SOLUTION: Hello, I am struggling with this complex numbers problem: If z=6√(3)(cos330°+i*sin330°) in trigonometric form, find z in standard form. √=radical Thank you for y

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Hello, I am struggling with this complex numbers problem: If z=6√(3)(cos330°+i*sin330°) in trigonometric form, find z in standard form. √=radical Thank you for y      Log On


   

Question 396771: Hello, I am struggling with this complex numbers problem:
If z=6√(3)(cos330°+i*sin330°) in trigonometric form, find z in standard form.
√=radical
Thank you for your time!
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

cos330° = sqrt%283%29%2F2

sin330° = -1/2

+z+=+6sqrt%283%29%28sqrt%283%29%2F2+%2B+i%28-1%2F2%29%29

+z+=+9+-+i%2A3sqrt%283%29

thought You might find this (cosx,sinx) summary helpful