# Why do you multiply a complex number by its complex conjugate to get rid of it in a fraction?

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?

-
but 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 $-iz$ –  Dennis Gulko Apr 22 '11 at 16:42
I 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! –  Mariano Suárez-Alvarez Apr 22 '11 at 17:55
@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]") –  Mark S. Nov 15 '13 at 2:48

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.
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 $\rm\ y\:z\ =\ s\:z'\:z\in \mathbb R\:.$
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 $r \exp(-i \theta) = \overline{z} = x - i y$.