |b| = |a|\,|e^{i\pi/3}| = \mathrm{OA} because a rotation preserves length. All three sides are equal, so OAB is an equilateral triangle.

[1]

(c) The side length squared is \mathrm{OA}^2 = 2^2 + 2^2 = 8.

[1]

Area = \dfrac{\sqrt{3}}{4} \times 8 = 2\sqrt{3}.

[1]

Final answer

(a) (1-\sqrt3)+(1+\sqrt3)i
(b) equilateral
(c) 2\sqrt3

Question 17

[8 marks]

The complex number z = x + iy satisfies |z - 3| = 2|z + 3|.

(a) Show that x^2 + y^2 + 10x + 9 = 0.

[4]

(b) Write down the centre and the radius of the circle with this equation.

[2]

(c) Find the greatest value of |z|.

[2]

Show solutionHide solution

(a) Square both sides: (x-3)^2 + y^2 = 4\bigl[(x+3)^2 + y^2\bigr].

x^2 - 6x + 9 + y^2 = 4x^2 + 24x + 36 + 4y^2.

0 = 3x^2 + 3y^2 + 30x + 27.

Dividing by 3: x^2 + y^2 + 10x + 9 = 0.

(b) (x+5)^2 + y^2 = 16, so the centre is (-5, 0), that is z = -5, and the radius is 4.

A1A1

(c) The origin is 5 from the centre, outside the circle. The point of the circle furthest from O is on the far side of the centre.

Greatest |z| = 5 + 4 = 9.

Final answer

(b) centre -5, radius 4
(c) 9

Question 18

[8 marks]

You may use the identity \cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1.

(a) Show that x = \cos\dfrac{\pi}{8} is a root of the equation 8x^4 - 8x^2 + 1 = 0.

[2]

(b) Hence find the exact value of \cos\dfrac{\pi}{8}.

[3]

(c) Show that \sec^2\dfrac{\pi}{8} + \sec^2\dfrac{3\pi}{8} = 8.

[3]

Show solutionHide solution

(a) Put \theta = \dfrac{\pi}{8}: \cos\dfrac{\pi}{2} = 0 = 8\cos^4\dfrac{\pi}{8} - 8\cos^2\dfrac{\pi}{8} + 1.

[1]

Hence x = \cos\dfrac{\pi}{8} satisfies 8x^4 - 8x^2 + 1 = 0.

[1]

(b) Let u = x^2: 8u^2 - 8u + 1 = 0, so u = \dfrac{8 \pm \sqrt{64 - 32}}{16} = \dfrac{2 \pm \sqrt{2}}{4}.

[2]

\cos^2\dfrac{\pi}{8} > \cos^2\dfrac{\pi}{4} = \dfrac{1}{2}, and only \dfrac{2 + \sqrt{2}}{4} is larger than \frac12 (the other root, \frac{2-\sqrt{2}}{4}, is smaller than \frac12). Since \cos\dfrac{\pi}{8} > 0, \cos\dfrac{\pi}{8} = \dfrac{\sqrt{2 + \sqrt{2}}}{2}.

[1]

(c) Putting \theta = \dfrac{3\pi}{8} gives \cos\dfrac{3\pi}{2} = 0, so \cos^2\dfrac{3\pi}{8} is also a root of 8u^2 - 8u + 1 = 0, and it is the other root. So u_1 = \cos^2\dfrac{\pi}{8} and u_2 = \cos^2\dfrac{3\pi}{8} are the two roots.

[1]

u_1 + u_2 = 1 and u_1 u_2 = \dfrac{1}{8}.

[1]

\sec^2\dfrac{\pi}{8} + \sec^2\dfrac{3\pi}{8} = \dfrac{1}{u_1} + \dfrac{1}{u_2} = \dfrac{u_1 + u_2}{u_1 u_2} = \dfrac{1}{1/8} = 8.

[1]

Final answer

(b) \cos\frac{\pi}{8}=\frac{\sqrt{2+\sqrt2}}{2}

Question 19

[8 marks]

Find all real values of a for which (a + i)^6 is purely imaginary. Give your answers in exact form.

[8]

Show solutionHide solution

Write a + i = \rho e^{i\varphi} with \rho = \sqrt{a^2 + 1}. Since \mathrm{Im}(a + i) = 1 > 0, \varphi \in (0, \pi), and \cot\varphi = a.

[1]

(a + i)^6 = \rho^6 e^{6i\varphi} is purely imaginary when 6\varphi = \dfrac{\pi}{2} + m\pi, i.e.; \varphi = \dfrac{\pi}{12} + \dfrac{m\pi}{6}.

[1]

For \varphi \in (0, \pi) this gives exactly six values (m = 0, 1, \dots, 5).

[1]

\varphi = \dfrac{\pi}{12},\ \dfrac{\pi}{4},\ \dfrac{5\pi}{12},\ \dfrac{7\pi}{12},\ \dfrac{3\pi}{4},\ \dfrac{11\pi}{12}.

[1]

Then a = \cot\varphi. Using \tan\dfrac{\pi}{12} = 2 - \sqrt{3} and \tan\dfrac{5\pi}{12} = 2 + \sqrt{3} (from \tan\left(\frac{\pi}{4} \mp \frac{\pi}{6}\right)):

[1]

\cot\dfrac{\pi}{12} = \dfrac{1}{2 - \sqrt{3}} = 2 + \sqrt{3} and \cot\dfrac{5\pi}{12} = 2 - \sqrt{3}.

[1]

\cot\dfrac{\pi}{4} = 1 and \cot\dfrac{3\pi}{4} = -1.

[1]

\cot\dfrac{7\pi}{12} = -\cot\dfrac{5\pi}{12} = -(2 - \sqrt{3}) and \cot\dfrac{11\pi}{12} = -(2 + \sqrt{3}).

So a \in \{1,\ -1,\ 2 - \sqrt{3},\ -(2 - \sqrt{3}),\ 2 + \sqrt{3},\ -(2 + \sqrt{3})\}.

[1]

(Check a = 1: (1 + i)^6 = (2i)^3 = -8i.)

Final answer

a=\pm1,\ \pm(2-\sqrt3),\ \pm(2+\sqrt3)

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 20

[5 marks]

The complex number z = 6 + qi satisfies z^2 - i z = p + 42i, where p, q \in \mathbb{R}.

Determine p and q.

[5]

Show solutionHide solution

z^2 = 36 - q^2 + 12q\mathrm{i} and -\mathrm{i}z = -6\mathrm{i} + q. So

z^2 - \mathrm{i}z = (36 - q^2 + q) + \mathrm{i}(12q - 6).

Imaginary parts: 12q-6=42\Rightarrow q=4.

Real parts: p=36-16+4=24.

p = 24, q = 4

[5]

Final answer

p=24, q=4

Question 21

[7 marks]

The complex numbers z and w satisfy

w = (2 + i)z \qquad \text{and} \qquad 2w - z^{*} = 9 - 13i.

Determine z and w, giving each in the form a + bi with a, b \in \mathbb{Z}.

[7]

Show solutionHide solution

Substitute w=(2+\mathrm{i})z and let z = x + y\mathrm{i}:

2(2+\mathrm{i})(x+y\mathrm{i}) - (x - y\mathrm{i}) = (3x - 2y) + \mathrm{i}(2x + 5y)

Equating with 9 - 13\mathrm{i}:

3x - 2y = 9, \qquad 2x + 5y = -13

Eliminating x: 19y = -57, so y=-3 and x=1.

z = 1 - 3\mathrm{i}

w = (2+\mathrm{i})(1-3\mathrm{i}) = 2 - 6\mathrm{i} + \mathrm{i} + 3 = 5 - 5\mathrm{i}

z = 1 - 3\mathrm{i}, w = 5 - 5\mathrm{i}

[7]

Final answer

z=1-3i, w=5-5i

Question 22

[6 marks]

Let z_1 = p + i and z_2 = 2 - pi, where p \in \mathbb{R}^{+}.

(a) Express z_1 z_2 in terms of p, in the form a + bi.

[3]

(b) It is given that \arg(z_1 z_2) = \dfrac{\pi}{4}. Determine the exact value of p.

[3]

Show solutionHide solution

(a) z_1z_2 = (p+\mathrm{i})(2-p\mathrm{i}) = 2p - p^2\mathrm{i} + 2\mathrm{i} - p\mathrm{i}^2 = 3p + (2-p^2)\mathrm{i}

[3]

(b) Since p>0, the real part 3p>0, so \arg(z_1z_2)=\dfrac{\pi}{4} requires the imaginary part to equal the real part:

2 - p^2 = 3p \implies p^2 + 3p - 2 = 0 \implies p = \frac{-3\pm\sqrt{17}}{2}

Because p>0:

p = \frac{\sqrt{17}-3}{2}

[3]

Final answer

(a) 3p+(2-p^2)i
(b) p=\frac{\sqrt{17}-3}{2}

Question 23

[12 marks]

(a) Find the square roots of -5 + 12i in the form a + bi.

[5]

(b) Solve z^2 - 2(1+i)z + 5 - 10i = 0, giving the roots in the form a + bi.

[4]

(c) The roots in part (b) are represented by the points P and Q on an Argand diagram. O is the origin. Find the area of triangle OPQ.

[3]

Show solutionHide solution

(a) Let (a+bi)^2 = -5 + 12i: a^2 - b^2 = -5 and 2ab = 12.

|-5 + 12i| = 13, so a^2 + b^2 = 13.

M1A1

a^2 = 4 and b^2 = 9.

ab = 6 > 0, so the signs agree: the square roots are \pm(2 + 3i).

(b) Complete the square: \bigl(z - (1+i)\bigr)^2 = (1+i)^2 - 5 + 10i = 2i - 5 + 10i = -5 + 12i.

M1A1

z - (1+i) = \pm(2 + 3i) by part (a).

z = 3 + 4i or z = -1 - 2i.

(c) P(3, 4) and Q(-1, -2).

Area = \tfrac{1}{2}\,|3\times(-2) - 4\times(-1)| = \tfrac{1}{2}\,|-6 + 4|.

Area = 1.

Final answer

(a) \pm(2+3i)
(b) 3+4i,
-1-2i
(c) 1

Question 24

[6 marks]

Let f(z) = z^3 - mz^2 + \dfrac{16}{m}z - 4m, where z \in \mathbb{C} and m \in \mathbb{R}^{+}.

It is given that z - 2i is a factor of f(z).

Find m, and hence solve f(z)=0.

[6]

Show solutionHide solution

f(2\mathrm{i}) = (2\mathrm{i})^3 - m(2\mathrm{i})^2 + \dfrac{16}{m}(2\mathrm{i}) - 4m = -8\mathrm{i} + 4m + \dfrac{32}{m}\mathrm{i} - 4m = \mathrm{i}\left(\dfrac{32}{m}-8\right).

Since f(2\mathrm{i})=0, \dfrac{32}{m}=8, so m=4.

[3]

Then f(z) = z^3 - 4z^2 + 4z - 16 = (z^2+4)(z-4) (real coefficients, so z+2\mathrm{i} is also a factor).

Roots: z = 2\mathrm{i},\ -2\mathrm{i},\ 4

[3]

Final answer

m=4; z=2i,\ -2i,\ 4

Question 25

[7 marks]

The quartic equation z^4 + pz^3 + z^2 + 6z - 40 = 0 has real coefficients, with p \in \mathbb{R}.

Three of its roots are 1 + 2i, \alpha and \alpha^2, where \alpha \in \mathbb{R}.

(a) Using the product of all four roots, determine \alpha.

[4]

(b) Hence determine p.

[3]

Show solutionHide solution

(a) Real coefficients imply 1-2\mathrm{i} is also a root, so the four roots are 1\pm2\mathrm{i}, \alpha, \alpha^2.

Product of the roots = -40 (the constant term, degree even):

(1+2\mathrm{i})(1-2\mathrm{i})\,\alpha\cdot\alpha^2 = 5\alpha^3 = -40 \implies \alpha^3=-8 \implies \alpha=-2

[4]

(b) The roots are 1\pm2\mathrm{i}, -2, 4. Sum of roots =-p:

(1+2\mathrm{i})+(1-2\mathrm{i})-2+4 = 4 \implies p=-4

[3]

Final answer

(a) \alpha=-2
(b) p=-4

Question 26

[5 marks]

Let z = \cos\theta + i\sin\theta, where z \neq -1.

Prove that \mathrm{Re}\left(\dfrac{1}{1+z}\right) = \dfrac{1}{2}.

[5]

Show solutionHide solution

Since |z|=1, z^{*} = \cos\theta - \mathrm{i}\sin\theta and

\frac{1}{1+z} = \frac{1+z^{*}}{(1+z)(1+z^{*})} = \frac{1+z^{*}}{1 + z + z^{*} + zz^{*}}.

Now zz^{*}=1 and z+z^{*}=2\cos\theta, so the denominator is 2+2\cos\theta = 2(1+\cos\theta), which is non-zero because z\neq-1.

The numerator has real part 1+\cos\theta. Hence

\mathrm{Re}\left(\frac{1}{1+z}\right) = \frac{1+\cos\theta}{2(1+\cos\theta)} = \frac12 \qquad \blacksquare

[5]

Final answer

Proof: multiply by \frac{1+z^*}{1+z^*}; the real part is \frac{1+\cos\theta}{2(1+\cos\theta)}.

Question 27

[12 marks]

The complex number z = x + iy satisfies |z - 6i| = 2|z - 3i|.

(a) Show that x^2 + y^2 - 4y = 0.

[5]

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

[2]

(c) Find the value of z that also satisfies \arg z = \dfrac{\pi}{4}.

[3]

(d) Find the least value of |z - 6|.

[2]

Show solutionHide solution

(a) |z - 6i|^2 = x^2 + (y-6)^2 and |z - 3i|^2 = x^2 + (y-3)^2.

Squaring the given equation: x^2 + (y-6)^2 = 4\bigl[x^2 + (y-3)^2\bigr].

x^2 + y^2 - 12y + 36 = 4x^2 + 4y^2 - 24y + 36.

0 = 3x^2 + 3y^2 - 12y.

x^2 + y^2 - 4y = 0.

(b) x^2 + (y-2)^2 = 4: centre (0, 2), that is z = 2i, radius 2.

A1A1

(c) \arg z = \tfrac{\pi}{4} means y = x with x > 0.

2x^2 - 4x = 0, so x = 0 or x = 2.

x = 0 gives z = 0, whose argument is undefined, so z = 2 + 2i.

(d) |z - 6| is the distance from (6, 0). The centre is \sqrt{6^2 + 2^2} = 2\sqrt{10} from (6, 0).

Least value = 2\sqrt{10} - 2.

Final answer

(b) centre 2i, radius 2
(c) 2+2i
(d) 2\sqrt{10}-2

Question 28

[8 marks]

Let z_1 = 1 + i and z_2 = -8\sqrt2.

(a) Write z_1 in the form re^{i\theta}, with r>0 and -\pi<\theta\le\pi.

[3]

The three cube roots of \dfrac{z_2}{z_1} are u, v and w.

(b) Calculate u, v and w, each in the form re^{i\theta}.

[5]

Show solutionHide solution

(a) |z_1|=\sqrt2, \arg z_1=\dfrac\pi4, so z_1=\sqrt2\,\mathrm{e}^{\mathrm{i}\pi/4}.

[3]

(b) z_2=8\sqrt2\,\mathrm{e}^{\mathrm{i}\pi}, so

\frac{z_2}{z_1}=\frac{8\sqrt2\,\mathrm{e}^{\mathrm{i}\pi}}{\sqrt2\,\mathrm{e}^{\mathrm{i}\pi/4}}=8\mathrm{e}^{\mathrm{i}3\pi/4}.

Cube roots: modulus 8^{1/3}=2, arguments \dfrac{3\pi/4+2k\pi}{3}=\dfrac\pi4+\dfrac{2k\pi}3 for k=0,1,-1:

2\mathrm{e}^{\mathrm{i}\pi/4},\qquad 2\mathrm{e}^{\mathrm{i}11\pi/12},\qquad 2\mathrm{e}^{-\mathrm{i}5\pi/12}

[5]

Final answer

(a) \sqrt2\,e^{i\pi/4}
(b) 2e^{i\pi/4},
2e^{i11\pi/12},
2e^{-i5\pi/12}

Question 29

[10 marks]

Let z = 1 + \cos2\theta - i\sin2\theta, where 0 < \theta < \dfrac{\pi}{2}.

(a) Prove that

(i) |z| = 2\cos\theta;

(ii) \arg z = -\theta.

[7]

(b) Find the value of \theta for which z^{3} is purely imaginary with negative imaginary part, and state z^3 for this \theta.

[3]

Show solutionHide solution

(a) Using 1+\cos2\theta=2\cos^2\theta and \sin2\theta=2\sin\theta\cos\theta:

z=2\cos^2\theta-2\mathrm{i}\sin\theta\cos\theta=2\cos\theta\,(\cos\theta-\mathrm{i}\sin\theta)

Since 0<\theta<\dfrac\pi2, 2\cos\theta>0, so this is already in modulus-argument form:

|z|=2\cos\theta,\qquad \arg z=-\theta

[7]

(b) z^3=8\cos^3\theta\,\mathrm{e}^{-3\mathrm{i}\theta}. For this to be purely imaginary with negative imaginary part we need -3\theta=-\dfrac\pi2+2m\pi; within 0<3\theta<\dfrac{3\pi}2 the only option is 3\theta=\dfrac\pi2, so

\theta=\frac\pi6

Then z^3=8\left(\dfrac{\sqrt3}{2}\right)^3\mathrm{e}^{-\mathrm{i}\pi/2}=3\sqrt3\,(-\mathrm{i})=-3\sqrt3\,\mathrm{i}.

[3]

Final answer

(b) \theta=\frac{\pi}{6},
z^3=-3\sqrt3\,i

Question 30

[22 marks]

Consider the equation (z+1)^3 = -8i, where z \in \mathbb{C}. Its roots are \omega_1, \omega_2 and \omega_3, where \omega_1 = -1 + 2i, \mathrm{Re}(\omega_2) > -1 and \mathrm{Re}(\omega_3) < -1.

(a) (i) Check that \omega_1 is a root of the equation.

(ii) Find \omega_2 and \omega_3 in the form -1 + 2e^{i\theta}, where -\pi < \theta \leq \pi.

[6]

Let A, B and C denote the Argand-plane points corresponding to \omega_1, \omega_2 and \omega_3 in turn.

(b) Give \omega_2 and \omega_3 in exact Cartesian form and state the exact coordinates of A, B and C on the Argand diagram.

[4]

(c) Calculate the length AC.

[3]

Now consider the equation (z+1)^3 = -8i\,z^3, where z \in \mathbb{C}.

(d) Use De Moivre's theorem to prove that \alpha = \dfrac{1}{2e^{-i\frac{\pi}{6}} - 1} is a root of this equation.

[3]

(e) Find the exact value of \mathrm{Re}(\alpha).

[6]

Show solutionHide solution

(a)(i) (\omega_1+1)^3 = (2\mathrm{i})^3 = 8\mathrm{i}^3 = -8\mathrm{i}. (verified)

(ii) -8\mathrm{i} = 8\mathrm{e}^{\mathrm{i}\left(-\frac{\pi}{2}+2k\pi\right)}, so z+1 = 2\mathrm{e}^{\mathrm{i}\left(-\frac{\pi}{6}+\frac{2k\pi}{3}\right)}: the arguments are -\dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{7\pi}{6}\equiv-\dfrac{5\pi}{6}. The argument \dfrac{\pi}{2} gives \omega_1. The root with \mathrm{Re}>-1 has \cos\theta>0:

\omega_2 = -1 + 2\mathrm{e}^{-\mathrm{i}\frac{\pi}{6}}, \qquad \omega_3 = -1 + 2\mathrm{e}^{-\mathrm{i}\frac{5\pi}{6}}

[6]

(b) \omega_2 = -1 + 2\left(\dfrac{\sqrt3}{2} - \dfrac{1}{2}\mathrm{i}\right) = (\sqrt3 - 1) - \mathrm{i} and \omega_3 = -1 + 2\left(-\dfrac{\sqrt3}{2} - \dfrac12 \mathrm{i}\right) = (-1-\sqrt3) - \mathrm{i}.

Coordinates: A(-1,\,2), B(\sqrt3-1,\,-1), C(-1-\sqrt3,\,-1).

[4]

(c) \mathrm{AC} = |\omega_1-\omega_3| = |(\sqrt3)+3\mathrm{i}| = \sqrt{3+9} = 2\sqrt3

[3]

(d) \alpha = \dfrac{1}{2\mathrm{e}^{-\mathrm{i}\pi/6}-1} gives \dfrac{1}{\alpha}+1 = 2\mathrm{e}^{-\mathrm{i}\pi/6}, i.e. \dfrac{\alpha+1}{\alpha}=2\mathrm{e}^{-\mathrm{i}\pi/6}. Cube both sides using de Moivre:

\left(\frac{\alpha+1}{\alpha}\right)^3 = 8\mathrm{e}^{-\mathrm{i}\pi/2} = -8\mathrm{i}

so (\alpha+1)^3 = -8\mathrm{i}\,\alpha^3.

[3]

(e) 2\mathrm{e}^{-\mathrm{i}\pi/6} = \sqrt3 - \mathrm{i}, so \alpha = \dfrac{1}{(\sqrt3-1) - \mathrm{i}} = \dfrac{(\sqrt3-1)+\mathrm{i}}{(\sqrt3-1)^2+1} = \dfrac{(\sqrt3-1)+\mathrm{i}}{5-2\sqrt3}.

\mathrm{Re}(\alpha) = \frac{\sqrt3-1}{5-2\sqrt3}\cdot\frac{5+2\sqrt3}{5+2\sqrt3} = \frac{5\sqrt3+6-5-2\sqrt3}{13} = \frac{3\sqrt3+1}{13}

[6]

Final answer

(a)(ii) \omega_2=-1+2e^{-i\pi/6},
\omega_3=-1+2e^{-i5\pi/6}
(b) \omega_2=(\sqrt3-1)-i, \omega_3=(-1-\sqrt3)-i; A(-1,2), B(\sqrt3-1,-1), C(-1-\sqrt3,-1)
(c) 2\sqrt3
(e) \frac{3\sqrt3+1}{13}

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Two squares times two squares

G1 Worked example

[11 marks]

(a) Expand (3+2i)(4+i) and (3+2i)(4-i). Hence, by considering moduli, write 221 as the sum of the squares of two positive integers in two different ways.

[4]

(b) Let z=m+ni, where m,n\in\mathbb{Z}^+ and m>n. By finding z^2 and |z^2|, show that (m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2.

[3]

(c) Use the method of part (a) with 5=2^2+1^2 and 13=3^2+2^2 to write 65 as a sum of two squares in two ways. Hence find two right-angled triangles with integer sides and hypotenuse 65 whose three sides have no common factor greater than 1.

[4]

Show solutionHide solution

(a) (3+2i)(4+i)=12+3i+8i+2i^2=10+11i

(3+2i)(4-i)=12-3i+8i-2i^2=14+5i

|zw|^2=|z|^2|w|^2, and |3+2i|^2\,|4\pm i|^2=13\times17=221

221=10^2+11^2=14^2+5^2

Note: the conjugate 4-i has the same modulus as 4+i, which is why it gives a second way.

(b) z^2=(m^2-n^2)+2mn\,i

|z^2|=|z|^2=m^2+n^2

|z^2|^2=(m^2-n^2)^2+(2mn)^2, so (m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2

A1AG

(c) (2+i)(3+2i)=4+7i and (2+i)(3-2i)=8-i, so 65=7^2+4^2=8^2+1^2

M1A1

(m,n)=(7,4) in (b): sides 33,56,65; ; (m,n)=(8,1): sides 63,16,65

M1A1

Note: 65=5\times13; neither 5 nor 13 divides 56 or 16, so neither triangle has a common factor. The other two integer triangles with hypotenuse 65, (39,52,65) and (25,60,65), are multiples of (3,4,5) and (5,12,13).

Final answer

(a) 221=10^2+11^2=14^2+5^2
(c) 65=7^2+4^2=8^2+1^2; triangles (33,56,65) and (16,63,65)

G1 Investigation

[12 marks]

For n\in\mathbb{Z}^+, let (2+i)^n=a_n+b_n\,i, where a_n,b_n\in\mathbb{Z}. In this question, the remainder of an integer N on division by 5 is the r\in\{0,1,2,3,4\} with N=5q+r for some integer q.

(a) Find a_n and b_n for n=1,2,3,4, and verify that a_n^2+b_n^2=5^n for these values.

[2]

(b) Prove that a_n^2+b_n^2=5^n for every n\in\mathbb{Z}^+.

[2]

(c) Show that a_{n+1}=2a_n-b_n and b_{n+1}=a_n+2b_n.

[1]

(d) Find the remainders of a_n and b_n on division by 5 for n=1,2,\dots,6. Conjecture a pattern, and prove it by mathematical induction.

[4]

(e) Hence show that, for every n\in\mathbb{Z}^+, there is a right-angled triangle with integer sides and hypotenuse 5^n whose three sides have no common factor greater than 1.

[3]

Show solutionHide solution

(a) (2+i)^1=2+i, (2+i)^2=3+4i, (2+i)^3=2+11i, (2+i)^4=-7+24i

2^2+1^2=5, 3^2+4^2=25, 2^2+11^2=125, 7^2+24^2=625

(b) a_n^2+b_n^2=|(2+i)^n|^2=|2+i|^{2n}

=\left(\sqrt5\right)^{2n}=5^n

A1AG

(c) (2+i)^{n+1}=(a_n+b_n i)(2+i)=(2a_n-b_n)+(a_n+2b_n)i; equate real and imaginary parts

A1AG

(d) a_5=-38, b_5=41, a_6=-117, b_6=44. Remainders of (a_n,b_n): (2,1),(3,4),(2,1),(3,4),(2,1),(3,4)

Conjecture: for odd n the remainders are 2 and 1; for even n they are 3 and 4

n=1: (2,1) is true. If a_n=5p+2, b_n=5q+1, then a_{n+1}=5(2p-q)+3 and b_{n+1}=5(p+2q)+4

If a_n=5p+3, b_n=5q+4, then a_{n+1}=5(2p-q)+2 and b_{n+1}=5(p+2q+2)+1. So each case gives the next, and the pattern holds for every n\in\mathbb{Z}^+ by induction

(e) Apply (b) to 2n: a_{2n}^2+b_{2n}^2=5^{2n}=(5^n)^2, so |a_{2n}|, |b_{2n}|, 5^n are the sides of a right-angled triangle

By (d), 5 divides neither a_{2n} nor b_{2n}; in particular both are non-zero

A common factor greater than 1 would divide 5^n, so it would be a multiple of 5 and would divide a_{2n}: impossible

Example: n=3 gives (2+i)^6=-117+44i and the triangle (117,44,125).

Final answer

(a) (2,1),(3,4),(2,11),(-7,24)
(d) odd n: remainders 2,1; even n: 3,4
(e) e.g. n=3: (117,44,125)

G2 Every third coefficient

G2 Worked example

[10 marks]

For n\in\mathbb{Z}^+, let T(n)=\binom n0+\binom n4+\binom n8+\cdots, the sum of every fourth binomial coefficient in row n, starting with \binom n0.

(a) Show that 1+i^j+(-1)^j+(-i)^j equals 4 when j is a multiple of 4, and equals 0 for every other j\in\mathbb{N}.

[3]

(b) By adding the expansions of (1+1)^n, (1+i)^n, (1-1)^n and (1-i)^n, show that 4T(n)=2^n+(1+i)^n+(1-i)^n.

[2]

(c) Hence show that T(n)=2^{n-2}+2^{\frac n2-1}\cos\dfrac{n\pi}{4}.

[3]

(d) Use (c) to find T(8), and check your answer against row 8 of Pascal's triangle.

[2]

Show solutionHide solution

(a) If j=4m, every term is 1 (for example i^{4m}=(i^4)^m=1), so the sum is 4

Otherwise let w=i^j\ne1. The terms are 1,w,w^2,w^3, a geometric series with sum \dfrac{1-w^4}{1-w}

w^4=i^{4j}=1, so the sum is 0

A1AG

(b) The four expansions add to \displaystyle\sum_{j=0}^{n}\binom nj\left(1+i^j+(-1)^j+(-i)^j\right)

By (a) only the terms with j a multiple of 4 survive, each as 4\binom nj; and (1-1)^n=0 for n\ge1. So 4T(n)=2^n+(1+i)^n+(1-i)^n

A1AG

(c) 1+i=\sqrt2\,e^{i\pi/4} and 1-i=\sqrt2\,e^{-i\pi/4}

(1+i)^n+(1-i)^n=2^{n/2}\left(e^{in\pi/4}+e^{-in\pi/4}\right)=2^{n/2}\cdot2\cos\dfrac{n\pi}{4}

T(n)=\dfrac{2^n+2^{\frac n2+1}\cos\frac{n\pi}{4}}{4}=2^{n-2}+2^{\frac n2-1}\cos\dfrac{n\pi}{4}

A1AG

(d) T(8)=2^6+2^3\cos2\pi=72

\binom80+\binom84+\binom88=1+70+1=72

Final answer

(d) T(8)=72

G2 Investigation

[13 marks]

For n\in\mathbb{Z}^+, let S(n)=\binom n0+\binom n3+\binom n6+\cdots, R(n)=\binom n1+\binom n4+\binom n7+\cdots and Q(n)=\binom n2+\binom n5+\binom n8+\cdots. Let \omega=e^{2\pi i/3}.

(a) Calculate S(n) for n=1,2,\dots,8, and check each value against the formula in the key result.

[2]

(b) Calculate R(n) for n=1,2,\dots,8. Compare each value with \dfrac{2^n}{3} and conjecture a formula for R(n) of the same shape as the one for S(n).

[3]

(c) Prove your formula by adding (1+1)^n, \omega^2(1+\omega)^n and \omega(1+\omega^2)^n term by term.

[4]

(d) Find a formula for Q(n), and show that no two of S(n), R(n) and Q(n) differ by more than 1.

[4]

Show solutionHide solution

(a) S(1),\dots,S(8)=1,1,2,5,11,22,43,85; for example S(5)=1+10=11 and \dfrac13\left(32+2\cos\dfrac{5\pi}{3}\right)=11

A1A1

(b) R(1),\dots,R(8)=1,2,3,5,10,21,43,86

R(n)-\dfrac{2^n}{3}=\dfrac13,\dfrac23,\dfrac13,-\dfrac13,-\dfrac23,-\dfrac13,\dfrac13,\dfrac23, which is \dfrac23\cos\dfrac{(n-2)\pi}{3}

R(n)=\dfrac13\left(2^n+2\cos\dfrac{(n-2)\pi}{3}\right)

Note: accept a period-6 description of the differences for the M1.

(c) The factor 1+\omega^{j-1}+\omega^{2(j-1)} is 3 when 3\mid j-1 and 0 otherwise, so \displaystyle\sum_{j=0}^{n}\binom nj\left(1+\omega^{j-1}+\omega^{2(j-1)}\right)=3R(n)

The left side is (1+1)^n+\omega^{-1}(1+\omega)^n+\omega^{-2}(1+\omega^2)^n, and \omega^{-1}=\omega^2, \omega^{-2}=\omega

1+\omega=e^{i\pi/3} and \omega^2=e^{-2\pi i/3}, so \omega^2(1+\omega)^n=e^{i(n-2)\pi/3}

The last term is its conjugate, so 3R(n)=2^n+2\cos\dfrac{(n-2)\pi}{3}

A1AG

(d) S+R+Q=2^n since every coefficient lies in exactly one class, so 3Q=2^n-2\cos\dfrac{n\pi}{3}-2\cos\dfrac{(n-2)\pi}{3}

\cos\dfrac{n\pi}{3}+\cos\dfrac{(n-2)\pi}{3}=2\cos\dfrac{(n-1)\pi}{3}\cos\dfrac{\pi}{3}=\cos\dfrac{(n-1)\pi}{3}=-\cos\dfrac{(n-4)\pi}{3}, so Q(n)=\dfrac13\left(2^n+2\cos\dfrac{(n-4)\pi}{3}\right)

Each of the three is \dfrac{2^n}{3} plus a number of size at most \dfrac23

so two of them differ by less than \dfrac43; they are integers, so they differ by at most 1

Final answer

(a) 1,1,2,5,11,22,43,85
(b) 1,2,3,5,10,21,43,86; R(n)=\frac13\left(2^n+2\cos\frac{(n-2)\pi}{3}\right)
(d) Q(n)=\frac13\left(2^n+2\cos\frac{(n-4)\pi}{3}\right)

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Polynomial roots	`APPS, PlySmlt2 (or graph and 2nd TRACE, 2:zero, for real roots)`
Radian mode	`MODE, Radian, ENTER`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 6 teaches complex numbers from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book