Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer

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 $\rm\ y\:z\ =\ s\:z'\:z\in \mathbb R\:.$

share|cite|improve this answer

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 $r \exp(-i \theta) = \overline{z} = x - i y$.

share|cite|improve this answer
I don't agree that this the "best" way to think about it because the polar form only works for complex numbers, but in fact the "conjugation" idea is both simpler and and much more general - it works in many rings (domains) - see my proof. – Bill Dubuque Apr 22 '11 at 18:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.