5

I'm wondering if the concept of "divisibility" also works in complex numbers.

For example, if I say $a+bi$ is divisible by $c+di$. Does this mean that there exists another complex number $\alpha+\beta i$ such that $$\begin{align*} (c+di)(\alpha+\beta i)=a+bi \end{align*}$$If this is true, can I say that $5i$ is divisible by $5$ and is divisible by $i$ since $$\begin{align*} 5i=(5+0i)(0+i) \end{align*}$$

Solomon Ucko
  • 245
M_k
  • 1,855
  • 3
    This could make sense if we assume that all numbers appearing are integers , otherwise there is always such a complex number unless $c+di=0$ – Peter Feb 10 '23 at 16:33
  • 1
    The definition of divisibility clearly works in any monoid, and many divisibility properties are purely multiplicative and can be studied at the monoid level. – Bill Dubuque Feb 10 '23 at 16:48
  • 1
    For $~0 + i(0) \neq z = x + iy, ~x,y \in \Bbb{R},~$ let $~\overline{z}~$ denote $~x - iy,~$ and let $~|z|~$ denote $~\displaystyle \sqrt{x^2 + y^2}.~$ Then, at first glance, your divisibility question becomes trivial because, given any $~w \in \Bbb{C},~$ you have that $$\frac{w}{z} = \frac{w \times \overline{z}}{|z|^2} \implies w = \frac{w \times \overline{z}}{|z|^2} \times z.$$ – user2661923 Feb 10 '23 at 16:49
  • 1
    The method in the prior comment (viewable similarly to rationalizing the denominator) works for division by any algebraic number - as explained at lenght here. – Bill Dubuque Feb 10 '23 at 16:59

1 Answers1

11

You certainly can do this. You can do this in any situation where multiplication makes sense -- although you have to watch out that in other settings, divisibility might not work the way you're used to.

If you use this definition for all complex numbers, you get something very boring: every complex number $z$ is divisible by every non-zero complex number $a$ because $$ z = \frac za \times a. $$ (You get something equally boring if you use all real numbers.)

However, if you limit yourself to complex numbers $a + bi$ where $a, b$ are integers, you get the Gaussian integers, and there divisiblity is certainly interesting.

Mees de Vries
  • 26,947