PureMath.pdf

# Similar to that used above that am1 ρ is unchanged

• 587

This preview shows pages 565–568. Sign up to view the full content.

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 ,

This preview has intentionally blurred sections. Sign up to view the full version.

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.
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.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.
• Fall '14

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern