How do you convert #2+(sqrt 3)i# to polar form?

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How do you convert #2+(sqrt 3)i# to polar form?

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Answer 1 · 2016-03-19T05:15:55

The polar coordinates are: #P(sqrt(7),.714)#

Explanation:

calculate the magnitude, r:
#r = sqrt(2^2+(sqrt(3))^2)=sqrt(4+3)=sqrt(7)#
Calculate the angle #theta#:
#theta=tan^-1 (sqrt(3)/2)=40.89^0=.714 " radians"#
The polar coordinates are: #P(sqrt(7),.714)#

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How do you convert #2+(sqrt 3)i# to polar form?

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