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Division of Complex Numbers with 2 Conjugates

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2i/2-i - 4i/1-2i = R/-5 Maybe I am just seeing this wrong but it is quite apparent that there are 2 conjugates, how em I to remove the imaginary divisors and solve for R? I know R = -6 given the answer key. I cannot however get to that point myself.

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When you have a complex divisor, there is a simple way to eliminate imaginary parts in divisor by multiplying both upper and lower parts by conjugate complex number. In this example, both divisors will be 5 ( (2-i)(2+i) = (1-2i)(1+2i) = 5)
E.g. (2i(2+i))/(4+1) - (4i(1+2i))/(1+4) = R/-5
E.g. (4i-2-4i+8)/5 = R/-5
E.g. -6 = R

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Ah, thank you very much.
So multiplying each seperatly by the conjugate does not disturb the equality.
I figured to eliminate one divisor by using the conjugate would mean I would have to multiply the whole equation by the conjugate, thus causing some very odd results.

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Guest Anonymous Poster
If the conjugate of the complex denominator is C, you're multiplying the fraction by C/C. This will always be 1. Multiplying a number by 1 results in that same number. It's a trick to write the same number in a different way that doesn't have imaginary numbers in the denominator.

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