Mjc 2010 H2 — Math Prelim

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).

(c) Find the exact area of the triangle formed by these three roots. Mjc 2010 H2 Math Prelim

Better: (16^1/3 = 2^4/3). But leave as (\sqrt[3]16 = 2\sqrt[3]2). So roots: [ z_0 = \sqrt[3]16 , e^i\pi/4,

So area = (\frac3\sqrt34 (16^2/3)). (16^2/3 = (2^4)^2/3 = 2^8/3 = 4 \cdot 2^2/3 = 4\sqrt[3]4). (c) Find the exact area of the triangle

I notice you’ve asked for "Mjc 2010 H2 Math Prelim" — but it seems you want me to , likely meaning a problem or solution from that paper .

Thus exact area = (\frac3\sqrt34 \cdot 4\sqrt[3]4 = 3\sqrt3 \cdot \sqrt[3]4). If you meant something else (e.g., a different question from MJC 2010 Prelim), just let me know the , and I’ll produce the exact problem and solution.

Derivation: The triangle formed by cube roots of a complex number is equilateral, area formula (\frac3\sqrt34 R^2).