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Homework Help: Complex Analysis: Using polar form to show arg(z1) - arg(z2) = 2n*pi

  1. Jan 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Given that z[itex]_{1}[/itex]z[itex]_{2}[/itex] ≠ 0, use the polar form to prove that
    Re(z[itex]_{1}[/itex][itex]\bar{z}[/itex][itex]_{2}[/itex]) = norm (z[itex]_{1}[/itex]) * norm (z[itex]_{2}[/itex]) [itex]\Leftrightarrow[/itex] θ[itex]_{1}[/itex] - θ[itex]_{2}[/itex] = 2n∏, where n is an integer, θ[itex]_{1}[/itex] = arg(z[itex]_{1}[/itex]), and θ[itex]_{2}[/itex] = arg(z[itex]_{2}[/itex]). Also, [itex]\bar{z}[/itex][itex]_{2}[/itex] is the conjugate of z[itex]_{2}[/itex].

    2. Relevant equations

    norm (z) = [itex]\sqrt{a^{2} + b^{2}}[/itex], where z = a +i*b.

    norm (z) = r, where r is the radius.

    z = r[cos θ + i*sin θ]

    3. The attempt at a solution

    Trying to prove the forward direction, I know the above formulas, and that arg(z[itex]_{1}[/itex]z[itex]_{2}[/itex]) = θ[itex]_{1}[/itex] + θ[itex]_{2}[/itex] +2n∏.
    I'm having trouble getting the first step. I know that norm (z[itex]_{1}[/itex]) * norm (z[itex]_{2}[/itex]) = r[itex]_{1}[/itex]r[itex]_{2}[/itex], but I don't know if this is how you begin.

    Thanks for any help!
  2. jcsd
  3. Jan 14, 2012 #2

    [tex]{z_1} = {r_1}{e^{{\theta _1}\pi i}},{z_2} = {r_2}{e^{{\theta _2}\pi i}},{{\bar z}_2} = {r_2}{e^{ - {\theta _2}\pi i}}[/tex]

    then the first condition is

    [tex]{\mathop{\rm Re}\nolimits} \left( {{r_1}{r_2}{e^{i\pi ({\theta _1} - {\theta _2})}}} \right) = {r_1}{r_2}[/tex]


    [tex]\cos ({\theta _1} - {\theta _2}) = 1[/tex]

    or [tex]{\theta _1} - {\theta _2} = 2n\pi [/tex]

    You can make this more general if you like.

    I'll let you take it from there.
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