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Complex numbers

Working with complex numbers reduces to being able to compute the signature of i, the imaginary unit. Sn(i) should satisfy the basic relation of Sn(i)2+1=0. This means that $S_n(i) = \sqrt{-1} \bmod n$. If n = 4k+1 is a prime, then -1is a non-residue, and we can find its square root. If n=n1 n2 ... is composite, then -1 should be a non-residue for all ni. As a consequence n has two possible forms, n=4k+1 or n=8k+2. Once that we have chosen n so that Sn(i) has a representation, we still have two choices (plus and minus). Any choice works provided that we use it consistently. This choice is not yet a problem, but will become one once that we introduce algebraic numbers in general. This will be discussed later.

Suppose we want to test whether (x+i)(x-i)-x2-1 is identically 0. We can work with n=13, where S(i)=5 is a possible choice. Selecting S(x)=2 as before we obtain

\begin{displaymath}7 \times (-3) - 4 - 1 = -26 = 0 \pmod{13} \end{displaymath}

When we are forced to compute signatures in a field or congruence which does not allow the representation of i, we will use field extensions. This will be described for the signatures of ex and trigonometric functions.


next up previous
Next: Square roots of integers Up: Modular mappings Previous: Arbitrary powers
Gaston Gonnet
1999-07-04