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13 February, 06:01

How do you write the quotient (9+i) over (9-i) in standard form? I need the answer but more importantly, I need to learn how to solve these kinds of problems.

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  1. 13 February, 07:24
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    Since they are the conjugates of each (9-a and 9+a are conjugates, or with any number replacing 9), we multiply it by (9+i) / (9+i) to get (9+i) ^2 / (9-i^2) = (81+2i+i^2) / (9 - (-1)) = (81-1+2i) / (10) = 80/10+2i/10=8+i/5. Conjugates work because they make i into i^2 (-1) without any of the sneaky i's in the middle!
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