2022 ASRJC P1 Q2

2022 ASRJC P1 Q2

6 marks
Prelims

The complex numbers, \({{z}_{1}}\), \({{z}_{2}}\) and \({{z}_{3}}\) are given by \({{z}_{1}}={{\left( 1-\sqrt{3}\mathbf{i} \right)}^{2}}\), \({{z}_{2}}={{\left[ \sqrt{2}\left( \cos \frac{\pi }{4}+\mathbf{i}\sin \frac{\pi }{4} \right) \right]}^{6}}\) and \({{z}_{3}}=-1+\sqrt{3}\,\mathbf{i}\).

  1. Using an algebraic method, find \(\frac{{{z}_{2}}}{{{z}_{1}}}\) in the form \(r{{\mathrm{e}}^{\mathbf{i}\theta }}\), where \(r>0\) and \(\theta \) is an exact real constant such that \(-\pi <\theta \le \pi \).[3]
  2. Hence find \(\frac{{{z}_{2}}}{{{z}_{1}}}+{{z}_{3}}\) in the form \(p{{\mathrm{e}}^{\mathbf{i}\alpha }}\), where both \(r\) and \(\theta \) are exact real constants such that \(r>0\) and \(-\pi <\theta \le \pi \).[3]

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Answer:\(\frac{z_2}{z_1} = 2e^{i\frac{\pi}{6}}\); \(\frac{z_2}{z_1} + z_3 = 2\sqrt{2}\, e^{i\frac{5\pi}{12}}\)

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