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.