APPM 4360/5360 Exam #2 Solutions Spring 2015
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1 APPM 436/536 Exam # Solutios Sprig 5 O the frot of your bluebook, write your ame ad make a gradig table. You re allowed oe sheet (letter-sized, frot ad back of otes. You are ot allowed to use textbooks, class otes, or a calculator. Problem # ( poits:fid ad classify all the sigular poits i the exteded complex plae for the followig fuctios: ( (a z si (b ez +z z z z z 4 (z + (c cos z si z si z (4z + π (d z(z π Solutio: (a ( z si z ( z z 3!z 3 + 5!z 5... z 3!z + 5!z 3 7!z It follows z is a essetial sigularity. The poit at ifiity is a simple pole of stregth. (b Sice the umerator ad deomiator are etire fuctios, the oly sigular poits are zeros of the deomiator, z ad z. About z e z +z z z z 4 (z + + z + z + (z + z / + z z z 4 (z + z z 4 + z +..., where... deote higher-degree terms i z, so z is a secod-order pole. About z deote z + t ad expad about t : e z +z z z z 4 (z + + t( + t + t ( + t / + t( + t ( + t 4 t t 6t + t , which meas that z is removable sigularity. It is clear from the series expasio z is a essetial sigularity. (c The oly sigular poit is z π/4 but ad cos z si z (cos z cos(π/4 si z si(π/4 cos(z + π/4 cos z si z (cos(z + π/4 (4z + π 6(z + π/4 (z + π/4 / (z + π/ , i.e. the sigularity is removable. It is clear from the series expasio z is a essetial sigularity.
2 (d Here the sigularities are at z ad z ±π. The umerator has zeros at z π for N ad so these sigularities are both removable. The poit at ifiity is a essetial sigularity of si z si z ad so is a essetial sigularity of z(z π. This is because it is a removable sigularity for the the deomiator. That is, t 3 ( t π t π t ad so t is a removable sigularity of the deomiator. Problem # ( poits: For which z does the series f (z coverge? Coverge uiformly? Justify your aswers. e z Solutio: This is a geometric series so it coverges whe e z <, i.e. e (x+i y e x y +i x y e x y <. This implies x y < i.e. a coe y > x aroud the i y-axis betwee the lies y x ad y x. Uiform covergece is achieved whe e x y r <, for ay such r, because the the series satisfies Weierstrass M-test: e z r r. Problem #3 ( poits: Fid the first four terms of the Lauret expasio about z π of the fuctio f (z z + az si (z b, where a ad b are some complex umbers. What are a ad b if the Lauret series about z π is i fact the Taylor series? Solutio: Let z π + u ad expad f (z about u ; si(z b si(π b + u si(π bcosu + cos(π bsiu si(π b( u / + u 4 /4!... + cos(π b(u u 3 /3! + u 5 /5!..., the further expasio depeds o whether b π, Z, or ot. If b π, the f (z π + aπ + (π + au + u ( π (u u 3 /3! + u 5 /5!... + aπ u + π + a + ( + u /3! u 4 /5! (u /3! +... u ( π + aπ u + π + a + + (π + aπ + u 3! (π + a 3! u Thus, it caot be a Taylor series whe b π, Z sice takig a π or a π kills oe egative power but ot both.
3 If b π, the si(π b ad f (z ( π + aπ (π + au si + (π b si (π b + u π + aπ + (π + au + u (si(π b( u / + u 4 /4! + cot(π b(u u 3 /3! + u 5 /5!... si (π b π + aπ si (π b + ( cot(π bu +( cot (π bu +4cot (π bu +... ( (π + aπcot(π b si + π + a (π b si (π b ( π + aπ + si (π b ( + (π + acot(π b 3cot (π b si + (π b which is always a Taylor series. Problem #4 ( poits: Compute f (zdz for z u+ si (π b u +..., (a f (z ez z 3, (b f (z z + z(z 4(z /4, (c f (z cosh(z z + z(6z 8z +, Solutio: (a For g (z e z, which is aalytic i the fiite complex plae, we use Cauchy s derivative formula for the pole at z. That is, g (z πi z 3 dz g ( (! z Computig the derivatives of g we have g (z ze z ad g (z ( + 4z e z ad so g (. Therefore, e z dz πi z3 z (b The sigular poits z ad z /4 are iside the cotour ad z 4 is outside. Therefore f (zdz πi ( Res(f ; + Res(f ;/4. Sice z is a simple pole, z Res(f ; lim z z f (z 4. For the double pole z /4, Res(f ;/4 d dz ((z /4 f (z d ( z + z/4 dz z(z 4 z/4 (z 4 z 4 z (z 4 z/ Thus, z ( f (zdz πi
4 (c The umerator is etire, the deomiator equals z(4z so the oly sigular poits iside are z ad z /4. cosh(z z + for z or z /4 so z is simple pole while z /4 is double pole. The Res(f ; lim z f (z cosh, z Res(f ;/4 d dz ((z /4 f (z d cosh(z z + z/4 dz 6z z/4 ( (z sih(z z + cosh(z z + 6z 6z sih(3/6 cosh(3/6, z/4 8 f (zdz πi ( Res(f ; + Res(f ;/4 ( πi cosh sih(3/6 cosh(3/6. 4 z Problem #5 ( poits: Let f (z be the fuctio (a Compute the residue of f (z at z. ez f (z z 3 z (b Compute the residue of f (z at z. (c Compute the residue of f (z at z. (d Compute the value of the itegral z 3 f (zdz Solutio: The fuctio f (z has sigularities at z, ad is aalytic everywhere else. (a Usig the formula from homework ad the book: Oe could also use Taylor expasios to see d Res(f (z; lim z dz (z f (z f (z e z z z ( z z z 3... z z ( e z d lim z dz z (z e z e z lim z (z ( z + z +! +... (b Here we have a simple pole at z ad ote that so the residue is e. e z lim(z f (z lim z z z e 4
5 (c The fuctio is aalytic at z so the residue is. (d Sice both the poles are eclosed i the circle z 3 we use the residue theorem to show that f (zdz πi ( + e z 3 Extra-Credit Problem #6 ( poits: Show that for π x π ( si(x x. You may wat to use Demoivre s theorem ad the fact that log( + z ( Solutio: Recall DeMoivre s Theorem: e iz cos(z + i si(z ad the Taylor expasio of log( + z, ( log( + z z. Puttig these two ideas together we have That is, log( + e i x ( eix ( cos(x + i ( si(x Im( log( + e i x Sice log( + e i x l + e i x + i Arg( + e i x we obtai ( si(x ( si(x Arg( + e i x ( ta si(x + cos(x ( si(x/cos(x/ ta cos (x/ ( si(x/ ta cos(x/ x This is true oly for π < x < π due to the brachig from the arcta fuctio. This fuctio is the sawtooth fuctio ad the sum is its fourier series. The plots below show the sum for,5 ad 5. The last plot zooms i at the discotiuity. What is iterestig to ote is that the oscillatios aroud the discotiuity do ot disappear as we icrease the umber of terms. This is because the series does ot coverge uiformly. I fact, As N icreases the amplitude of the oscillatios will always be about % of the jump across the discotiuity. 5 z
6 .5.5 First terms of sawtooth series.5 - First 5 terms of sawtooth series x (a First terms (b First 5 terms.5.5 First 5 terms of sawtooth series.5 - First 5 terms of sawtooth series aroud x π x x (a First 5 terms (b First 5 terms aroud the discotiuity x π However, the regio i which they appear go as O(N. The fact that these oscillatios appear ear the discotiuities of the sum of a fourier series is kow as Gibbs pheomeo. 6
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