1 z i 5 at i 3 z 12 z 1 10 at 1 solution 1 we see

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1 ( z + i ) 5 at - i 3. z 12 ( z - 1) 10 at 1 Solution. 1. We see that the function f ( z ) = z + 1 z sin (1 /z ) has a pole of order 2 at z = 0 . We need to prove that lim z 0 z 2 f ( z ) = 0 Indeed, we have z 2 z + 1 z sin (1 /z ) ≤ | z ( z + 1) | We know that lim z 0 z ( z + 1) = 0 , so we obtain the proof. 2. We see that the function has a pole of order 6 at z = - i . Indeed, we have lim z →- i ( z + i ) 6 1 ( z + i ) 5 = lim z →- i ( z + i ) = 0 3. In a similar fashion, the function has a pole of order 11 at z = 1 . We see this as follows 178
6 RESIDUES AND THEIR USE IN INTEGRATION lim z 1 ( z - 1) 11 z 12 ( z - 1) 10 = lim z 1 z 12 ( z - 1) = 0 Exercise 6.9. (E. 27, E. 31 P. 358) Use residues to evaluate the following integrals. Use the principal branch of multivalued func- tions. 1. dz sin z around | z - 6 | = 4 2. e 1 /z z 2 - 1 dz around | z - 1 | = 3 / 2 Solution. 179
6 RESIDUES AND THEIR USE IN INTEGRATION 6.4 EVALUATION OF REAL INTEGRALS WITH RESIDUE CALCULUS I The reader can see exercises 3, 4, 7, 8 for examples of residue calculus I. Exercise 6.10. (E. 3, E. 4 P. 364) Using residue calculus, establish the following identities. 1. ˆ π - π a + b cos θ = 2 π a 2 - b 2 for a > b 0 . 2. ˆ 3 π 2 - π 2 cos θ a + b cos θ = 2 π b 1 - a a 2 - b 2 for a > b > 0 . Solution. 1. We put t = π + θ and see that ˆ π - π a + b cos θ = ˆ 2 π 0 a - b cos θ The next is the denominator is definited because of a > b 0 . Then we apply the method of evaluation of real integrals with residue calculus 1. We get z = e , so that = dz iz cos θ = z + z - 1 2 Then the integral becomes ˆ 2 π 0 a - b cos θ = | z | =1 2 dz 2 iaz - ibz 2 - ib We need to face with the equation ibz 2 - 2 iaz + ib = 0 and have 2 roots easily. z = a ± a 2 - b 2 b But the only point, z = a - a 2 - b 2 b , is in the unit circle, the reader can verify themselves. Now we only compute the residue at this point. ˆ 2 π 0 2 2 iaz - ibz 2 - ib = 4 πiRes 1 2 iaz - ibz 2 - ib ; a - a 2 - b 2 b ! = 2 π a 2 - b 2 Note that, we only get the result in the case a > b > 0 , but do not worry, it is very easy to get the proof for the cases a > b = 0 . 180
6 RESIDUES AND THEIR USE IN INTEGRATION 2. Like the exercise above, we put t = θ + π 2 and see that ˆ 3 π 2 - π 2 cos θ a + b cos θ = ˆ 2 π 0 sin θ a + b sin θ The next is the denominator is definited because of a > b > 0 . Then we apply the method of evaluation of real integrals with residue calculus 1. We get z = e , so that = dz iz sin θ = z - z - 1 2 i Then the integral becomes ˆ 2 π 0 sin θ a + b sin θ = | z | =1 ( z 2 - 1 ) dz iz ( bz 2 + 2 iaz - b ) We need to face with the equation bz 2 + 2 iaz - b = 0 and have 2 roots easily. z = - ia ± i a 2 - b 2 b But the only point, z = - ia + i a 2 - b 2 b , is in the unit circle, the reader can verify them- selves. Now we only compute the residue at this point and the point z = 0 . ˆ 2 π 0 sin θ a + b sin θ = 2 πi " Res z 2 - 1 bz 2 + 2 iaz - b ; - ia + i a 2 - b 2 b ! + Res z 2 - 1 bz 2 + 2 iaz - b ; 0 # = 2 π b 1 - a a 2 - b 2 Exercise 6.11. (E. 7, E. 8, E. 9 P. 364) Using residue calculus, establish the following identities. 1. ˆ 2 π 0 ( a + b sin θ ) 2 = 2 πa a 2 - b 2 3 for a > b 0 . 2. ˆ 2 π 0 a + sin 2 θ = 2 π p a ( a + 1) for a > 0 .

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