SOLUTION: Whats is ( (SQRT3) + i ) ^ 9 or in other words the quantity of the square root of three plus "i" to the power of nine in a + bi form

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What's ( (SQRT3) + i ) ^ 9
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Convert to trig form:
r = sqrt((sqrt3)^2+1^2) = 2
theta = tan^-1(1/sqrt3)= 30 degrees
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[2cis(30)]^9 = (2^9)cis(270)
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Convert back to complex form:
= 512(cos(270)+isin(270)
=512(0 + i (-1))
= -512i
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Cheers,
Stan H.

You can put this solution on YOUR website!
Start with the given expression


Replace and with and . Then place a 2 outside the parenthesis. Notice how the 2 on the outside will distribute back to . In other words,


Notice how and .

Now replace with and replace with .

Now use De Moivre's Thereom to expand. Remember, De Moivre's Thereom states:




Expand using De Moivre's Thereom


Simplify


Now evaluate and


Distribute


So the answer is now in form where and


So