anonymous
  • anonymous
simplify: 3i/2+5i
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\frac{ 3i}{ 2+5i }\]
ganeshie8
  • ganeshie8
Multiply both numerator and denominator by the conjugate of denominator
anonymous
  • anonymous
\[\frac{ 3i (2-5i) }{ 2+5i(2-5i) }\] kind of like that right? Just not all over the fraction bar if you know what i mean...

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ganeshie8
  • ganeshie8
Yes, if you mean : \(\dfrac{ 3i (2-5i) }{ (2+5i)(2-5i) }\)
anonymous
  • anonymous
you're multiplying them right?
ganeshie8
  • ganeshie8
right
ganeshie8
  • ganeshie8
for denominator you may use the identity: \((a+bi)(a-bi)=a^2+b^2\)
anonymous
  • anonymous
ok, i understand how to get the top part. you multiply 3i and 2 to get 6i, and 3i and 5i to get 15, and because of the i on both, im pretty sure that it ccancels out. the bottom is where i always get lost
anonymous
  • anonymous
oh i see thank you

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