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Help with complex number derivation

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data
    (a) Suppose the segment connecting (a,b) to (0,0) has length r[itex]_{1}[/itex] and forms an angle [itex]\theta[/itex][itex]_{1}[/itex] with the positive side of the x-axis. Suppose the segment connecting (c,d) to (0,0) has a length r[itex]_{2}[/itex] and forms an angle [itex]\theta[/itex][itex]_{2}[/itex] with the positive side of the x-axis. Now let (a+bi)(c+di)=x+yi. Show that the length of the segment connecting (x,y) to the origin is r[itex]_{1}[/itex]r[itex]_{2}[/itex] and the angle formed is [itex]\theta[/itex][itex]_{1}[/itex]+[itex]\theta[/itex][itex]_{2}[/itex].

    (b) Use the result from (a) to find a complex number z[itex]\in[/itex]C such that z^2=i.

    2. Relevant equations



    3. The attempt at a solution
    (a+bi)(c+di)=x+yi
    ac+adi+bci+bd(i[itex]^{2}[/itex])=x+yi
    ac+adi+bci-bd=x+yi
    (ac-bd)+(ad+bc)=x+yi
    x=(ac-bd), y=(ad+bc)

    I'm not sure where to go from here. Just looking for some help. Thanks!
     
  2. jcsd
  3. Sep 6, 2011 #2

    eumyang

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    Remember the formulas for going from standard form of a complex number to trig form. If z = a + bi = r(cos θ + i sin θ), then
    [itex]r = \sqrt{a^2 + b^2}[/itex],
    a = r cos θ and b = r sin θ, and
    [itex]\tan \theta = \frac{b}{a}[/itex].

    If x=(ac-bd) and y=(ad+bc), then what is
    [itex]\sqrt{x^2 + y^2}[/itex]? You'll need to make it equal to r1r2.

    For the angle, you'll need the tangent of a sum identity.
    [itex]\tan (\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2}[/itex]
     
    Last edited: Sep 6, 2011
  4. Sep 7, 2011 #3
    I'm honestly not getting very far with this. Could you help me out a little more?
     
  5. Sep 7, 2011 #4
    Ok, I got the first part. Now I need to prove the angle part. Any help?
     
  6. Sep 7, 2011 #5

    eumyang

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    Start with this. Plug in what x and y equals into the square root and expand the radicand. Show us what you get.

    EDIT: Never mind. You said that you got this part.


    [itex]\tan \theta_1 = \frac{b}{a}[/itex]
    and
    [itex]\tan \theta_2 = \frac{d}{c}[/itex]
    Plug these into the formula above and simplify. Somehow you'll have to simplify to y/x. (Remember that you had x and y can be expressed in terms of a, b, c, and d.)
     
  7. Sep 7, 2011 #6
    Hmmm...

    After I plug those in, what methods can I use to simply that expression?
     
  8. Sep 7, 2011 #7

    eumyang

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    After plugging those in, try multiplying the numerator and denominator of this complex fraction by ac.

    (Going to bed now... :zzz:)
     
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