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3+2i/1-2i I know I'm lame, I ended up with the answer -1+8i/5 I know that can't be right, can someone please help
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idea is to multiply top and bottom by the conjugate of the denominator the conjugate of \(a+bi\) is \(a-bi\) so the conjugate of \(1-2i\) is \(1+2i\)
this works because \((a+bi)(a-bi)=a^2+b^2\) a real number in your case \((1+2i)(1-2i)=1^2+2^2=1+4=5\) that will be your denominator
the work looks like \[\frac{3+2i}{1-2i}\times \frac{1+2i}{1+2i}=\frac{(3+2i)(1+2i)}{5}\] so now it is left to multiply out up top
ok so u end up with a numerator of 3+8i+4i^2?
so you end up with a real number answer of -9/5?
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