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State the equivalent form of
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\[\frac{2}{3+i}\]
\[a) \frac{3-i}{4}\] \[b) \frac{3-i}{5}\] \[c)\frac{4-i}{4}\] \[d) \frac{4-i}{5}\]
I said the answer was a but I got it wrong. It says that it suppose to be b can anyone explain?
\[\frac{2}{3+i} = \frac{2}{3+i} \frac{3-i}{3-i} = \frac{2(3-i)}{3^2 -i^2} = \frac{2(3-i)}{9-(-1)} = \frac{3-i}{5}\]
because i is a radical... you cannot have a radical in the denominator so you need to rationalize the denominator so that the radical goes away from the denom..... so you multiply everything times its conjugate and simplify
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