SOLUTION: [16(cos300+isin300)]/[8(cos60+isin60)]

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Question 420873: [16(cos300+isin300)]/[8(cos60+isin60)]
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%2816%28cos%28300%29%2Bi%2Asin%28300%29%29%29%2F%288%28cos%2860%29%2Bi%2Asin%2860%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:
%2816%2F8%29%28cos%28300-60%29+%2B+i%2Asin%28300-60%29%29
which simplifies to
%282%29%28cos%28240%29+%2B+i%2Asin%28240%29%29
which is the answer in polar form. For standard form, a+bi, we replace the cos and sin with their values. 240 is a special angle so we don't need a calculator:
%282%29%28%28-1%2F2%29+%2B+i%2A%28-sqrt%283%29%2F2%29%29
Distributing the 2 we get:
-1+%2B+i%2A%28-sqrt%283%29%29
or
-1+%2B+%28-sqrt%283%29%29i