else we use the range of angles method for 1 z 2 i z 1 z 1 If we use angle

# Else we use the range of angles method for 1 z 2 i z

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else, we use the range of angles method for 1 - z 2 = i z - 1 z + 1. If we use angle range (0 , 2 π ) for z - 1 we get a branch cut on [1 , ). Angle range (0 , 2 π ) Page 1
MATH 301, Homework 5 Do not hand in for z + 1 gives a branch cut on (1 , ). This function has no branch point at infinity, and the branch cuts on (1 , ) cancel. Checking the angles we find that this gives the negative square root on the upper lip, so we reverse the sign (or use angle range (2 π, 4 π ) for one of the two). In short, the branch is - i ( z - 1) 1 / 2 ( z + 1) 1 / 2 z 2 + 1 , where the square roots are both using range of angles (0 , 2 π ). (b) Consider the two contours C and Γ in the figure. C Γ If we connect the two contours we can get a simple contour that retraces the connection between them, and so the sum of the integrals is given by the residue theorem. I Γ f ( z ) dz + I C f ( z ) dz = 2 πi [Res( f ; i ) + Res( f ; - i )] , The residues are easy to compute, as the poles are simple: Res( f ; i ) = - i ( i - 1) 1 / 2 ( i + 1) 1 / 2 2 i = - i 2 , Res( f ; - i ) = - i ( - i - 1) 1 / 2 ( - i + 1) 1 / 2 - 2 i = - i 2 . (c) We compute the integral along the large circle by using the residue at infinity. The residue at infinity is lim z →∞ zf ( z ), which is - i , and so the integral around the large circle is 2 πi · ( - i ) = 2 π . (d) On the right small circle, if | z - 1 | = ε then | f ( z ) C ε , and so the integral on the circle is at most of order ε 3 / 2 , and tends to 0 as ε 0. The same applies to the left small circle. (e) On the upper lip we get I , and on the lower lip the function is the negative of the value on the upper lip, and the integral is in the opposite direction, so the integral is also I . Summing up in the limit

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• Spring '14
• Staff
• Methods of contour integration, Complex Plane, upper lip, Pole