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?
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$\begingroup$For any complex number $z$, multiplying by the conjugate always gives a nonnegative real number: $$(a+bi)(a-bi) = a^2+b^2.$$ While sometimes you can multiply a complex number by some other complex number to get a real (e.g., you can multiply a purely imaginary number by $i$), the conjugate always works.
$\endgroup$ $\begingroup$Precisely the real multiples of the conjugate suffice to rationalize a denominator $\rm\:z\in \mathbb C\:.\:$ Proof: if $\rm\ z\ne0\ $ and $\rm\ y\:z\ =\ r \in \mathbb R\ $ then $\rm\: y\:z\:z'\: =\ r\:z'\: $ so $\rm\ y\ =\ z'\:r/(z\:z')\ =\ s\:z',\ \ s\: =\ r/(z\:z')\in \mathbb R\:.\:$ Conversely, if $\rm\ y\ =\ s\:z'\:,\ s\in\mathbb R\ $ then $\rm\ y\:z\ =\ s\:z'\:z\in \mathbb R\:.$
$\endgroup$ $\begingroup$This is best thought of in terms of the polar representation of a complex number. Let $z = r \exp (i \theta)$. What do we have to multiply by to turn this into a real number? $\exp (- i \theta)$, of course. Any real multiple of it also works, and we have one readily at hand. Given $z = x + i y = r \exp(i \theta)$, we can write $r \exp(-i \theta) = \overline{z} = x - i y$.
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