Let \(\omega = \cos \theta  + {\text{i}}\sin \theta \) . Find, in terms of \(\theta \) , the modulus and argument of \({(1 - {\omega ^2})^ * }\) .

 

 

 

METHOD 1 

\({(1 - {\omega ^2})^ * } = {(1 - {\text{cis}}\,2\theta )^ * } = {\left( {(1 - \cos 2\theta ) - {\text{i}}\sin 2\theta } \right)^ * }\)     M1A1

\( = (1 - \cos 2\theta ) + {\text{i}}\sin 2\theta \)     A1

\(\left| {{{(1 - {\omega ^2})}^ * }} \right| = \sqrt {{{(1 - \cos 2\theta )}^2} + {{\sin }^2}2\theta } \left( { = \sqrt {{{(2{{\sin }^2}\theta )}^2} + {{(2\sin \theta \cos \theta )}^2}} } \right)\)     M1

\( = \left| {2\sin \theta } \right|\)     A1

\(\arg \left( {{{(1 - {\omega ^2})}^ * }} \right) = \alpha  \Rightarrow \tan \alpha  = \cot (\theta )\)     M1

\(\alpha  = \frac{\pi }{2} - \theta \)     A1

therefore:

modulus is \(2\left| {\sin \theta } \right|\) and argument is \(\frac{\pi }{2} - \theta {\text{ or }}\frac{\pi }{2} - \theta  \pm \pi \) 

Note: Accept modulus is 2sinθ and argument is \(\frac{\pi }{2} - \theta \)

 

 

METHOD 2

EITHER

\({(1 - {\varpi ^2})^ * } = {(1 - {\text{cis}}\,2\theta )^ * } = {\left( {(1 - \cos 2\theta ) - {\text{i}}\sin 2\theta } \right)^ * }\)     M1A1

\( = (1 - \cos 2\theta ) + {\text{i}}\sin 2\theta \)     A1

\( = (1 - 1 + 2{\sin ^2}\theta ) + 2{\text{i}}\sin \theta \cos \theta \)     M1

OR

\({(1 - {\varpi ^2})^ * } = {\left( {1 - {{(\cos \theta  + {\text{i}}\sin \theta )}^2}} \right)^ * }\)     M1A1

\( = {(1 - {\cos ^2}\theta  + {\sin ^2}\theta - 2{\text{i}}\sin \theta \cos \theta )^ * }\)     A1

\( = 2{\sin ^2}\theta  + 2{\text{i}}\sin \theta \cos \theta \)     M1

THEN

\( = 2\sin \theta (\sin \theta  + {\text{i}}\cos \theta )\)     (M1)

\( = 2\sin \theta \left( {\cos \left( {\frac{\pi }{2} - \theta } \right) + {\text{i}}\sin \left( {\frac{\pi }{2} - \theta } \right)} \right)\)     A1A1

\( = 2\sin \theta \,{\text{cis}}\left( {\frac{\pi }{2} - \theta } \right)\)

therefore:

modulus is \(2\left| {\sin \theta } \right|\) and argument is \(\frac{\pi }{2} - \theta {\text{ or }}\frac{\pi }{2} - \theta  \pm \pi \)

Note: Accept modulus is 2sinθ and argument is \(\frac{\pi }{2} - \theta \) .

 

[7 marks]

This was the most challenging question in part A with just a few candidates scoring full marks. This question showed that many candidates have difficulties with algebraic manipulations, application of De Moivre’s theorem and use of trigonometric identities. Although some candidates managed to calculate the square of a complex number, many failed to write down its conjugate or made algebraic errors which lead to wrong results in many cases. Just a few candidates were able to calculate the modulus and the argument of the complex number.