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?


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. $\sin(-x) = -\sin x.$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.