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anonymous

  • one year ago

Find the quotient. Write your answer in standard form. 3 + i / 3 - i A. -1 B. 1 - i C. 3/5 + 4/5 i D. 4/5 + 3/5 i

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  1. Nnesha
    • one year ago
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    multiply top and bottom both by the conjugate of the denominator (3-i)

  2. Nnesha
    • one year ago
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    do you know the definition of conjugate ?

  3. anonymous
    • one year ago
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    no. ): @Nnesha

  4. Nnesha
    • one year ago
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    conjugate example just change the sign of imaginary number a+bi conjugate is a-bi bi imaginary a= real so 3-i conjugate is what ?

  5. anonymous
    • one year ago
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    3+i ?? @Nnesha

  6. anonymous
    • one year ago
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    is it \[\frac{3+i}{3-i}\]?

  7. anonymous
    • one year ago
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    if so, multiply top and bottom by the conjugate of the denominator the conjugate of \(a+bi\) is \(a-bi\) and this works because \[(a+bi)(a-bi)=a^2+b^2\] a real number

  8. anonymous
    • one year ago
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    first step is \[\frac{3+i}{3-1}\times \frac{3+i}{3+1}\] which gives you \[\frac{(3+i)(3+i)}{3^2+1^2}\]

  9. anonymous
    • one year ago
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    the denominator is evidently \(10\) and the numerator is whatever you get when you multiply that out

  10. anonymous
    • one year ago
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    you good from there?

  11. anonymous
    • one year ago
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    sorry typo there i meant "first step is \[\frac{3+i}{3-i}\times \frac{3+i}{3+i}\]

  12. anonymous
    • one year ago
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    I got (9 + 6i + i^2) / (9 - i) I don't know where to go from there? @satellite73

  13. anonymous
    • one year ago
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    ok lets back up, your denominator is wrong

  14. anonymous
    • one year ago
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    Is it 9-i^2?

  15. anonymous
    • one year ago
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    it is not \(9-i\) but rather \(9+1=10\) \[3-i\] has real part \(3\) and imaginary part \(-1\) when you multiply \[(3+i)(3-i)\] you get \[3^2+(-1)^2\] or \(10\)

  16. anonymous
    • one year ago
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    you can write it that way if you want, but don't forget \(i^2=-1\) so \(9-i^2=9+1=10\)

  17. anonymous
    • one year ago
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    So is i by itself 1?

  18. anonymous
    • one year ago
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    you should really just be thinking that \[(a+bi)(a-bi)=a^2+b^2\] just like in pythagoras

  19. anonymous
    • one year ago
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    no \(i\neq 1\) but \(i^2=-1\) and so \(-i^2=+1\)

  20. anonymous
    • one year ago
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    probably the confusing part is how to interpret \(a+bi\) when you have \(3-i\) in that case \(a=3,b=-1\) of course when you square, you can ignore the minus sign

  21. anonymous
    • one year ago
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    your numerator of \[9+6i+i^2\]is correct, but not complete since \(i^2=-1\) then \[9+6i+i^2=9+6i-1=8+6i\]

  22. anonymous
    • one year ago
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    giving you an almost final answer of \[\frac{8+6i}{10}\] now cancel the common factor of 2 top and bottom

  23. anonymous
    • one year ago
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    So it's D!! I UNDERSTAND!! Thank you so much. (:

  24. anonymous
    • one year ago
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    yes it is D, and you are welcome

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