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amarab Group Title

simplify the expression? would like help, not an answer. -6+i/-5+i

  • 2 years ago
  • 2 years ago

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  1. etemplin Group Title
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    http://www.sparknotes.com/math/algebra2/complexnumbers/section3.rhtml

    • 2 years ago
  2. asnaseer Group Title
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    do you know what the conjugate of a complex number \((a+ib)\) is?

    • 2 years ago
  3. amarab Group Title
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    a-bi?

    • 2 years ago
  4. asnaseer Group Title
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    correct. now notice what happens if you multiple a complex number by its conjugate:\[(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2\]

    • 2 years ago
  5. asnaseer Group Title
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    i.e. the imaginary parts of a complex number disappear when you multiply it by its conjugate

    • 2 years ago
  6. amarab Group Title
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    okay thanks! just what i needed

    • 2 years ago
  7. asnaseer Group Title
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    so the idea here is to multiply your fraction with something that would result in the denominator losing its imaginary part.

    • 2 years ago
  8. asnaseer Group Title
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    ok - glad you got it! :)

    • 2 years ago
  9. amarab Group Title
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    okay so i did... -6+i(-5-i) over -5+i(-5-i) then got 30+6i+-5i+-i^2 or 30+2i^2 but to the bottom... i ended up with 25+2i^2 and i don't know what to do after that??

    • 2 years ago
  10. amarab Group Title
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    it looks like 30+2i^2/25+2i^2 but i know it's wrong

    • 2 years ago
  11. irene22988 Group Title
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    i^2=-1

    • 2 years ago
  12. asnaseer Group Title
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    \[\frac{-6+i}{-5+i}=\frac{-6+i}{-5+i}\times\frac{-5-i}{-5-i}=\frac{(-6+i)(-5-i)}{(-5+i)(-5-i)}\]\[\qquad=\frac{30+6i-5i-i^2}{25+5i-5i-i^2}=\frac{30+i-i^2}{25-i^2}\]Then (as suggested by @irene22988) use the fact that \(i^2=-1\) to simplify further

    • 2 years ago
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