Show that $\left| {{{\text{e}}^{{\text{i}}\theta }}} \right| = 1$.

Consider the geometric series $1 + \frac{1}{3}{{\text{e}}^{{\text{i}}\theta }} + \frac{1}{9}{{\text{e}}^{2{\text{i}}\theta }} +  \ldots {\text{ .}}$

Write down the common ratio, z, of the series, and show that $\left| z \right| = \frac{1}{3}$.

Find an expression for the sum to infinity of this series.

Hence, show that $\sin \theta  + \frac{1}{3}\sin 2\theta  + \frac{1}{9}\sin 3\theta  +  \ldots  = \frac{{9\sin \theta }}{{10 - 6\cos \theta }}$.

$\left| {{{\text{e}}^{{\text{i}}\theta }}} \right|{\text{ }}\left( { = \left| {\cos \theta  + {\text{i}}\sin \theta } \right|} \right) = \sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } = 1$     M1AG

[1 mark]

$z = \frac{1}{3}{{\text{e}}^{{\text{i}}\theta }}$     A1

$\left| z \right| = \left| {\frac{1}{3}{{\text{e}}^{{\text{i}}\theta }}} \right| = \frac{1}{3}$     A1AG

[2 marks]

${S_\infty } = \frac{a}{{1 - r}} = \frac{1}{{1 - \frac{1}{3}{{\text{e}}^{{\text{i}}\theta }}}}$     (M1)A1

[2 marks]

EITHER

${S_\infty } = \frac{1}{{1 - \frac{1}{3}\cos \theta  - \frac{1}{3}{\text{i}}\sin \theta }}$     A1

$= \frac{{1 - \frac{1}{3}\cos \theta + \frac{1}{3}{\text{i}}\sin \theta }}{{\left( {1 - \frac{1}{3}\cos \theta - \frac{1}{3}{\text{i}}\sin \theta } \right)\left( {1 - \frac{1}{3}\cos \theta + \frac{1}{3}{\text{i}}\sin \theta } \right)}}$     M1A1

$= \frac{{1 - \frac{1}{3}\cos \theta + \frac{1}{3}{\text{i}}\sin \theta }}{{{{\left( {1 - \frac{1}{3}\cos \theta } \right)}^2} + \frac{1}{9}{{\sin }^2}\theta }}$     A1

$= \frac{{1 - \frac{1}{3}\cos \theta + \frac{1}{3}{\text{i}}\sin \theta }}{{1 - \frac{2}{3}\cos \theta + \frac{1}{9}}}$     A1

OR

${S_\infty } = \frac{1}{{1 - \frac{1}{3}{{\text{e}}^{{\text{i}}\theta }}}}$

$= \frac{{1 - \frac{1}{3}{{\text{e}}^{ - {\text{i}}\theta }}}}{{\left( {1 - \frac{1}{3}{{\text{e}}^{{\text{i}}\theta }}} \right)\left( {1 - \frac{1}{3}{{\text{e}}^{ - {\text{i}}\theta }}} \right)}}$     M1A1

$= \frac{{1 - \frac{1}{3}{{\text{e}}^{ - {\text{i}}\theta }}}}{{1 - \frac{1}{3}({{\text{e}}^{{\text{i}}\theta }} + {{\text{e}}^{ - {\text{i}}\theta }}) + \frac{1}{9}}}$     A1

$= \frac{{1 - \frac{1}{3}{{\text{e}}^{ - {\text{i}}\theta }}}}{{\frac{{10}}{9} - \frac{2}{3}\cos \theta }}$     A1

$= \frac{{1 - \frac{1}{3}(\cos \theta - {\text{i}}\sin \theta )}}{{\frac{{10}}{9} - \frac{2}{3}\cos \theta }}$     A1

THEN

taking imaginary parts on both sides

$\frac{1}{3}\sin \theta + \frac{1}{9}\sin 2\theta +  \ldots = \frac{{\frac{1}{3}\sin \theta }}{{\frac{{10}}{9} - \frac{2}{3}\cos \theta }}$     M1A1A1

$= \frac{{\sin \theta }}{{\frac{{10}}{9} - \frac{2}{3}\cos \theta }}$

$\Rightarrow \sin \theta + \frac{1}{3}\sin 2\theta  + \ldots = \frac{{9\sin \theta }}{{10 - 6\cos \theta }}$     AG

[8 marks]

Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.

Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.

Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.

Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates.