(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): (\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
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),
Better: (16^1/3 = 2^4/3). But leave as (\sqrt[3]16 = 2\sqrt[3]2). Better: (16^1/3 = 2^4/3)
The complex number (z) satisfies the equation [ z^3 = -8\sqrt2 + 8\sqrt2 i. ]