Mjc 2010 H2 Math Prelim Online

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.

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

(b) On a single Argand diagram, sketch the three roots. Modulus of (z^3): [ |z^3| = \sqrt(-8\sqrt2)^2 +

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 [ z^3 = 16 e^i(3\pi/4 + 2k\pi)

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).