Question

How do you convert #(cos120+isin120) to its rectangular form?

Answer 1

You evaluate the `cos` and `sin` and multiply the modulus (in this case `1`):
You have:

`1[cos(120°)+isin(120°)]=`

`=cos(120°)+isin(120°)=`

`=-1/2+isqrt(3)/2`

which is in rectangular form: `a+ib`

Hope it helps