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2-3i/1-2i
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conjugate
\[\frac{2-3i}{1-2i}\times \frac{1+2i}{1+2i}\] \[\frac{(2-3i)(1+2i)}{1^2+2^2}\] then multiply in the numerator
(2-3i/1-2i)*(1+2i/1+2i)
\[\frac{8+i}{5}=\frac{8}{5}+\frac{1}{5}i\]
got that far then once I multiplied it out I'm coming up with 2+9i/1+4i when the book says 8/5+1/5i
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how did you get the 8
I am coming up with 2+7i/1+4i where did the 8 come from?
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