Mjc 2010 H2 Math Prelim Official

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).

(b) On a single Argand diagram, sketch the three roots.

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)).

Better: (16^1/3 = 2^4/3). But leave as (\sqrt[3]16 = 2\sqrt[3]2). Mjc 2010 H2 Math Prelim

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).

(c) Find the exact area of the triangle formed by these three roots. So area = (\frac3\sqrt34 (16^2/3))

Modulus of (z^3): [ |z^3| = \sqrt(-8\sqrt2)^2 + (8\sqrt2)^2 = \sqrt128 + 128 = \sqrt256 = 16. ] Argument of (z^3): [ \tan\theta = \frac8\sqrt2-8\sqrt2 = -1. ] Point is in 2nd quadrant (negative real, positive imag), so [ \arg(z^3) = \pi - \frac\pi4 = \frac3\pi4. ] Thus [ z^3 = 16 e^i(3\pi/4 + 2k\pi). ] Taking cube roots: [ z = \sqrt[3]16 ; e^i\left(\frac\pi4 + \frac2k\pi3\right), \quad k=0,1,2. ] (\sqrt[3]16 = 16^1/3 = 2^4/3 = 2\sqrt[3]2) but wait — check carefully: Actually (16^1/3 = (2^4)^1/3 = 2^4/3). Yes. But sometimes they keep as (2\sqrt[3]2). We’ll keep exact.

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 .

(a) Find the modulus and argument of (z^3), hence find the three roots of the equation in the form (r e^i\theta) where (r>0) and (-\pi < \theta \le \pi). Thus: For (k=0): (\theta = \pi/4) For (k=1):

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).

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