$Complex Numbers$

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$Complex Numbers$

Represent $(\sqrt{3} -i)^{-8} $ as a+bi without the use of a calculator. Show all of your work. Hint: use the polar form $re^{i\theta } $

Find the three cube roots of 1 or unity. In other words, find the solution set $z^{3} -1=0$

Complex Numbers

Wcramsay

73

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There are 2 independent problems here.
Problem 1:

$(\sqrt{3} -1)^{-8} = (2(cos(-\pi/6) + i \; sin(-\pi/6)))^{-8} = (2e^{i (-\pi/6)})^{-8} = 2^{-8} e^{i (8\pi/6)} $
$ = 2^{-8} (cos(4\pi/3) + i \; sin(4\pi/3)) = 2^{-8} (-1/2 + i(-\sqrt{3}/2)) = -2^{-9} - 2^{-9}\sqrt{3} \; i$

Problem 2.
$z^{3} - 1 = 0 \; \Rightarrow\; (z-1)(z^{2}+z+1)=0$
$z-1 = 0 \Rightarrow z=1$
$z^{2}+z+1 = 0 \Rightarrow z = (-1 \pm \sqrt{1-4})/2 \Rightarrow z=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \; i$

Shadmehr

29

$Complex Numbers$

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