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**Unformatted text preview: **MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 6 SOLUTIONS 1. Prove the Fundamental Theorem of Algebra using Cauchy’s theorem as follows. Suppose we have a polynomial with complex coefficients p ( z ) = ∑ n k =0 a k z k that has no zeros in C . (a) Consider the polynomial q ( z ) = ∑ n k =0 a k z k whose coefficients are conjugates of the corre- sponding coefficients of p ( z ). Show that f ( z ) = 1 p ( z ) q ( z ) is an entire function. Solution. All we need to show is that if p ( z ) has no zeroes in C , then neither does q ( z ). This is obvious since if α ∈ C is such that q ( α ) = 0, then taking conjugate, we get 0 = q ( α ) = n X k =0 a k α k = p ( α ) and so α is a zero of p ( z ), contradicting our assumption. It follows that the product p ( z ) q ( z ) is always non-zero on C and therefore its reciprocal is an entire function. (b) Let R > 0. Apply Cauchy’s theorem to the line integral along the closed curve L in Figure 1b to show that Z R- R dx | p ( x ) | 2 + i Z π Re iθ p ( Re iθ ) q ( Re iθ ) dθ = 0 . Figure 1. L is a closed curve in the counter clockwise direction along the semicircle { Re it | ≤ t ≤ π } and the line segment { z | - R ≤ Re z ≤ R, Im z = 0 } . Solution. Since f is analytic, Cauchy’s theorem yields Z L dz p ( z ) q ( z ) = 0 . Date : November 14, 2009 (Version 1.0). 1 Let L be the line segment L 1 , parameterized as z : [- R,R ] → C , z ( t ) = t , and the semi- circular arc L 2 , parameterized as z : [0 ,π ] → C , z ( t ) = Re it . Then Z L dz p ( z ) q ( z ) = Z L 1 dz p ( z ) q ( z ) + Z L 2 dz p ( z ) q ( z ) = Z R- R dt | p ( t ) | 2 + i Z π Re it p ( Re...

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