3+2i / 3-2i leave answer in a+bi form the answer is 5/13 + 12/13i , but i keep getting a different answer please help me!!

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3+2i / 3-2i leave answer in a+bi form the answer is 5/13 + 12/13i , but i keep getting a different answer please help me!!

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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so you have 3+2i/3-2i or do you mean (3+2i)/(3-2i)?
i think you mean the latter \[\text{ so we have } \frac{3+2i}{3-2i}\] do you know to put it in a+bi form the first step you must take is to multiply both top and bottom by bottom's conjugate
do you know what the conjugate of 3-2i is?

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multipy numerator and denominator with 3+2i
@myininaya yes i multiplied the top and bottom by 3+2i and got 9+6i+6i+4i^2 over 9+6i-6i-4i^2 and got the final answer as 9+12i / 9 but its the wrong answer
so how did you get 9+4i^2 is 9?
recall i^2 is -1
so you got it right so far to say \[\frac{9+6i+6i+4i^2}{9+6i-6i-4i^2}\]
and you know again i^2 is -1 so you have \[\frac{9+12i+4(-1)}{9-4(-1)}\] wherever there was an i^2 i just replaced it with -1
simplify this and you will see you got the answer as it should be written
Thanks so much @myininaya
I think this means you got it so gj

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