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simplify: 3i/2+5i
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\[\frac{ 3i}{ 2+5i }\]
Multiply both numerator and denominator by the conjugate of denominator
\[\frac{ 3i (2-5i) }{ 2+5i(2-5i) }\] kind of like that right? Just not all over the fraction bar if you know what i mean...
Yes, if you mean : \(\dfrac{ 3i (2-5i) }{ (2+5i)(2-5i) }\)
you're multiplying them right?
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right
for denominator you may use the identity: \((a+bi)(a-bi)=a^2+b^2\)
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
oh i see thank you
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