How do you convert #3 sqrt2 - 3 sqrt 2i# to polar form?

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How do you convert #3 sqrt2 - 3 sqrt 2i# to polar form?

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Answer 1 · 2016-04-17T11:47:12

In Polar form the equation is #6(cos(-0.785)+isin(-0.785))#

Explanation:

#Z=3sqrt2-3sqrt2i# so Modulas #Z=sqrt((3sqrt2)^2+(-3sqrt2)^2) =6#
Argument #z = theta = tan^-1(-(3sqrt2)/(3sqrt2))=tan_1(-1)=-0.785 radian# So in Polar form the equation is #6(cos(-0.785)+isin(-0.785))#

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How do you convert #3 sqrt2 - 3 sqrt 2i# to polar form?

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