# Imaginary numbers: polar form division: switching of signs

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. $\sin(-x) = -\sin x.$
• @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. – tilper Aug 3 '16 at 11:26
• @tilper and why go to the latter form, why not just compute $$cos(2π/3) + i sin(2π/3)$$ ? – Lagastic Aug 3 '16 at 13:33