Question 462738
Determine (-2+i3)^5 in the polar form 
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r = sqrt(2^2+3^2) = sqrt(13)
theta = tan^-1(-3/2) = 303.69 degrees
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(-2+3i)^5 = 13^(5/2)[cos(303.69)+isin(303.69)
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and in cartesian form
(-2+3i)^5 = x + yi
x = 13^(5/2)*cos(303.69) = 338
y = 13^(5/2)*isin(303.69) = -507i
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Cheers,
Stan H.
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