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Homework Help: Problem involving square / square root of a complex number

  1. Jan 17, 2012 #1
    The problem statement, all variables and given/known data
    [itex]z = (n + i)^{2}[/itex]

    [itex]n[/itex] is a positive real number, and [itex]arg(z) = \frac{\pi}{3}[/itex]

    Find the value of [itex]n[/itex].

    The attempt at a solution

    I am reviewing old problem sets from years past, and came across this problem that appears pretty simple. I have my old answer as [itex]n=\sqrt{3}[/itex], which I can verify numerically as the correct answer. The problem is that I no longer remember the solution, and think I am missing a simple solution to this. We know the following:

    [itex]z^{1/2} = n + i[/itex]
    [itex]arg(n + i) = arctan(\frac{1}{n})[/itex]
    [itex]tan(arg(z)) = \sqrt{3}[/itex]

    I think there must be some way to compare the arguments of the 2 sides and use the given fact that [itex]arg(z) = \frac{\pi}{3}[/itex] for a simple solution, but I don't see what it is. I am not sure how to deal with the square in the original problem statement, or the square root in my first equation.
     
  2. jcsd
  3. Jan 17, 2012 #2

    Dick

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    If x is a complex number, how is arg(x) related to arg(x^2)? exp(it)^2=exp(2it), right?
     
  4. Jan 17, 2012 #3

    ehild

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    Expand the right-hand side of the equation and compare the tangent of the arguments of both sides.

    ehild
     
  5. Jan 17, 2012 #4
    So in my case:
    [itex]arg(z^{1/2})=\frac{1}{2}arg(z)[/itex]
    Then if we take [itex]z^{1/2}=n+i[/itex] and set the argument of each side equal, we have:
    [itex]arg(z^{1/2})=arg(n+i)[/itex]
    [itex]\frac{1}{2}\frac{\pi}{3}=arctan(\frac{1}{n})[/itex]
    [itex]\frac{\pi}{6} = arctan(\frac{1}{n})[/itex]
    Taking the tangent of both sides:
    [itex]\frac{1}{\sqrt{3}}=\frac{1}{n}[/itex]
    Thus:
    [itex]n=\sqrt{3}[/itex]
     
  6. Jan 18, 2012 #5

    Dick

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    That's it. We can be a little bit sloppy about 2*pi factors in the arg because we know that 0<arg(1+n)<pi/2.
     
  7. Jan 18, 2012 #6
    Thanks for the help!
     
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