Question

How do you write this complex number in rectangular form `sqrt8 "cis"(-pi/3)`?

Answer 1

`sqrt8"cis"(-pi/3)=sqrt2-sqrt6i`

Explanation:

Assuming `"cis"-=cos+sin`, then

`sqrt8"cis"(-pi/3)=sqrt8(cos(-pi/3)+sin(-pi/3)i)=2sqrt2(1/2-sqrt3/2i)=sqrt2-sqrt2sqrt3i=sqrt2-sqrt6i`