nayfie
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I have solved part a, I just have no idea how to go about doing part b. If anybody could point me in the right direction, that would be greatly appreciated!
a. Express z = \frac{1 + \sqrt{3}i}{-2 -2i} in the form rcis\theta
b. What is the smallest positive integer n such that z^{n} is a real number? Find z^{n} for this particular n.
z^{n} = r^{n}cis(n\theta)
\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}}cis(\theta_{1} - \theta_{2})
Part A
z_{1} = 1 + \sqrt{3}i
r = |z_{1}| = \sqrt{4} = 2
\theta = arccos(\frac{1}{2}) = \frac{\pi}{3}
z_{1} = 2cis(\frac{\pi}{3})
z_{2} = -2 -2i
r = |z_{2}| = \sqrt{8} = 2\sqrt{2}
\theta = arccos(\frac{-2}{2\sqrt{2}}) = \frac{3\pi}{4}
z_{2} = 2\sqrt{2}cis(\frac{3\pi}{4})
\frac{z_{1}}{z_{2}} = \frac{2}{2\sqrt{2}}cis(\frac{\pi}{3} - \frac{3\pi}{4}) = \frac{1}{\sqrt{2}}cis(\frac{-5\pi}{12})
Part B
No idea :(
Homework Statement
a. Express z = \frac{1 + \sqrt{3}i}{-2 -2i} in the form rcis\theta
b. What is the smallest positive integer n such that z^{n} is a real number? Find z^{n} for this particular n.
Homework Equations
z^{n} = r^{n}cis(n\theta)
\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}}cis(\theta_{1} - \theta_{2})
The Attempt at a Solution
Part A
z_{1} = 1 + \sqrt{3}i
r = |z_{1}| = \sqrt{4} = 2
\theta = arccos(\frac{1}{2}) = \frac{\pi}{3}
z_{1} = 2cis(\frac{\pi}{3})
z_{2} = -2 -2i
r = |z_{2}| = \sqrt{8} = 2\sqrt{2}
\theta = arccos(\frac{-2}{2\sqrt{2}}) = \frac{3\pi}{4}
z_{2} = 2\sqrt{2}cis(\frac{3\pi}{4})
\frac{z_{1}}{z_{2}} = \frac{2}{2\sqrt{2}}cis(\frac{\pi}{3} - \frac{3\pi}{4}) = \frac{1}{\sqrt{2}}cis(\frac{-5\pi}{12})
Part B
No idea :(