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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Question 117942: 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
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
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.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28sqrt%283%29%2Bi%29%5E9 Start with the given expression


%282%28sqrt%283%29%2F2%2Bi%281%2F2%29%29%29%5E9 Replace sqrt%283%29 and i with sqrt%283%29%2F2 and i%281%2F2%29. Then place a 2 outside the parenthesis. Notice how the 2 on the outside will distribute back to sqrt%283%29%2Bi. In other words, 2%28sqrt%283%29%2F2%2Bi%281%2F2%29%29=sqrt%283%29%2Bi


Notice how cos%28pi%2F6%29=sqrt%283%29%2F2 and sin%28pi%2F6%29=1%2F2.

%282%28cos%28pi%2F6%29%2Bi%2Asin%28pi%2F6%29%29%29%5E9 Now replace sqrt%283%29%2F2 with cos%28pi%2F6%29 and replace 1%2F2 with sin%28pi%2F6%29.

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

%28r%28cos%28x%29%2Bi%2Asin%28x%29%29%29%5En=r%5En%2A%28cos%28nx%29%2Bi%2Asin%28nx%29%29


2%5E9%28cos%289%28pi%2F6%29%29%2Bi%2Asin%289%28pi%2F6%29%29%29 Expand using De Moivre's Thereom


512%28cos%283pi%2F2%29%2Bi%2Asin%283pi%2F2%29%29 Simplify


512%280%2Bi%2A%28-1%29%29 Now evaluate cos%283pi%2F2%29 and sin%283pi%2F2%29


0-512i Distribute


So the answer is now in a%2Bbi form where a=0 and b=-512


So %28sqrt%283%29%2Bi%29%5E9=0-512i