For example, in $\displaystyle{\frac{a+bi}{n+zi}}$, you would multiply both by the complex conjugate of the denominator, $n-zi$, to get rid of the complex number in the denominator. Wouldn't multiplying both by $i$ to get $i^2$ on the bottom and top get rid of the complex numbers?
Why do you multiply a complex number by its complex conjugate to get rid of it in a fraction?
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algebra-precalculus
complex-numbers
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4but then you would get $(n+zi)i=ni-z$, which is still non-real if $n\neq 0$, ad if $n=0$ then the conjugate is $-iz$ – 2011-04-22
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14I don't understand the question... *Have you actually tried multiplying by $i$ to see what happens?* One of the differences between physics and maths is that to carry out experiments we do not need to build multi-billion dollar particle accelerators to see what if our guess that multiplying by $i$ is enough! – 2011-04-22
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0@MarianoSuárez-Alvarez, I think you're misinterpreting the meaning of "wouldn't" in this context. I don't think Neal meant "wouldn't" as in "in the hypothetical situation where you multiplied by $i/i$ which I haven't bothered to try...", but rather "did I make a calculation error when I multiplied the top and bottom by $i$ and it made everything real (because that seems like a solution to me)?." (to which the answer is "yes [Dennis Gulko's comment]") – 2013-11-15