Mjc 2010 H2 Math Prelim ★ Limited & Newest

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)).

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 . 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):

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). Mjc 2010 H2 Math Prelim