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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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    $\begingroup$ Are you familiar with the identities for $\cos(\theta_1+\theta_2)$ and $\sin(\theta_1+\theta_2)$? $\endgroup$ – Gerry Myerson May 14 '12 at 0:45
  • $\begingroup$ @GerryMyerson No I was not. I plugged them in, and arrived at the correct equation. Thanks! $\endgroup$ – user26649 May 14 '12 at 0:50
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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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    $\begingroup$ Well since that's exactly what I was halfway through typing, I certainly deserves my vote! $\endgroup$ – The Chaz 2.0 May 14 '12 at 0:47
  • $\begingroup$ My thoughts, precisely. $\endgroup$ – Cameron Buie May 14 '12 at 0:57
  • $\begingroup$ @Froggie Thanks! $\endgroup$ – user26649 May 14 '12 at 1:05
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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.

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