# Get Analytic Functions of Several Complex Variables PDF

By Robert C. Gunning, Hugo Rossi

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Extra resources for Analytic Functions of Several Complex Variables

Example text

The condition of the theorem is by (1) precisely the condition y(x2 x(x2 + Y2 - 1) + y + 1) 2 = _Y:r - a which for y ¢ 0 reduces to (3). Under the conditions of Theorem 1 the force in the half-plane x > 0 due to the particles at +i and - i has a component toward the right, and in the halfplane x < 0 has a component toward the left. If we have a > 0, then on the arc of the circle (3) in the half-plane x > 0 the force is directed toward z = a and on the arc in the half-plane x < 0 is directed away from a.

For application of Theorem 1 we must determine the final locus of critical points of p(z) as a set of intervals whose initial points are the critical points of p(z) when all the Otk are as far to the left as possible and whose terminal points are the critical points of p(z) when all the Otk are as far to the right as possible. 3) to the case of circular regions rather than line intervals. 5). I; CoROLLARY 4. 1 (z - {j2)-k 2+1 (z - {j3)-kaH times the derimtive of (z - a 1)k 1 (z - {j 2)k 2 (z - {j3t 3• An interval I; disjo-int from the other interml I~ , from a 1 , and from the intervals I2 and I 3 contains precisely one critical point of p(z).

99 > 97. 2. Sufficient conditions for real critical points. +i and -i, 2k)/l;l] 112 , n zeros 2k)/k2] 112 , and with no other whose abscissas are less than or_equal to -[n(n zeros. :terior to all other Jensen circles for p(z). :n the unit circle C. THEORE~I 2. -)( 2 • 34 CHAPTER II. REAL POLYNOMIALS In either of these two cases we make the convention that only one zero of p'(z) shall be assigned to the interval I. Henceforth we exclude these degenerate cases. The force at the point [m/(m 2k)] 112 is directed toward the right, for otherwise the force at that point due to the particles at +i and -i is less than the force at that point due to the m particles, all coincident at [m(m 2k)/k2] 112 : + + m 2k ( m m + 2k ) 112 + (m + 2~~) 112 < (m(m + 2k)) 1' 2 m k m 2 _ ( m m + 2k ) 1' 2 ' < m, which is absurd.