Midterm Examination #2

Transcription

1 Anthony Austin MATH 47 April, 9 AMDG Midterm Examination #. The integrand has poles of order whenever 4, which occurs when is equal to i, i,, or. Since a R, a >, the only one of these poles that lies inside the circle a a is the pole at, as the following illustration makes apparent: The residue of the integrand at is Res 4 lim( 4 ] ( + i( i( + ( + i( i( + 4. Therefore, by the Residue Theorem a a 4 d πi 4 i π.. Since b lies inside the contour, by auchy s Integral Formula for the derivatives, e ] πi d d ( b! d e ] ] d πi d e + e ] πi e + e ] b πie b (b +. b b

2 . The integrand has a pole of order at and a pole of order at. The residues of the integrand corresponding to each of these poles are Res e e ( lim ( e ] ( e ( and Res e ( d (! lim d e ]] ( d lim d e ] ] d lim e + e ]] d lim e + e + e e ] ( e + e + e e e. There are then four cases, depending on the status of the points and with respect to the contour. (For simplicity, we assume that the winding number of about each pole is. For full generality, each residue will need to be multiplied by the winding number corresponding to its associated pole. ase : Both and Inside : In this case, e d πi ( ( e. ase : Inside, but Outside : Now, e d πi. ( ase : Inside, but Outside : Here, e d πie. ( ase 4: Both and Outside : In this case, the integrand is actually analytic in the region bounded by, and so

3 e ( d. 4. Set f( + (. This function has poles of order at and. The corresponding residues are: Res f( lim f( ] + +. and Res f( lim( f( ] tan sin cos has poles wherever cos. From Problem # on Ahlfors, pg. 44 (worked in Homework #, we know that cos cos R( cosh I( i sin R( sinh I(. Thus, cos will be ero if and only if cos R( cosh I( and sin R( sinh I( are both simultaneously ero. cosh I( is never ero, so we must have cos R(, but since sine and cosine have no common eros, this implies that sin R(. Hence, cos if and only if cos R( and sinh I(, but sinh I( if and only if I(. Hence, the eroes of the complex cosine function are the same as those of the real cosine function, i.e., cos if and only if (k + π for some integer k. Thus, tan has poles at (k + π, where k is any integer. The corresponding residues are Res (k+ π tan lim ( (k + π (k+ π tan lim (k+ π lim (k+ π sin ( (k + π sin ( (k + π. ( (k + π sin cos ( (k + π cos + sin sin (L Hopital s Rule ( The expression cos has an essential isolated singularity at. We recall the Taylor expansion

4 cos! + 4 4!, and so for, ( cos!( + 4!( 4. Viewing this as an expansion of the form ( cos n A n ( n, i.e., a Laurent expansion about, we see that the coefficient A. Hence, ( 6. The expression cos we have ( Res cos. has an essential isolated singularity at. From the preceding problem, ( cos!( + 4!( 4. for. Furthermore, we can Taylor expand the function f( about to obtain ] + d d ] ( +! d d 8 + ( ( + (6 ( + 6( ( + 6( + (. ] ( + d! d ] ( Therefore, ( cos!( + 4!( ( + 6( + ( 8 + ( + 6( + ( ( + 6( + ( 4!( 4.!( From this, we see that the term with a factor of ( is

5 ! ( + 4! ( 4 4 (. Thus, ( Res cos For the first integral, the integrand has an essential isolated singularity enclosed by the contour at but is analytic everywhere else in the plane. We recall the Taylor expansion sin! + 5 5!. Hence, sin (! + (5! 5. The coefficient of the term is. Hence, Res sin. The Residue Theorem therefore implies (r sin ( d πi πi. For the second integral, observe that cos (! + (4 4! + 4. Hence, sin cos 4 4 +, and so

6 As there is no term, we conclude that sin Res sin, whence (r sin ( d. 8. Since x x + y (x + y, the circle may be described by the set of all points satisfying. The integrand has poles wherever 4 +, or wherever ( 4. Explicitly, the integrand has four poles of order : p + i p i p i p 4 + i. Of these, only p and p lie inside the circle. The corresponding residues are Res p 4 + (p p (p p (p p 4 i ( + i i ( + i Res p 4 + (p p (p p (p p 4 i ( i i ( i

7 Therefore, d 4 + ( πi i ( + i i ( i π ( ( i ( + i ( + i( i iπ. 9. We have π e cos ϕ cos(nϕ sin ϕdϕ π π π π e cos ϕ ( e i(nϕ sin ϕ + e i(nϕ sin ϕ dϕ. e cos ϕ e inϕ e i sin ϕ dϕ + e cos ϕ i sin ϕ e inϕ dϕ + e e iϕ ( e iϕ n dϕ + π π π e cos ϕ e inϕ e i sin ϕ dϕ e cos ϕ+i sin ϕ e inϕ dϕ e eiϕ ( e iϕ n dϕ. Let e iϕ. Then, d i dϕ, and the above becomes π e cos ϕ cos(nϕ sin ϕdϕ i e n d + i e n d. Recall that e + +! +! +. Hence, e + + (! + (! +, and so we have n e n + n + n+! + n+! + as well as n e n + n + n! + n 4! +.

8 There are now two cases, depending on the value of n. ase : n : In this case, both n e and n e series expansions from above, we find that have an isolated singularity at. Using the Res n e n! and Hence, Res n e n!. π e cos ϕ cos(nϕ sin ϕdϕ i πi n! + i πi n! π n!. ase : n < : Now, only n e has an isolated singularity at, while n e is analytic on the whole plane. Looking at the series expansion for n e, we see that if n <, there is no term, and so the residue at is ero! Hence, π e cos ϕ cos(nϕ sin ϕdϕ.