SOLUTION: [24(cos150+isin150)]/[2(cos30+isin30)]

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Question 420875: [24(cos150+isin150)]/[2(cos30+isin30)]
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
%2824%28cos%28150%29%2Bi%2Asin%28150%29%29%29%2F%282%28cos%2830%29%2Bi%2Asin%2830%29%29%29
Dividing complex numbers written in polar form is fairly simple because there is a formula that makes it easy. For two complex numbers:
z%5B1%5D+=+r%5B1%5D%28cos%28x%5B1%5D%29+%2B+i%2Asin%28x%5B1%5D%29%29
and
z%5B2%5D+=+r%5B2%5D%28cos%28x%5B2%5D%29+%2B+i%2Asin%28x%5B2%5D%29%29
we can divide them using the formula:
(as long as z%5B2%5D is not zero).

Using this formula to divide your complex numbers we get:
%2824%2F2%29%28cos%28150-30%29+%2B+i%2Asin%28150-30%29%29
which simplifies to
%2812%29%28cos%28120%29+%2B+i%2Asin%28120%29%29
which is the answer in polar form. For standard form, a+bi, we replace the cos and sin with their values. 120 is a special angle so we don't need a calculator:
%2812%29%28%28-1%2F2%29+%2B+i%2Asqrt%283%29%2F2%29
Distributing the 12 we get:
-6+%2B+i%2A%286sqrt%283%29%29
or
-6+%2B+%286sqrt%283%29%29i