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amyna

  • 4 months ago

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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  1. myininaya
    • 4 months ago
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    so you have 3+2i/3-2i or do you mean (3+2i)/(3-2i)?

  2. myininaya
    • 4 months ago
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    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

  3. myininaya
    • 4 months ago
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    do you know what the conjugate of 3-2i is?

  4. spandana
    • 4 months ago
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    multipy numerator and denominator with 3+2i

  5. amyna
    • 4 months ago
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    @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

  6. myininaya
    • 4 months ago
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    so how did you get 9+4i^2 is 9?

  7. myininaya
    • 4 months ago
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    recall i^2 is -1

  8. myininaya
    • 4 months ago
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    so you got it right so far to say \[\frac{9+6i+6i+4i^2}{9+6i-6i-4i^2}\]

  9. myininaya
    • 4 months ago
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    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

  10. myininaya
    • 4 months ago
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    simplify this and you will see you got the answer as it should be written

  11. amyna
    • 4 months ago
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    Thanks so much @myininaya

  12. myininaya
    • 4 months ago
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    I think this means you got it so gj

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