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The Gaussian Möbius function on nonzero Gaussian integers is zero when a square of a Gaussian prime divides the argument, and otherwise equals $(-1{)}^r$, where $r$ is the number of nonassociate prime factors. It takes value one on units and is unchanged by multiplying its argument by a unit.

To sum over divisors without counting associates four times, one may define it on nonzero principal ideals, or choose one generator from each associate class. Then ${\sum }_{\mathfrak{d}\mid \mathfrak{a}}\mu (\mathfrak{d})$ is one for the unit ideal and zero otherwise: for a product of $r$ distinct prime ideals the sum is $(1-1{)}^r$. This gives Möbius inversion in Gaussian arithmetic.

A nonzero Gaussian integer is squarefree if no square of a Gaussian prime divides it. Equivalently, every exponent in its Gaussian prime factorization is at most one, disregarding multiplication by units.

The Gaussian integer $2$ is not squarefree, because $2=-i(1+i{)}^2$. By contrast $5=(2+i)(2-i)$ is squarefree: its factors are nonassociate Gaussian primes. Thus squarefreeness in $\mathbb{Z}[i]$ need not agree with squarefreeness in $\mathbb{Z}$. This distinction is essential when using a Gaussian Möbius function to exclude repeated prime factors.