IB DiplomaIB-MAA-23-P1-TZ2• Non-Calculator

May 2023 Paper 1 Time Zone 2 (HL)

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Total Marks: 110

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INTERNATIONAL BACCALAUREATE (IB)IB-MAA-23-P1-TZ2• Official Candidate Question Paper

May 2023 Paper 1 Time Zone 2 (HL) • Mathematics: Analysis and Approaches (HL)

base

[5 Marks]

Complex Numbers & De Moivre’s Theorem

Worth 5 Marks • Challenging Level

Let

z = \sqrt{3} + i

. Using polar form

z = r e^{i\theta}

and De Moivre's theorem, find the exact value of

z^6

Formula Guide:

|z| = \sqrt{(\sqrt{3})^2 + 1^2} = 2

\operatorname{Arg}(z) = \arctan(1/\sqrt{3}) = \frac{\pi}{6}

. By De Moivre:

z^6 = 2^6 \left(\cos(6 \cdot \frac{\pi}{6}) + i \sin(6 \cdot \frac{\pi}{6})\right) = 64(\cos\pi + i\sin\pi) = 64(-1 + 0) = -64

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