amarab
  • amarab
simplify the expression? would like help, not an answer. -6+i/-5+i
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
http://www.sparknotes.com/math/algebra2/complexnumbers/section3.rhtml
asnaseer
  • asnaseer
do you know what the conjugate of a complex number \((a+ib)\) is?
amarab
  • amarab
a-bi?

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asnaseer
  • asnaseer
correct. now notice what happens if you multiple a complex number by its conjugate:\[(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2\]
asnaseer
  • asnaseer
i.e. the imaginary parts of a complex number disappear when you multiply it by its conjugate
amarab
  • amarab
okay thanks! just what i needed
asnaseer
  • asnaseer
so the idea here is to multiply your fraction with something that would result in the denominator losing its imaginary part.
asnaseer
  • asnaseer
ok - glad you got it! :)
amarab
  • amarab
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??
amarab
  • amarab
it looks like 30+2i^2/25+2i^2 but i know it's wrong
anonymous
  • anonymous
i^2=-1
asnaseer
  • asnaseer
\[\frac{-6+i}{-5+i}=\frac{-6+i}{-5+i}\times\frac{-5-i}{-5-i}=\frac{(-6+i)(-5-i)}{(-5+i)(-5-i)}\]\[\qquad=\frac{30+6i-5i-i^2}{25+5i-5i-i^2}=\frac{30+i-i^2}{25-i^2}\]Then (as suggested by @irene22988) use the fact that \(i^2=-1\) to simplify further

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