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

[I Love Math]

Homework Statement

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

Homework Equations

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

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

z = r[cos θ + i*sin θ]

The Attempt at a Solution

Trying to prove the forward direction, I know the above formulas, and that arg(z_{1}z_{2}) = θ_{1} + θ_{2} +2n∏.
I'm having trouble getting the first step. I know that norm (z_{1}) * norm (z_{2}) = r_{1}r_{2}, but I don't know if this is how you begin.

Thanks for any help!

 

[Hariraumurthy]

Homework Statement

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

Homework Equations

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

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

z = r[cos θ + i*sin θ]

The Attempt at a Solution

Trying to prove the forward direction, I know the above formulas, and that arg(z_{1}z_{2}) = θ_{1} + θ_{2} +2n∏.
I'm having trouble getting the first step. I know that norm (z_{1}) * norm (z_{2}) = r_{1}r_{2}, but I don't know if this is how you begin.

Thanks for any help!

remember

{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}}

then the first condition is

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

i.e.

\cos ({\theta _1} - {\theta _2}) = 1

or {\theta _1} - {\theta _2} = 2n\pi

You can make this more general if you like.

I'll let you take it from there.