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Vectors and imaginary numbers

  1. Aug 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Consider a vector z defined by the equation z=z1z2, where z1=a+ib, z2=c+id.
    (a) show that the length of z is the product of the lengths of z1 and z2.
    (b) show that the angle between z and the x axis is the sum of the angles made by z1 and z2 separately.

    3. The attempt at a solution
    (a) i want to just do regular multiplication. (a+ib)(c+id)= ac-bd+i(ad+bc) however i dont see how that would show the length of z is the product of z1 and z2, all i did was multiply.

    my next idea would be to take the magnitudes of z1 and z2 and multiply them. so, (√[a^2+(ib)^2]) * (√[c^2+(id)^2]) = (√[a^2-b^2])(√[c^2-d^2]).

    (b)this would depend in part on which attempt of part (a) is correct. this is because depending on the correct way vector z is represented with its components the angle is going to be different.

    Thank you in advance.
     
  2. jcsd
  3. Aug 27, 2013 #2
    I believe for part (a) you are trying to do

    [itex] |z| = |z_1||z_2| [/itex] so you can verify that based on what you have you know that if [itex] z = x + iy [/itex] then [itex]|z| = \sqrt{x^2 + y^2} [/itex]

    (b) Use the fact that [itex] z = r e^{i \theta} [/itex] I think.
     
  4. Aug 28, 2013 #3

    vela

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    Let ##r_i = |z_i|## and ##\theta_i## = angle between ##z_i## and the x-axis. Using basic trig, you should be able to see that you can write
    \begin{align*}
    a &= r_1\cos\theta_1 \\
    b &= r_1\sin\theta_1 \\
    c &= r_2\cos\theta_2 \\
    d &= r_2\sin\theta_2
    \end{align*} Try plugging that into your expression for the product.

    You don't want to include the factor of ##i## when squaring. The magnitude of a complex number ##z## is given by ##|z| = \sqrt{z z^*}## where ##z^*## is the conjugate. If you work that out, you'll see the ##i## drops out.

     
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