# Why can't there be complex numbers in the denominator?

Is this rule/convention something mathematicians simply agreed upon, like always rationalizing the denominator when it contains a radical expression, or is there a logical/mathematical reason for this?

My current understanding of math is only up to pre-calculus, so any explanation beyond that will most likely go over my head.

• It's just convention. And a convention that isn't really followed after high school math, in my experience. – Joe Aug 21 '18 at 21:10
• Who says there cannot be complex numbers in a denominator? – Henning Makholm Aug 21 '18 at 21:11
• I can't say I speak for all mathematicians, but a division by a real number is easy to visualize. A division by a nontrivial complex number is not. – GBQT Aug 21 '18 at 21:11
• Dividing a complicated expression by an integer is probably considered easier to grasp than dividing by a complicated expression, whatever complicated means in the respective context. In the end, it is a convention similar to the one to give fractions in lowest terms, e.g., $\frac12$ instead of $\frac 36$. Then again, we say $25\%$, which is literally $\frac{25}{100}$ instead of $\frac14$ ... – Hagen von Eitzen Aug 21 '18 at 21:12
• See for example $\frac{1}{2\pi i}$ in Wikipedia's statement of Cauchy's integral formula. Or Euler's formula $$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$$ – Henning Makholm Aug 21 '18 at 21:12

There can be complex non-real numbers in the denominator. It just happens that if you have an expression of the type$$\frac{a+bi}{c+di},$$with $d\neq0$, if you convert it to$$\frac{(a+bi)(c-di)}{(c+di)(c-di)}=\frac{ac+bd+i(bc-ad)}{c^2+d^2},$$you get, in general, a more readable expression.
But you are not required to do this. Complex Analysis has lots of formulas that begin with $\frac1{2\pi i}$ and, as far as I know, nobody ever suggested that one should use $-\frac i{2\pi}$ instead.
• Importantly, this computation shows that the quotient equals $\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i$, so the complex numbers are closed under division (except by $0$)! – Henning Makholm Aug 21 '18 at 21:18
• It's a small point but worth mentioning, for those that don't know that, given $c+di$, $c-di$ is referred to as the conjugate. So that the product of any complex number with its conjugate is real. The conjugate is indicated with a bar so if $z=c+di$ then $\overline{z}=c-di$. So $z.\overline{z}$ is always real. – Simon Terrington Aug 21 '18 at 21:22