You can still find divisors for Gaussian integers. If x, y, and xy are all Gaussian integers, which will be trivially fulfilled for any x when y=1, then x, y both divide xy.
You can then generalize the \sigma function by summing over all the divisors of z and dividing by |z|.
The resulting number \sigma(z) lies in C (or maybe Q + iQ), not just Q, but it's perfectly well defined.
If you sum over all the divisors of z, the result is perfectly well defined; however, it's 0. Whenever x divides z, so does -x.
Over the integers, this is solved by summing over all positive divisors. However, there's no canonical choice of what divisors to consider positive in the case of Gaussian integers, and making various arbitrary choices (like summing over all divisors in the upper half-plane) leads to unsatisfying results.
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