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 \(a,{\text{ }}b \in \mathbb{R}\).
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.
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]
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]