| Date | May 2008 | Marks available | 5 | Reference code | 08M.1.hl.TZ1.1 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Express | Question number | 1 | Adapted from | N/A |
Question
Express \(\frac{1}{{{{(1 - {\text{i}}\sqrt 3 )}^3}}}{\text{ in the form }}\frac{a}{b}{\text{ where }}a,{\text{ }}b \in \mathbb{Z}\) .
Markscheme
METHOD 1
\(r = 2,{\text{ }}\theta = - \frac{\pi }{3}\) (A1)(A1)
\(\therefore {(1 - {\text{i}}\sqrt 3 )^{ - 3}} = {2^{ - 3}}{\left( {\cos \left( { - \frac{\pi }{3}} \right) + {\text{i}}\sin \left( { - \frac{\pi }{3}} \right)} \right)^{ - 3}}\) M1
\( = \frac{1}{8}(\cos \pi + {\text{i}}\sin \pi )\) (M1)
\( = - \frac{1}{8}\) A1
[5 marks]
METHOD 2
\((1 - \sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = 1 - 2\sqrt 3 {\text{i}} - 3\,\,\,\,\,( = - 2 - 2\sqrt 3 {\text{i}})\) (M1)A1
\(( - 2 - 2\sqrt 3 {\text{i}})(1 - \sqrt 3 {\text{i}}) = - 8\) (M1)(A1)
\(\therefore \frac{1}{{{{(1 - \sqrt 3 {\text{i}})}^3}}} = - \frac{1}{8}\) A1
[5 marks]
METHOD 3
Attempt at Binomial expansion M1
\({(1 - \sqrt 3 {\text{i}})^3} = 1 + 3( - \sqrt 3 {\text{i}}) + 3{( - \sqrt 3 {\text{i}})^2} + {( - \sqrt 3 {\text{i}})^3}\) (A1)
\( = 1 - 3\sqrt 3 {\text{i}} - 9 + 3\sqrt 3 {\text{i}}\) (A1)
\( = - 8\) A1
\(\therefore \frac{1}{{{{(1 - \sqrt 3 {\text{i}})}^3}}} = - \frac{1}{8}\) M1
[5 marks]
Examiners report
Most candidates made a meaningful attempt at this question using a variety of different, but correct methods. Weaker candidates sometimes made errors with the manipulation of the square roots, but there were many fully correct solutions.
