<PRE><B>Question 101254</B>
Write in trig form:  -6 + 6sqrt3i
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Conversion formulas:
r = sqrt(a^2+b^2)
theta = tan^-1(b/a)
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Your Problem:
r = sqrt((-6)^2+(6sqrt3)^2)= sqrt(36+108)= 12
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theta = tan^-1(6sqrt3/-6) = tan^-1(-sqrt3)=(2/3)pi
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Trig form:
-6+6sqrt3i = 12(cos((2/3)pi) + isin((2/3)pi)
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Cheers,
Stan H.
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