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Homework Help: Complex number question

  1. Jul 16, 2006 #1
    Hi, how to solve this question?

    Find the square roots fo the complex number -40-42i.
    (i) Find the square roots of the complex number 40+42i,
    (ii) solve the equation (z+1)^2 + 160 + 168i = 0 for all complex roots.

    I don't know how to start solving this question.
  2. jcsd
  3. Jul 16, 2006 #2


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    Hello Jack,

    do you know the following representation of a complex number [itex]z[/itex] in the complex plane?

    [tex]z=x+iy=r(\cos\phi+i\,\sin\phi)=r e^{i\phi}[/tex]

    Here's a sketch to clarify what [itex]r[/itex] and [itex]\phi[/itex] are meant to be.

    http://upload.wikimedia.org/wikipedia/en/c/c2/Complex.png [Broken]


    Last edited by a moderator: May 2, 2017
  4. Jul 17, 2006 #3
    A couple of useful formulas:

    K e^{i\psi} = a + ib \quad
    then \quad
    K = \sqrt{a^2 + b^2} \quad
    \psi = tan^{-1} b/a \quad
    That way you can convert from one form to another. Since taking a square root is easy in the alternative form, you should have no problem.

    This is what nazzard said, but maybe in a manner that is a bit clearer to a beginner in complex numbers. It is really just the application of the pythagorean theorum to the chart above. Compare what I wrote to nazzard's picture. You'll see,

    Last edited: Jul 17, 2006
  5. Jul 17, 2006 #4


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    Thank you tony. I want to point out that one has to be very careful with using arctan to calculate [tex]\phi[/tex], the so called complex argument of z. Remember: case differentiation for different values of b and a (or y and x in my post).


  6. Jul 17, 2006 #5


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    Of course, there's a (possibly) more intuitive and direct way of doing it.

    If sqrt(-40-42i) = (a+bi), then just solve for a and b by squaring both sides (you should be able to fnid two independent equations to break it up into)
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