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Homework Help: Complex number in polar form

  1. Jun 15, 2011 #1
    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!

    1. The problem statement, all variables and given/known data

    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].

    2. Relevant 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]

    3. 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 :(
  2. jcsd
  3. Jun 15, 2011 #2


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    Science Advisor

    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?
  4. Jun 15, 2011 #3
    Oh right. So they want me to find the smallest value of [itex]n[/itex] such that [itex]arg(z)[/itex] = [itex]k\pi[/itex] (so that the imaginary part = 0)?
  5. Jun 15, 2011 #4
    Okay it turns out that's the way to solve the question.

    Thank you for the help!
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