In the ring Z[i], the units (divisors of 1) are +1, -1, +i, -i.  Thus if d is a divisor of 3+i, so are -d, id, -id.  In taking the sum of divisors, how do you make a canonical choice among the associates of each divisor?  If you take them all, the sum is zero.  In the rational integers, one chooses the positive divisor.   --  Gene

________________________________ From: Hans Havermann <pxp@rogers.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Wednesday, June 8, 2011 11:50 AM Subject: [math-fun] Some Sum

What is the sum of the aliquot (proper) divisors of 3+i?

a) The divisors are {1, 1+i, 1+2i, 3+i}. The sum of the first three numbers is 3+3i.

b) Robert Spira (1961) defined a "complex sum of divisors" that is incorporated into Mathematica's DivisorSigma function:

In[1]:= DivisorSigma[1,3+I] Out[1]= 2+6 I

Of course that answer includes the divisor 3+i, so we subtract that from 2+6i to get -1+5i.

c) Neither of the above. Spira's sum-of-divisors function is (I think) a product. So subtracting the 3+i after calculating the "sum" is a leap of faith.