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Recently I have started studying imaginary numbers. I've come at the division of complex numbers in polar form. However when I do the exercise I get almost exactly the same answer as in my textbook except in the textbook they switch the signs for reasons I don't quite understand.

The following image is the example from the textbook image

division of z1 and z2

$$ z_1= \frac12 \cos⁡(3\pi / 4)+ i \sin⁡(3\pi/4) $$ $$ z_2 =4 \cos⁡(11\pi / 6)+ i \sin⁡(11π/6) $$

At the last step of the example the signs are switched but I don't understand why. Why do they do that?

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The first $-$ signs come from actually computing the differences. Then use the fact that $\cos$ is an even function, i.e. $\cos(-x)=\cos(x),$ and $\sin$ is odd, i.e. $\sin(-x) = -\sin x.$

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  • $\begingroup$ Thank you, that explains it. And any idea why it switches and divides by 2 in multplication of complex numbers here (i.imgur.com/BfeZPvA.png) ? $\endgroup$ – Lagastic Aug 3 '16 at 10:28
  • $\begingroup$ @Lagastic, that looks wrong. They're using facts from the unit circle and reference angles, but what it should be is $\cos(2\pi/3) = -\cos(\pi/3)$ and $\sin(2\pi/3) = \sin(\pi/3)$. Check the unit circle to verify and read up on reference angles for more detailed info. $\endgroup$ – tilper Aug 3 '16 at 11:26
  • $\begingroup$ @tilper and why go to the latter form, why not just compute $$cos(2π/3) + i sin(2π/3)$$ ? $\endgroup$ – Lagastic Aug 3 '16 at 13:33
  • $\begingroup$ @tilper Edit: Is it because z1 and z2 are in kwadrant 1, consequently the product of z1 and z2 should also lie in kwadrant 1. π/3 lies in kwadrant 1 while 2π/3 does not ? $\endgroup$ – Lagastic Aug 3 '16 at 13:41
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    $\begingroup$ Oh ok -3 + 3sqrt(3)i was my solution at first, didn't think there would be a mistake in the textbook so that confused me quite a bit. Thanks. $\endgroup$ – Lagastic Aug 3 '16 at 15:18

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