Mjc 2010 H2 Math | Prelim

For now, here’s a in the style of MJC 2010 H2 Math Prelim Paper 1: Question (Complex Numbers)

Thus: For (k=0): (\theta = \pi/4) For (k=1): (\theta = \pi/4 + 2\pi/3 = 3\pi/12 + 8\pi/12 = 11\pi/12) For (k=2): (\theta = \pi/4 + 4\pi/3 = 3\pi/12 + 16\pi/12 = 19\pi/12) But (19\pi/12 = 19\pi/12 - 2\pi = 19\pi/12 - 24\pi/12 = -5\pi/12) (to fit (-\pi<\theta\le\pi)). Mjc 2010 H2 Math Prelim

Derivation: The triangle formed by cube roots of a complex number is equilateral, area formula (\frac3\sqrt34 R^2). For now, here’s a in the style of

So roots: [ z_0 = \sqrt[3]16 , e^i\pi/4, \quad z_1 = \sqrt[3]16 , e^i11\pi/12, \quad z_2 = \sqrt[3]16 , e^-i5\pi/12. ] Argand diagram: points on circle radius (\sqrt[3]16 \approx 2.52), arguments (\pi/4) (45°), (165°), (-75°). (c) Area of triangle = (\frac3\sqrt34 R^2) where (R = \sqrt[3]16). ] Argand diagram: points on circle radius (\sqrt[3]16