How do you convert #4 (sqrt(3) +i)# to polar form?

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How do you convert #4 (sqrt(3) +i)# to polar form?

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Answer 1 · 2016-04-18T03:48:29

#8 e^(i(pi/3))=8 ( cos (pi/6) + i sin (pi/6))#

Explanation:

#x=4sqrt3=y#
#x+ i y = r ( cos theta + i sin theta )#

#r=sqrt(x^2+y^2)=8#
#cos theta = x/r = sqrt3/2, sin theta = y/r=1/2#, Both are positive. #theta# is in the 1st quadrant.
#theta=pi/6#
So, #4( sqrt3 + i )=8 (cos pi/6 + i sin pi/6 )#

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How do you convert #4 (sqrt(3) +i)# to polar form?

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