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Date November 2015 Marks available 11 Reference code 15N.1.hl.TZ0.11
Level HL only Paper 1 Time zone TZ0
Command term Calculate, Find, Hence, and Write Question number 11 Adapted from N/A

Question

Solve the equation \({z^3} = 8{\text{i}},{\text{ }}z \in \mathbb{C}\) giving your answers in the form \(z = r(\cos \theta  + {\text{i}}\sin \theta )\) and in the form \(z = a + b{\text{i}}\) where \(a,{\text{ }}b \in \mathbb{R}\).

[6]
a.

Consider the complex numbers \({z_1} = 1 + {\text{i}}\) and \({z_2} = 2\left( {\cos \left( {\frac{\pi }{2}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6}} \right)} \right)\).

(i)     Write \({z_1}\) in the form \(r(\cos \theta  + {\text{i}}\sin \theta )\).

(ii)     Calculate \({z_1}{z_2}\) and write in the form \(z = a + b{\text{i}}\) where \(a,{\text{ }}b \in \mathbb{R}\).

(iii)     Hence find the value of \(\tan \frac{{5\pi }}{{12}}\) in the form \(c + d\sqrt 3 \), where \(c,{\text{ }}d \in \mathbb{Z}\).

(iv)     Find the smallest value \(p > 0\) such that \({({z_2})^p}\) is a positive real number.

[11]
b.

Markscheme

Note:     Accept answers and working in degrees, throughout.

 

\({z^3} = 8\left( {\cos \left( {\frac{\pi }{2} + 2\pi k} \right) + {\text{i}}\sin \left( {\frac{\pi }{2} + 2\pi k} \right)} \right)\)     (A1)

attempt the use of De Moivre’s Theorem in reverse     M1

\(z = 2\left( {\cos \left( {\frac{\pi }{6}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6}} \right)} \right);{\text{ }}2\left( {\cos \left( {\frac{{5\pi }}{6}} \right) + {\text{i}}\sin \left( {\frac{{5\pi }}{6}} \right)} \right);\)

\(2\left( {\cos \left( {\frac{{9\pi }}{6}} \right) + {\text{i}}\sin \left( {\frac{{9\pi }}{6}} \right)} \right)\)     A2

 

Note:     Accept cis form.

 

\(z =  \pm \sqrt 3  + {\text{i}},{\text{ }} - 2{\text{i}}\)     A2

 

Note:     Award A1 for two correct solutions in each of the two lines above.

[6 marks]

a.

Note:     Accept answers and working in degrees, throughout.

 

(i)     \({z_1} = \sqrt 2 \left( {\cos \left( {\frac{\pi }{4}} \right) + {\text{i}}\sin \left( {\frac{\pi }{4}} \right)} \right)\)     A1A1

(ii)     \(\left( {{z_2} = \left( {\sqrt 3  + {\text{i}}} \right)} \right)\)

\({z_1}{z_2} = (1 + {\text{i}})\left( {\sqrt 3  + {\text{i}}} \right)\)     M1

\( = \left( {\sqrt 3  - 1} \right) + {\text{i}}\left( {1 + \sqrt 3 } \right)\)     A1

(iii)     \({z_1}{z_2} = 2\sqrt 2 \left( {\cos \left( {\frac{\pi }{6} + \frac{\pi }{4}} \right) + {\text{i}}\sin \left( {\frac{\pi }{6} + \frac{\pi }{4}} \right)} \right)\)     M1A1

 

Note:     Interpret “hence” as “hence or otherwise”.

 

\(\tan \frac{{5\pi }}{{12}} = \frac{{\sqrt 3  + 1}}{{\sqrt 3  - 1}}\)     A1

\( = 2 + \sqrt 3 \)     M1A1

 

Note:     Award final M1 for an attempt to rationalise the fraction.

 

(iv)     \({z_2}^p = {2^p}\left( {{\text{cis}}\left( {\frac{{p\pi }}{6}} \right)} \right)\)     (M1)

\({z_2}^p\) is a positive real number when \(p = 12\)     A1

 

Note:     Accept a solution based on part (a).

[11 marks]

Total [17 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 1 - Core: Algebra » 1.6 » Modulus–argument (polar) form \(z = r\left( {\cos \theta + {\text{i}}\sin \theta } \right) = r{\text{cis}}\theta = r{e^{{\text{i}}\theta }}\)
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