Is there a reason why a complex number to a power $n$ equals $r ^ n(\cos \theta+i\sin \theta)?$

Question

I saw a question where it asked for what $z^7$ equalled to and the answer was equal to the $r^n(\cos \theta+i\sin \theta)$

I was wondering why?

Answer 1

Recall that if $z=re^{i\theta},$ then $$z=r\left(\cos\theta+i\sin\theta\right),$$ using Euler's formula. Hence, $$z^7=r^7e^{7i\theta}=r^7\left(\cos (7\theta)+i\sin (7\theta)\right).$$

Answer 2

I think you meant $$[r(\cos\theta+i\sin\theta)]^7=r^7(\cos 7\theta+i\sin 7\theta).$$

Well, this is an application of the result known as De Moivre's Theorem. That is, given a complex number $r(\cos\theta+i\sin\theta)$ and positive integer $n,$ then $$[r(\cos\theta+i\sin\theta)]^n=r^n(\cos n\theta+i\sin n\theta).$$ This follows by induction from the fact that when you multiply two complex numbers in this trigonometric form, the moduli are multiplied and the arguments are added. That is, given $z=r(\cos a+i\sin a)$ and $w=s(\cos b+i\sin b),$ then we have $$zw=rs(\cos(a+b)+i\sin(a+b).$$ You can verify this by performing the multiplication on the left and using the trigonometric identities for a sum of arcs. Of course, $r,s,a,b$ are real, and $r,s$ cannot be negative.