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Could someone explain why you multiply the lengths and add the angles when multiplying polar coordinates?

I tried multiplying the polar forms ($r_1\left(\cos\theta_1 + i\sin\theta_1\right)\cdot r_2\left(\cos\theta_2 + i\sin\theta_2\right)$), and expanding/factoring the result, and end up multiplying the lengths but can't seem to come to an equation where you add the angles.

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    @GerryMyerson No I was not. I plugged them in, and arrived at the correct equation. Thanks!2012-05-14

2 Answers 2

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By multiplying things out as usual, you get

$[r_1(\cos\theta_1 + i\sin\theta_1)][r_2(\cos\theta_2 + i\sin\theta_2)] = r_1r_2(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2 + i[\sin\theta_1\cos\theta_2 + \sin\theta_2\cos\theta_1]).$

Now you want to use the trig identities $\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2 = \cos(\theta_1 + \theta_2)$ and $\sin\theta_1\cos\theta_2 + \sin\theta_2\cos\theta_1 = \sin(\theta_1 + \theta_2)$ to conclude that this is in fact $r_1r_2[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)].$

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    @Froggie Thanks!2012-05-14
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It might be useful to write the numbers as

$z_1= \nu e^{i\theta}$

$z_2= \mu e^{i\psi}$

Then one has

$z_1\cdot z_2 =\nu \mu \cdot e^{(\psi+\theta)i}$

This representation stems from Euler's formula

$e^{i \theta}=\cos \theta+i\sin \theta$

which I suspect you haven't been told about yet.