Modulo of complex numbers

E

erroneum

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Jul 6, 2012
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Hello,
I am familiar with the modulo (division remainder) operator on real numbers, but I am needing to implement (in software) modulo on complex numbers, and am wondering if there is a general, universally accepted method for getting the same thing, and if so, what it is.
 
Nonzero complex numbers always divide evenly into each other. By "evenly" of course I mean for any complex x and y (y not 0) there always exists a unique complex number z such that zy=x. This is true of any field. So the remainder is always zero...

If you mean the Gaussian Integers, you'll first need to define your norm. Typically |x+iy| = sqrt(x^2+y^2). Then if a+bi and x+iy are Gaussian Integers, then there always exists unique r,q in Z such that (a+bi) = (x+yi)*q + r, where 0<= |r| < |x+yi|. See here for the division algorithm: http://fermatslasttheorem.blogspot.com/2005/06/division-algorithm-for-gaussian.html
 
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Interesting. I went to WolframAlpha, and entered "(6+4i)mod(3+5i)" and was given the answer "-2+2i." Assuming what you said to be true, I don't know how that answer was generated, but it is more like what I am looking for than 0 for any complex number.

To explain why I ask this, I am writing a calculator. It uses postfix notation, and operates on arbitrary precision complex numbers. One of the operations that is defined is modulus, and I would like it to work for all inputs, rather than just all real inputs. Since WolframAlpha does not tell me how it arrived at it's result, I can only hope that someone here can.
 
I think the problem is that the MOD operation is only defined in Integer field. If you consider all the real numbers - then there is no residue. Similar situation is in complex field. Now if you restrict the field such that (a,b) are integers only, then you'll get modulo result.
 
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I have seen mod extended to support all real numbers, and WolframAlpha will compute it for all complex numbers as well. As I have seen it in software, the modulus of non-integer real numbers a and b is something like a-floor(a/b)*b.

Actually, now that I look at the extended list of alternate representations, it appears that WolframAlpha uses the method that I saw in software.
 
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