Question 72133
{{{5/(6-2i)}}}
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to get this to the form a + bi begin by multiplying the given term by:
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{{{(6 + 2i)/(6+2i)}}} 
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Note that this is equivalent to multiplying the given term by 1 because {{{(6 + 2i)/(6+2i)}}} 
equals 1.
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The numerator multiplication of the 5 times the (6 + 2i) results in 30 + 20i.
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Then the denominator multiplication of {{{(6 - 2i)*(6 + 2i)}}} results in:
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{{{ 36 - 12i + 12i -4*i^2}}}
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The -12i and + 12i cancel each other out. Then recall that {{{i^2}}} by definition is -1.
Substituting -1 for {{{i^2}}} leads to:
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{{{36 -4*(-1) = 36 + 4 = 40}}}
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This is the denominator ... +40.  From above the numerator is 30 + 20i. So the answer is:
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{{{(30+20*i)/40 = (3/4) + (1/2)*i}}}
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The answer to your problem is {{{(3/4) + (1/2)*i}}} where {{{a = (3/4)}}} and {{{b= (1/2)}}}
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Hope this helps you to understand complex numbers. Notice how you can eliminate complex
numbers in the denominator by multiplying the denominator by the same complex number with 
a change in signs between the real and imaginary parts. This converts the denominator
to a real number. 
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