SOLUTION: My question is to solve the equation ------9 plus 3i over 2i

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Question 71508: My question is to solve the equation ------9 plus 3i over 2i
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
%289+%2B+3i%29%2F2i
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We can do this problem by converting the denominator to a real number. Do this by multiplying
the entire fraction by %28-2i%29%2F%28-2i%29. This converts the problem to:
.
%28%28-2i%29%2A%289%2B3i%29%29%2F%28%28-2i%29%2A%282i%29%29
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Let's work on the denominator first. When you multiply -2i times +2i the answer is -4i%5E2.
But recall that by definition i%5E2+=+-1. Therefore, we can replace i%5E2 by -1 and
the denominator becomes -4i%5E2+=+-4%2A%28-1%29+=+4. Substitute 4 for the entire denominator
and the problem becomes:
.
%28%28-2i%29%2A%289%2B3i%29%29%2F4
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Next let's do the multiplication in the numerator. The first numerator multiplication
is %28-2i%29%2A%289%29+=+-18i and the second multiplication is %28-2i%29%2A%283i%29=-6i%5E2.
Again i%5E2+=+-1 and substituting this results in %28-6%29%2A%28-1%29+=+6. Putting the
results of these two multiplications into the numerator makes the problem become:
.
%28-18i%2B6%29%2F4
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Reverse the two terms in the numerator to the more standard form of 6 - 18i.
Then divide
each of these two terms by the denominator 4 to get
.
%286%2F4%29-18i%2F4+=+1.5+-+4.5i
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There's the answer and how to work it out. The common way to deal with "i" terms in the
denominator is to multiply them by the same term with an opposite sign. And as a preview
of things to come, if the denominator is a + bi, multiply the numerator and denominator
by a - bi. Try an example. For example multiply 2 + 3i by 2 - 3i and see what would happen
to the denominator.
.
Hope this helps you to develop a little more understanding of dealing with complex
numbers.