- #1
nayfie
- 50
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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 [itex]z = \frac{1 + \sqrt{3}i}{-2 -2i}[/itex] in the form rcis[itex]\theta[/itex]
b. What is the smallest positive integer [itex]n[/itex] such that [itex]z^{n}[/itex] is a real number? Find [itex]z^{n}[/itex] for this particular [itex]n[/itex].
[itex]z^{n} = r^{n}cis(n\theta)[/itex]
[itex]\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}}cis(\theta_{1} - \theta_{2})[/itex]
Part A
[itex]z_{1} = 1 + \sqrt{3}i[/itex]
[itex]r = |z_{1}| = \sqrt{4} = 2[/itex]
[itex]\theta = arccos(\frac{1}{2}) = \frac{\pi}{3}[/itex]
[itex]z_{1} = 2cis(\frac{\pi}{3})[/itex]
[itex]z_{2} = -2 -2i[/itex]
[itex]r = |z_{2}| = \sqrt{8} = 2\sqrt{2}[/itex]
[itex]\theta = arccos(\frac{-2}{2\sqrt{2}}) = \frac{3\pi}{4}[/itex]
[itex]z_{2} = 2\sqrt{2}cis(\frac{3\pi}{4})[/itex]
[itex]\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})[/itex]
Part B
No idea :(
Homework Statement
a. Express [itex]z = \frac{1 + \sqrt{3}i}{-2 -2i}[/itex] in the form rcis[itex]\theta[/itex]
b. What is the smallest positive integer [itex]n[/itex] such that [itex]z^{n}[/itex] is a real number? Find [itex]z^{n}[/itex] for this particular [itex]n[/itex].
Homework Equations
[itex]z^{n} = r^{n}cis(n\theta)[/itex]
[itex]\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}}cis(\theta_{1} - \theta_{2})[/itex]
The Attempt at a Solution
Part A
[itex]z_{1} = 1 + \sqrt{3}i[/itex]
[itex]r = |z_{1}| = \sqrt{4} = 2[/itex]
[itex]\theta = arccos(\frac{1}{2}) = \frac{\pi}{3}[/itex]
[itex]z_{1} = 2cis(\frac{\pi}{3})[/itex]
[itex]z_{2} = -2 -2i[/itex]
[itex]r = |z_{2}| = \sqrt{8} = 2\sqrt{2}[/itex]
[itex]\theta = arccos(\frac{-2}{2\sqrt{2}}) = \frac{3\pi}{4}[/itex]
[itex]z_{2} = 2\sqrt{2}cis(\frac{3\pi}{4})[/itex]
[itex]\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})[/itex]
Part B
No idea :(