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HW5hint - Hints to Complex Analysis Homework 5 October 3...

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Hints to Complex Analysis Homework 5 October 3, 2011 Chapter 5 (in Greene & Krantz’s book) 8. You want to compare the zeros of f and g inside D ( P, r ), so, you want | f - g | < | f | on ∂D ( P, r ), obviously, | f - g | < min z ∂D ( P,r ) | f ( z ) | will do it (here since f has no zeros on ∂D ( P, r ), min z ∂D ( P,r ) | f ( z ) | > 0). So here the meaning g is uniformly suffciently close to f on ∂D ( P, r ) means we require that, on ∂D ( P, r ), | f - g | < min z ∂D ( P,r ) | f ( z ) | . 10 (b) f ( z ) = z 3 - 3 z 2 + 2 on D (0 , 1). Solution Method 1 (not very pretty) : Observe that z = 1 is a solution, so we can write f ( z ) = ( z - 1)( z 2 - 2 z - 2), and we only need to consider the zero of z 2 - 2 z - 2 on D (0 , 1). by the quadaritic formula, z 2 - 2 z - 2 = 0 has two roots 1 + 3 and 1 - 3, and 1 - 3 is inside D (0 , 1). So f has only one zero in D (0 , 1) Method 2 (Use Rouche’s theorem) : Note that f has at most three zeros on D (0 , 1), so we can take r (with 1 > r > 3 - 1 such that all possible three zeros are contained in D (0 , r ). let g ( z ) = - z 2 + 2 z - 2 (Observe that z = 1 is a solution, so we can write f ( z ) = ( z - 1)( z 2 - 2 z - 2)). Then, for z
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