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.Com
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)   (Show Source): You can put this solution on YOUR website!


Hi

cos330° = 

sin330° = -1/2





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

 

RELATED QUESTIONS


Hi, I am stuggling very much on these two questions on complex numbers. If you can help,... (answered by josmiceli)

Hi, can anyone help with this complex numbers problem?

In these complex numbers... (answered by stanbon)

How would I use DeMoiver's Theorem to find 2 square roots of 16(cos330 + j sin330)in... (answered by ikleyn)

Hi, I have a big test on complex numbers tomorrow and I am having a very hard time with... (answered by josmiceli,Edwin McCravy,austin92,richard1234)

Express these complex numbers in standard form correct to one decimal place:

If... (answered by stanbon)

Find all complex numbers $z$ such that $|z-1|=|z+3|=|z-i|$.

Express each answer in the  (answered by Fombitz,ikleyn)

Hello!

I am stuck! I am to find all the solutions in the complex number set of Z^5=-32
 (answered by Alan3354)

Write the complex number z = -6 -2i in trigonometric form (sometimes called polar form).... (answered by Fombitz,Alan3354)

Let 
z = 5 + 5sqrt(3)i,

Write z in polar form.

Find the complex number... (answered by greenestamps)