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Solve z^3 = -i

  1. Apr 6, 2012 #1
    1. The problem statement, all variables and given/known data
    First off convert to z=(-i)1/3.
    Here is where I am stuck, how do I convert the above to this form: e? I am mainly stuck because I don't know how to solve for theta.

    I know that θ = arctan(y/x) where y is the coefficient in front of i and x is the real value but in this case there is no x... So you would be dividing by zero which doesn't make sense.
     
    Last edited: Apr 6, 2012
  2. jcsd
  3. Apr 6, 2012 #2
    The "trick" is to write [itex](-i)^{1/3}[/itex] in the form [itex]re^{i\theta}[/itex] where you must figure out what [itex]r,\theta[/itex] are (this part should be very easy.)

    Now you are trying to compute [itex](re^{i\theta})^{1/3} = r^{1/3}(e^{i\theta})^{1/3}[/itex]. Do you see how to do this? Now, there is actually going to be three different solutions, do you see this? Do you see what they are?
     
  4. Apr 6, 2012 #3
    Alternatively, you can rewrite [itex]z=(-i)^{\frac{1}{3}}[/itex] as [itex] z=e^{log((-i)^\frac{1}{3})}[/itex] and use properties of complex logs to arrive at the correct answer.
     
  5. Apr 6, 2012 #4

    vela

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    Think about where the point z=-i lies in the complex plane. Forget using the formula. What are the polar coordinates of that point?
     
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