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Simplify 1+i over 1-i I'll draw it in the answer section.
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|dw:1336515978944:dw|
\[\frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}\] multiply by the conjugate of the denominator
I did that and got a fraction and the answer is a whole number. I don't have the work with me anymore though
reason this works is that \[(a+bi)(a-bi)=a^2+b^2\] a real number so your denominator will be \[1^2+1^2=2\] your numerator is whatever you get when you multiply
numerators is \[(1+i)(1+i)=1+2i-1=2i\] so answer is \[\frac{2i}{2}=i\]
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