Finding the Smallest n for Real z^n in Complex Numbers

AI Thread Summary
The discussion revolves around solving a complex number problem involving the expression of z in polar form and determining the smallest positive integer n for which z^n is a real number. Part a has been successfully solved, expressing z as 2cis(π/3) divided by 2√2cis(3π/4), resulting in a simplified form. The challenge lies in part b, where the user seeks guidance on finding the smallest n that makes the imaginary component of z^n equal to zero. A hint suggests focusing on the argument of z, indicating that n should be chosen such that the argument is a multiple of π. This approach clarifies the path to the solution.
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!

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 :(
 
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nayfie said:
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!

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 :(

Hint: What property does a real number have (what is its imaginary component equal to)?

Based on an argand diagram (a two dimensional graph with real and imaginary part), what is the argument (angle) equal to?
 
Oh right. So they want me to find the smallest value of n such that arg(z) = k\pi (so that the imaginary part = 0)?
 
Okay it turns out that's the way to solve the question.

Thank you for the help!
 
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