609S09Ex2solns

# Then by the cauchy integral formula applied to the

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of these disks. Then by the Cauchy Integral Formula applied to the larger disk | ζ - a | < r/ 2, for any point z in this smaller disk f n ( z ) = 1 2 πi | ζ - a | <r/ 2 f n ( ζ ) ( ζ - z ) 2 But since | ζ - z | > r/ 2 and the f n ’s converge uniformly, we can pass limit under the integral and conclude that the f n s converge uniformly in this disk: f n ( z ) 1 2 πi | ζ - a | <r/ 2 f ( ζ ) ( ζ - z ) 2 = f ( z ) . Because this finite collection of smaller disks cover K , we conclude that the convergence is uniform in K . B6. Find a conformal map from the unbounded region outside the disks {| z +1 | ≤ 1 }∪{| z - 1 | ≤ 1 } to the upper half plane. Solution Step 1: Writing w = u + iv , the map w = 1 /z maps this region to the vertical strip - 1 / 2 < u < 1 / 2. [In greater detail, it maps the real axis to itself, the imaginary axis to itself, the origin to infinity, and hence the circles | z ± 1 | = 1 to the vertical straight lines u = ± 1 / 2.] Step 2: By a translation, rotation, and stretching we can map this strip to the horiontal strip -∞ < s < , 0 < t < π in the ζ = s + it plane. Step 3: Then e ζ maps this strip to the upper half-plane. B7. Consider the family of polynomials p ( z ; t ) = z n + a n - 1 ( t ) z n - 1 + · · · a 1 ( t ) z + a 0 ( t ) , where the coefficients a j ( t ) depend continuously on the parameter t [0 , 1]. Assume that at t = 0 the polynomial p ( z ; 0) has k zeroes (counted with their multiplicity) in the disk | z - c | < R and has no zeroes on the circle | z - c | = R . Show that for all sufficiently small t the polynomial p ( z ; t ) also has k zeroes in | z - c | < R . Solution Write p ( z ; t ) = p ( z ; 0) + [ p ( z ; t ) - p ( z ; 0)] .
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• Fall '09
• Math, Cauchy Integral Formula, Short Answer Problems, Entire function, Jerry L. Kazdan

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