# Is there a complex variant of Möbius' function?

When you're dealing with arithmetic functions, you might have come across the classical Möbius' function

$$\mu(n)=\begin{cases} (-1)^{\omega(n)}=(-1)^{\Omega(n)} &\mbox{if }\; \omega(n) = \Omega(n)\\ 0&\mbox{if }\;\omega(n) < \Omega(n).\end{cases},$$ where $ω(n)$ is the number of distinct primes dividing the number $n$ and $Ω(n)$ is the number of prime factors of $n$, counted with multiplicities.

Is there a complex analogon $\mu^{\Bbb C}(z)$ that additionally takes into account that primes of the form $z=4n+1$ might be factored as well, e.g. $5=(1+2i)(1-2i)$?

So a number $z_n$ that contains a prime of that form would give

1. $\mu^{\Bbb C}(z_n)=0$, when think of e.g. $5=(1+2i)(1-2i)$ as a square
2. or $\mu^{\Bbb C}(z_n)=1$ when think of it as product of two Gaussian primes.

A more general question, but I'm not sure how this is related, is: How does the concept of factoring natural numbers carry over to complex natural numbers?

Anything to read on that topic would be nice...

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$5$ is the product of two distinct Gaussian primes. It would seem to me that $\mu^{\mathbb{C}}(5)=1$ (instead of $-1$ as with $\mu$). – robjohn Jan 28 '13 at 23:31
@robjohn I thought of it more like a square kind of thing, but ... ok – draks ... Jan 28 '13 at 23:33
One interesting property is that if $n>1$ and $\mu^{\mathbb{C}}(n)=1$, then $n$ is the sum of two non-zero squares. – robjohn Jan 28 '13 at 23:36
@robjohn the motivation for this came from the comments to Raymond's (great) answer... – draks ... Jan 28 '13 at 23:48
I think it would be better to call it a number field variant, rather than a complex variant. This doesn't make sense for transcendental elements, after all. – anon Feb 9 '13 at 21:17

There is slight trouble with choice of unit, using ideals instead of Gaussian primes as factors resolves this.

As always the Möbius function encodes the inclusion-exclusion principle.

To invert $$F(\mathfrak n) = (f*1)(\mathfrak n) = \sum_{\mathfrak d|\mathfrak n} f(\mathfrak d)$$ we will use $$\mu(\mathfrak p_1^{r_1} \mathfrak p_2^{r_2} \cdots \mathfrak p_k^{r_k}) = \begin{cases} (-1)^{k} \; \text{ when each r_i=1} \\ 0 \; \text{ otherwise}\end{cases}.$$

Then simplifying a bit $$(F*\mu)(\mathfrak p_1^{r_1} \mathfrak p_2^{r_2} \cdots \mathfrak p_k^{r_k}) = \sum_{\epsilon = \{0,1\}^k} (-1)^{\epsilon_1 + \epsilon_2 + \cdots + \epsilon_r} F(\mathfrak p_1^{r_1-\epsilon_1} \mathfrak p_2^{r_2-\epsilon_2} \cdots \mathfrak p_k^{r_k-\epsilon_k}) = f(\mathfrak p_1^{r_1} \mathfrak p_2^{r_2} \cdots \mathfrak p_k^{r_k}).$$

It may be worth checking that we get the same correspondence with Dirichlet series/zeta functions as in the natural numbers case.

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+1 sounds promising, the motivation for this came from the comments to Raymond's (great) answer... – draks ... Jan 28 '13 at 23:49
@draks..., I would try to put it to use on some simple things (like chapter 1 of Apostol) first to see if it's the correct generalization but I can't do that now. – user58512 Jan 28 '13 at 23:51
Forgot to say something about multiplicativity.. that needs checked more than anything else. – user58512 Jan 28 '13 at 23:52
Need to compute the complex Eulers totient. – user58512 Jan 29 '13 at 13:32
$\displaystyle\varphi(z) =z \prod_{\mathfrak p \mid z} \left(1-\frac{1}{\mathfrak p}\right)$, with $\mathfrak p$ being all (Gaussian) primes, why not? – draks ... Jan 29 '13 at 16:25