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Divide and express the result in standard form: 9i ----- 6-8i
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multiply top and bottom by the conjugate of the denominator. the conjugate of \(a+bi\) is \(a-bi\) this works because \(a+bi)(a-bi)=a^2+b^2\) a real number start with \[\frac{9i}{6-8i}=\frac{9i}{6-8i}\times \frac{6+8i}{6+8i}\]
\[(36-27i)/-14\]
the denominator will be \(6^2+8^2=100\) and the numerator will be whatever you get when you multiply
be careful, because it is not the answer given above
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