1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Homework Helper

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

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


    User Avatar
    Science Advisor
    Homework Helper

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook