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E. 10), then we arrive at the same answer. The general case can be handled in the same way. 1 with R(z) equal to z1k . 1) we nd the coe cients j of the polynomial R(z1). 2). 2) all variables except z1 . A MODIFIED ELIMINATION METHOD 43 The principal di culty of this method is the determination of the matrix A. We now discuss some ways of calculating A. 1. The general method for nding A is the method of undetermined coe cients. Since we have an estimate of the degree Nj , we can write down ajk (z) with unknown coe cients, then perform arithmetic operations, equating the coe cients at equal degrees of z, and hence get a system (in general under determined) of linear equations.

We assume that a0 6 0 and b0 6 0. Consider the resultant R(P Q) of these polynomials (in the variable z). It will be some polynomial R(w). 1 R( ) = 0. e. 5) has the solution ( )) or is a zero of the polynomials a0(w) and b0 (w). 5). 2. Consider the system P (z w) = wz2 + 3wz + 2w + 3 = 0 Q(z w) = 2zw ; 2z + 2w + 3 = 0: First we re-write P and Q in the form P = w z2 + 3w z + (2w + 3) Q = 2(w ; 1) z + (2w + 3): Then we get w 3w 2w + 3 0 R(P Q) = 2(w ; 1) 2w + 3 = 0 2(w ; 1) 2w + 3 w 3w w+3 + = (2w + 3) 2(w0; 1) 2(2w 2(w ; 1) 2w + 3 = ; 1) = (2w + 3)(4w2 ; 8w + 4 + 2w2 + 3w ; 6w2 + 6w) = = (2w + 3)(w + 4): 3 The zeros of R(w) are w1 = ; 2 and w2 = ;4.

Hence by the one-dimensional theorem on logarithmic residues (see Sec. 2) Z ;3z2 + 4abz3 Z 1 1 ;3 + 4abz2 2 2 J=2 i dz 2= 4 3 2 2 i jz2j="4 (a2 bz2 ; 1)z2 dz2 = 3 jz2 j="4 a bz2 ; z2 because the root z2 = 1=(a2b) does not lie in the disk jz2j < "4 for su ciently small " > 0. 6. The classical scheme for elimination of unknowns The results presented in this section one can nd in 97, 142]. First we consider a classical result of Sylvester. Take two polynomials: P (z) = a0zn + a1 zn;1 + : : : + an Q(z) = b0 zs + b1 zs;1 + : : : + bs z 2 C, aj , bj 2 C and a0 6= 0 b0 6= 0.

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