Given that \(z = \cos \theta + {\text{i}}\sin \theta \) show that

(a)     \(\operatorname{Im} \left( {{z^n} + \frac{1}{{{z^n}}}} \right) = 0,{\text{ }}n \in {\mathbb{Z}^ + }\);

(b)     \(\operatorname{Re} \left( {\frac{{z - 1}}{{z + 1}}} \right) = 0,{\text{ }}z \ne  - 1\).

(a)     using de Moivre’s theorem

\({z^n} + \frac{1}{{{z^n}}} = \cos n\theta + {\text{i}}\sin n\theta + \cos n\theta - {\text{i}}\sin n\theta {\text{ }}( = 2\cos n\theta )\), imaginary part of which is 0     M1A1

so \(\operatorname{Im} \left( {{z^n} + \frac{1}{{{z^n}}}} \right) = 0\)     AG

(b)     \(\frac{{z - 1}}{{z + 1}} = \frac{{\cos \theta + {\text{i}}\sin \theta - 1}}{{\cos \theta + {\text{i}}\sin \theta + 1}}\)

\( = \frac{{(\cos \theta - 1 + {\text{i}}\sin \theta )(\cos \theta + 1 - {\text{i}}\sin \theta )}}{{(\cos \theta + 1 + {\text{i}}\sin \theta )(\cos \theta + 1 - {\text{i}}\sin \theta )}}\)     M1A1

Note: Award M1 for an attempt to multiply numerator and denominator by the complex conjugate of their denominator.

\( \Rightarrow \operatorname{Re} \left( {\frac{{z - 1}}{{z + 1}}} \right) = \frac{{(\cos \theta - 1)(\cos \theta + 1) + {{\sin }^2}\theta }}{{{\text{real denominator}}}}\)     M1A1

Note: Award M1 for multiplying out the numerator.

\( = \frac{{{{\cos }^2}\theta + {{\sin }^2}\theta - 1}}{{{\text{real denominator}}}}\)     A1

\( = 0\)     AG

[7 marks] 

Part(a) - The majority either obtained full marks or no marks here.

Part(b) - This question was algebraically complex and caused some candidates to waste their efforts for little credit.