IBDP Maths analysis and approaches Topic: AHL 1.12 Complex numbers HL Paper 1

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Question:

Consider the complex numbers z₁ = 1 + bi and z₂ = (1 – b²) – 2bi, where b ∈ R, b ≠ 0.

(a) Find an expression for z₁z₂ in terms of b.

▶️Answer/Explanation

Ans: z₁z₂ = (1 + bi) ((1-b²) – (2b)i)

= (1-b² – 2i²b²) + i (-2b + b – b³)

= (1 + b²) + i (-b – b³)

Note: Award A1 for 1+ b² and A1 for − bi – b³i .

(b) Hence, given that arg (z₁ z₂) = \(\frac{\pi }{4}\) , find the value of b.

▶️Answer/Explanation

Ans: arg (z₁z₂) = arctan \(\left ( \frac{-b-b^{3}}{1 + b^{2}} \right )= \frac{\pi }{4}\)

EITHER

arctan (-b) = \(\frac{\pi }{4}\) (since 1+b² ≠, 0 for b ∈ R)

-b – b³ = 1 + b² (or equivalent)

THEN

b =−1

Question:

Consider the complex numbers z₁ = 1 + bi and z₂ = (1 – b²) – 2bi, where b ∈ R, b ≠ 0.

(a) Find an expression for z₁z₂ in terms of b.

▶️Answer/Explanation

Ans: z₁z₂ = (1 + bi) ((1-b²) – (2b)i)

= (1-b² – 2i²b²) + i (-2b + b – b³)

= (1 + b²) + i (-b – b³)

Note: Award A1 for 1+ b² and A1 for − bi – b³i .

(b) Hence, given that arg (z₁ z₂) = \(\frac{\pi }{4}\) , find the value of b.

▶️Answer/Explanation

Ans: arg (z₁z₂) = arctan \(\left ( \frac{-b-b^{3}}{1 + b^{2}} \right )= \frac{\pi }{4}\)

EITHER

arctan (-b) = \(\frac{\pi }{4}\) (since 1+b² ≠, 0 for b ∈ R)

-b – b³ = 1 + b² (or equivalent)

THEN

b =−1

Question

Consider the complex numbers Z₁ = cos \(\frac{11\pi }{12}+sin\frac{11\pi }{12}\) and Z2 = cos \(\frac{\pi }{6}+isin\frac{\pi }{6}\)

(a) (i) Find \(\frac{Z_1}{Z_2}\)

▶️Answer/Explanation

Ans: \(\frac{z_{1}}{z_{2}}= cos(\frac{11\pi }{12}-\frac{\pi }{6})+i sin (\frac{11\pi }{12}-\frac{\pi }{6}) = cos \frac{3\pi }{4}+i sin \frac{3\pi }{4}\)

(ii) Find \(\frac{Z_2}{Z_1}\) [3]

▶️Answer/Explanation

Ans: \(\frac{z_{2}}{z_{1}} = cos\frac{3\pi }{4}-isin \frac{3\pi }{4}\)

(b) \(\frac{Z_1}{Z_2}\) and \(\frac{Z_2}{Z_1}\) bare represented by three points O, A and B respectively on an Argand diagram. Determine the area of the triangle OAB. [2]

▶️Answer/Explanation
Ans: valid attempt to calculate area of their triangle (angle between OA and OB is \(\frac{\pi }{2}\Rightarrow area (=\frac{1}{2}\times 1\times 1)=\frac{1}{2}\)

Question

If \({z_1} = a + a\sqrt 3 i\) and \({z_2} = 1 – i\), where a is a real constant, express \({z_1}\) and \({z_2}\) in the form \(r\,{\text{cis}}\,\theta \), and hence find an expression for \({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6}\) in terms of a and i.

▶️Answer/Explanation

Markscheme

\({z_1} = 2a{\text{cis}}\left( {\frac{\pi }{3}} \right){\text{, }}{z_2} = \sqrt 2 {\text{ cis}}\left( { – \frac{\pi }{4}} \right)\) M1 A1 A1

EITHER

\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = \frac{{{2^6}{a^6}{\text{cis(0)}}}}{{{{\sqrt 2 }^6}{\text{cis}}\left( {\frac{\pi }{2}} \right)}}\left( { = 8{a^6}{\text{cis}}\left( { – \frac{\pi }{2}} \right)} \right)\) M1 A1 A1

\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^6} = {\left( {\frac{{2a}}{{\sqrt 2 }}{\text{cis}}\left( {\frac{{7\pi }}{{12}}} \right)} \right)^6}\) M1 A1

\( = 8{a^6}{\text{cis}}\left( { – \frac{\pi }{2}} \right)\) A1

THEN

\( = – 8{a^6}{\text{i}}\) A1

Note: Accept equivalent angles, in radians or degrees.

Accept alternate answers without cis e.g. \({\text{ = }}\frac{{8{a^6}}}{{\text{i}}}\)

[7 marks]

Question

Given that z is the complex number \(x + {\text{i}}y\) and that \(\left| {\,z\,} \right| + z = 6 – 2{\text{i}}\) , find the value of x

and the value of y .

▶️Answer/Explanation

Markscheme

\(\sqrt {{x^2} + {y^2}} + x + y{\text{i}} = 6 – 2{\text{i}}\) (A1)

equating real and imaginary parts M1

\(y = – 2\) A1

\(\sqrt {{x^2} + 4} + x = 6\) A1

\({x^2} + 4 = {(6 – x)^2}\) M1

\( – 32 = – 12x \Rightarrow x = \frac{8}{3}\) A1

[6 marks]

Question

Given that \((4 – 5{\text{i}})m + 4n = 16 + 15{\text{i}}\) , where \({{\text{i}}^2} = – 1\), find m and n if

a. m and n are real numbers;[3]

▶️Answer/Explanation

Ans: attempt to equate real and imaginary parts M1

equate real parts: \(4m + 4n = 16\); equate imaginary parts: \( -5m = 15\) A1

\( \Rightarrow m = -3,{\text{ }}n = 7\) A1

[3 marks]

b. m and n are conjugate complex numbers.[4]

▶️Answer/Explanation

let \(m = x + {\text{i}}y,{\text{ }}n = x – {\text{i}}y\) M1

\( \Rightarrow (4 – 5{\text{i}})(x + {\text{i}}y) + 4(x – {\text{i}}y) = 16 + 15{\text{i}}\)

\( \Rightarrow 4x – 5{\text{i}}x + 4{\text{i}}y + 5y + 4x – 4{\text{i}}y = 16 + 15{\text{i}}\)

attempt to equate real and imaginary parts M1

\(8x + 5y = 16,{\text{ }} -5x = 15\) A1

\( \Rightarrow x = -3,{\text{ }}y = 8\) A1

\(( \Rightarrow m = -3 + 8{\text{i}},{\text{ }}n = -3 – 8{\text{i}})\)

[4 marks]

Question

Consider the complex numbers

a. \({z_1} = 2\sqrt 3 {\text{cis}}\frac{{3\pi }}{2}\) and \({z_2} = – 1 + \sqrt 3 {\text{i }}\) .

(i) Write down \({z_1}\) in Cartesian form.

▶️Answer/Explanation

Ans: \({z_1} = 2\sqrt 3 {\text{cis}}\frac{{3\pi }}{2} \Rightarrow {z_1} = – 2\sqrt 3 {\text{i}}\) A1

(ii) Hence determine \({({z_1} + {z_2})^ * }\) in Cartesian form.[3]

▶️Answer/Explanation

Ans: \({z_1} + {z_2} = – 2\sqrt 3 {\text{i}} – 1 + \sqrt 3 {\text{i}} = – 1 – \sqrt 3 {\text{i}}\) A1

\({({z_1} + {z_2})^ * } = – 1 + \sqrt 3 {\text{i}}\) A1

[3 marks]

b. (i) Write \({z_2}\) in modulus-argument form.

▶️Answer/Explanation

Ans: \(\left| {{z_2}} \right| = 2\)

\(\tan \theta = – \sqrt 3 \) (M1)

\({z_2}\) lies on the second quadrant

\(\theta = \arg {z_2} = \frac{{2\pi }}{3}\)

\({z_2} = 2{\text{cis}}\frac{{2\pi }}{3}\) A1A1

(ii) Hence solve the equation \({z^3} = {z_2}\) .[6]

▶️Answer/Explanation

Ans: attempt to use De Moivre’s theorem M1

\(z = \sqrt[3]{2}\,{\text{cis}}\frac{{\frac{{2\pi }}{3} + 2k\pi }}{3},{\text{ }}k = 0{\text{, 1 and 2}}\)

\(z = \sqrt[3]{2}\,{\text{cis}}\frac{{2\pi }}{9},\,\sqrt[3]{2}\,{\text{cis}}\frac{{8\pi }}{9},\,\sqrt[3]{2}\,{\text{cis}}\frac{{14\pi }}{9}\left( { = \sqrt[3]{2}\,{\text{cis}}\left( {\frac{{ – 4\pi }}{9}} \right)} \right)\) A1A1

Note: Award A1 for modulus, A1 for arguments.

Note: Allow equivalent forms for z .

[6 marks]

c. Let \(z = r\,{\text{cis}}\theta \) , where \(r \in {\mathbb{R}^ + }\) and \(0 \leqslant \theta < 2\pi \) . Find all possible values of r and \(\theta \) ,

(i) if \({z^2} = {(1 + {z_2})^2}\);

▶️Answer/Explanation

Ans: METHOD 1

\({z^2} = {\left( {1 – 1 + \sqrt 3 {\text{i}}} \right)^2} = – 3\left( { \Rightarrow z = \pm \sqrt 3 {\text{i}}} \right)\) M1

\(z = \sqrt 3 \,{\text{cis}}\frac{\pi }{2}{\text{ or }}{z_1} = \sqrt 3 \,{\text{cis}}\frac{{3\pi }}{2}\left( { = \sqrt 3 \,{\text{cis}}\left( {\frac{{ – \pi }}{2}} \right)} \right)\) A1A1

so \(r = \sqrt 3 {\text{ and }}\theta = \frac{\pi }{2}{\text{ or }}\theta = \frac{{3\pi }}{2}\left( { = \frac{{ – \pi }}{2}} \right)\)

Note: Accept \(r\,{\text{cis}}(\theta )\) form.

METHOD 2

\({z^2} = {\left( {1 – 1 + \sqrt 3 {\text{i}}} \right)^2} = – 3 \Rightarrow {z^2} = 3{\text{cis}}\left( {(2n + 1)\pi } \right)\) M1

\({r^2} = 3 \Rightarrow r = \sqrt 3 \) A1

\(2\theta = (2n + 1)\pi \Rightarrow \theta = \frac{\pi }{2}{\text{ or }}\theta = \frac{{3\pi }}{2}{\text{ (as }}0 \leqslant \theta < 2\pi )\) A1

Note: Accept \(r\,{\text{cis}}(\theta )\) form.

(ii) if \(z = – \frac{1}{{{z_2}}}\).[6]

▶️Answer/Explanation

Ans: METHOD 1

\(z = – \frac{1}{{2{\text{cis}}\frac{{2\pi }}{3}}} \Rightarrow z = \frac{{{\text{cis}}\pi }}{{2{\text{cis}}\frac{{2\pi }}{3}}}\) M1

\( \Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}\)

so \(r = \frac{1}{2}{\text{ and }}\theta = \frac{\pi }{3}\) A1A1

METHOD 2

\({z_1} = – \frac{1}{{ – 1 + \sqrt 3 {\text{i}}}} \Rightarrow {z_1} = – \frac{{ – 1 – \sqrt 3 {\text{i}}}}{{\left( { – 1 + \sqrt 3 {\text{i}}} \right)\left( { – 1 – \sqrt 3 {\text{i}}} \right)}}\) M1

\(z = \frac{{1 + \sqrt 3 {\text{i}}}}{4} \Rightarrow z = \frac{1}{2}{\text{cis}}\frac{\pi }{3}\)

so \(r = \frac{1}{2}{\text{ and }}\theta = \frac{\pi }{3}\) A1A1

[6 marks]

d. Find the smallest positive value of n for which \({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} \in {\mathbb{R}^ + }\) .[4]

▶️Answer/Explanation
Ans: \(\frac{{{z_1}}}{{{z_2}}} = \sqrt 3 \,{\text{cis}}\frac{{5\pi }}{6}\) (A1)

\({\left( {\frac{{{z_1}}}{{{z_2}}}} \right)^n} = {\sqrt 3 ^n}{\text{cis}}\frac{{5n\pi }}{6}\) A1

equating imaginary part to zero and attempting to solve M1

obtain n = 12 A1

Note: Working which only includes the argument is valid.

[4 marks]

Question

(z + 2i) is a factor of 2z³– 3z² + 8z – 12. Find the other two factors.

▶️Answer/Explanation

If \((z + 2i)\) is a factor then \((z – 2i)\) is also a factor. \((z+2i)(z-2i)=(z^{2}+4)\)

The other factor is (2z³ – 3z² + 8z – 12) ÷ (z² + 4) = (2z – 3)
The other two factors are (z – 2i) and (2z–3).

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