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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!!

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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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