Similar to that used above that am1 ρ is unchanged

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similar to that used above, that am(1 + ρ ) is unchanged as z describes γ in the positive sense, while am z n on the other hand is increased by 2 . Hence am Z is increased by 2 , and the proof that Z = 0 has a root is completed. We have assumed throughout the argument that neither Γ, nor any of the smaller contours into which it is resolved, passes through the origin. This assumption is obviously legitimate, for to suppose the contrary, at any stage of the argument, is to admit the truth of the theorem. We leave it as an exercise to the reader to infer, from the discussion which precedes and that of § 43 , that when z describes any contour γ in the positive sense the increment of am Z is 2 , where k is the number of roots of Z = 0 inside γ , multiple roots being counted multiply. There is another proof, proceeding on different lines, which is often given. It depends, however, on an extension to functions of two or more variables of the results of §§ 102 et seq. We define, precisely on the lines of § 102 , the upper and lower bounds of a function f ( x, y ), for all pairs of values of x and y corresponding to any point of any region in the plane of ( x, y ) bounded by a closed curve. And we can prove, much as in § 102 , that a continuous function f ( x, y ) attains its upper and lower bounds in any such region. Now | Z | = | P ( x + iy ) | is a positive and continuous function of x and y . If m is its lower bound for points on and inside γ , then there must be a point z 0 for which | Z | = m ,
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APPENDIX I 550 and this must be the least value assumed by | Z | . If m = 0, then P ( z 0 ) = 0, and we have proved what we want. We may therefore suppose that m > 0. The point z 0 must lie either inside or on the boundary of γ : but if γ is a circle whose centre is the origin, and whose radius R is large enough, then the last hypothesis is untenable, since | P ( z ) | → ∞ as | z | → ∞ . We may therefore suppose that z 0 lies inside γ . If we put z = z 0 + ζ , and rearrange P ( z ) according to powers of ζ , we obtain P ( z ) = P ( z 0 ) + A 1 ζ + A 2 ζ 2 + · · · + A n ζ n , say. Let A k be the first of the coefficients which does not vanish, and let | A k | = μ , | ζ | = ρ . We can choose ρ so small that | A k +1 | ρ + | A k +2 | ρ 2 + · · · + | A n | ρ n - k < 1 2 μ. Then | P ( z ) - P ( z 0 ) - A k ζ k | < 1 2 μρ k , and | P ( z ) | < | P ( z 0 + A k ζ k | + 1 2 μρ k . Now suppose that z moves round the circle whose centre is z 0 and radius ρ . Then P ( z 0 ) + A k ζ k moves k times round the circle whose centre is P ( z 0 ) and radius | A k ζ k | = μρ k , and passes k times through the point in which this circle is intersected by the line joining P ( z 0 ) to the origin. Hence there are k points on the circle described by z at which | P ( z 0 ) + A k ζ k | = | P ( z 0 ) | - μρ k and so | P ( z ) | < | P ( z 0 ) | - μρ k + 1 2 μρ k = m - 1 2 μρ k < m ; and this contradicts the hypothesis that m is the lower bound of | P ( z ) | . It follows that m must be zero and that P ( z 0 ) = 0.
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APPENDIX I 551 EXAMPLES ON APPENDIX I 1. Show that the number of roots of f ( z ) = 0 which lie within a closed contour which does not pass through any root is equal to the increment of { log f ( z ) } / 2 πi when z describes the contour.
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