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M2sols

# M2sols - Math 185 Solutions to Midterm II 1(a The...

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Math 185 – Solutions to Midterm II 1. (a) The polynomial z 4 + 1 factors as a product ( z - a 1 ) · · · ( z - a 4 ) where a 1 , . . . , a 4 are its four roots. These are easy to find in polar coordinates since, if z = re ,we have z 4 = - 1 iff r 4 e 4 = e so r = 1 and 4 θ = π + 2 for k Z . As a result the four roots are: e i π 4 , e i 3 π 4 , e i 5 π 4 , e i 7 π 4 and the polynomial can be factored as z 4 + 1 = ( z - e i π 4 )( z - e i 3 π 4 )( z - e i 5 π 4 )( z - e i 7 π 4 ) (b) The distance between any two of the above roots is at least 2 so the circles C 1 and C 2 contain only the roots e i π 4 and e i 3 π 4 respectively. Now we will study the integrals of 1 z 4 +1 around C 1 and around C 2 independently. Define g ( z ) = 1 ( z - e i 3 π 4 )( z - e i 5 π 4 )( z - e i 7 π 4 ) . This is an analytic function at all points where the denominator does not vanish so in particular it is analytic in C 1 and its interior. As a result, we can apply the Cauchy integral formula and obtain C 1 1 z 4 + 1 dz = C 1 g ( z ) ( z - e i π 4 ) dz = 2 πig ( e i π 4 ) Similarly we can define h ( z ) = 1 ( z - e i π 4 )( z - e i 5 π 4 )( z - e i 7 π 4 ) and using the same reasoning as in the last bullet conclude that C 2 1 z 4 + 1 dz = C 2 h ( z ) ( z - e i 3 π 4 ) dz = 2 πih ( e i 3 π 4 ) 2. (a) The only points where the function 1 z 4 +1 is not analytic lie in the unit circle. As a result,

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M2sols - Math 185 Solutions to Midterm II 1(a The...

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