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Date November 2011 Marks available 7 Reference code 11N.2.hl.TZ0.6
Level HL only Paper 2 Time zone TZ0
Command term Find Question number 6 Adapted from N/A

Question

The complex numbers \({z_1}\) and \({z_2}\) have arguments between 0 and \(\pi \) radians. Given that \({z_1}{z_2} = - \sqrt 3  + {\text{i}}\) and \(\frac{{{z_1}}}{{{z_2}}} = 2{\text{i}}\), find the modulus and argument of \({z_1}\) and of \({z_2}\).

Markscheme

METHOD 1

\({\text{arg(}}{z_1}{z_2}) = \frac{{5\pi }}{6}\,\,\,\,\,(150^\circ )\)     (A1)

\(\arg \left( {\frac{{{z_1}}}{{{z_2}}}} \right) = \frac{\pi }{2}\,\,\,\,\,(90^\circ )\)     (A1)

\( \Rightarrow \arg ({z_1}) + \arg ({z_2}) = \frac{{5\pi }}{6};{\text{ }}\arg ({z_1}) - \arg ({z_2}) = \frac{\pi }{2}\)     M1

solving simultaneously

\(\arg ({z_1}) = \frac{{2\pi }}{3}{\text{ }}(120^\circ ){\text{ and }}\arg ({z_2}) = \frac{\pi }{6}{\text{ }}(30^\circ )\)     A1A1

Note: Accept decimal approximations of the radian measures.

 

\(\left| {{z_1}{z_2}} \right| = 2 \Rightarrow \left| {{z_1}} \right|\left| {{z_2}} \right| = 2;{\text{ }}\left| {\frac{{{z_1}}}{{{z_2}}}} \right| = 2 \Rightarrow \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}} = 2\)     M1

solving simultaneously

\(\left| {{z_1}} \right| = 2;{\text{ }}\left| {{z_2}} \right| = 1\)     A1

[7 marks]

METHOD 2

\({z_1} = 2i{z_2}\,\,\,\,\,2iz_2^2 = - \sqrt 3  + i\)     (M1)

\(z_2^2 = \frac{{ - \sqrt 3  + i}}{{2i}}\)     A1

\({z_2} = \sqrt {\frac{{ - \sqrt 3  + i}}{{2i}}} = \frac{{\sqrt 3 }}{2} + \frac{1}{2}i{\text{ or }}{e^{\frac{\pi }{6}i}}\)     (M1)(A1)

(allow \(0.866 + 0.5i\) or \({{\text{e}}^{0.524{\text{i}}}}\))

\({z_1} = - 1 + \sqrt 3 i{\text{ or }}2{{\text{e}}^{\frac{{2\pi }}{3}i}} - \) (allow −1 + 1.73i or \(2{{\text{e}}^{2.09{\text{i}}}}\))     (A1)

\({z_1}\,\,\,\,\,\)modulus = 2, argument \( = \frac{{2\pi }}{3}\)     A1

\({z_2}\,\,\,\,\,\)modulus = 1, argument \( = \frac{\pi }{6}\)     A1

Note: Accept degrees and decimal approximations to radian measure.

 

[7 marks]

Examiners report

Candidates generally found this question challenging. Many candidates had difficulty finding the arguments of \({z_1}{z_2}\) and \({z_1}/{z_2}\). Among candidates who attempted to solve for \({z_1}\) and \({z_2}\) in Cartesian form, many had difficulty with the algebraic manipulation involved.

Syllabus sections

Topic 1 - Core: Algebra » 1.5 » Complex numbers: the number \({\text{i}} = \sqrt { - 1} \) ; the terms real part, imaginary part, conjugate, modulus and argument.
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