Futioetheorie, SS 204 Solutios Übugsblatt 2 Aufgabe : Es sei g eie meromorphe Futio auf C mit höchstes eifache Pole. Wir ehme a, dass das Residuum a jedem Pol vo g eie gae Zahl ist. Zeige Sie: a) Es existiert eie meromorphe Futio f 0 auf C, so dass f /f = g. b) Falls h 0 eie weitere meromorphe Futio mit h /h = g ist, da ist h/f ostat. Solutio: a) Sice g is meromorphic with at most simple poles, ad sice the residues at these poles are itegers, the primitive g)d is defied modulo 2πiZ. This maes f) = exp g)d) well-defied o the etire C. Moreover, f is holomorph ad f) 0 for all C. b) Let w 0, w C so that f0) = e w0 ad h0) = e w. Defie ad L f ad L h are holomorphic. Moreover f ζ) L f ) = w 0 + 0 fζ) dζ h ζ) L h ) = w + 0 hζ) dζ. expl f )) = f) ad expl h )) = h). Sice f /f = g = h /h, the differece L f L h is costat; it is i fact equal to w 0 w. Therefore f)/h) = expl f ) L h )) = expw 0 w ) for all C.
Futioetheorie, SS 204 2 Aufgabe 2: Ma defiiert die Beroulli-Zahle {B } N durch die Potereihe Zeige Sie: e = B!. ) =0 a) 2 Pute) Es gilt B 0!! + B )!! +... + B! )! = falls = 0 falls >. Hiweise: Betrachte Sie die Potereihe-Etwiclug vo beute Sie Propositio.7 2). e e b) Put) B = 0 falls > ud ugerade ist, ud B Q für alle N. c) Put) Für alle N \ {0} gerade gilt es 2πi) 2ζ) = B.! Hiweise: Sete Sie i ) = 2πiw ud vergleiche Sie die Koeffiiete. ud Solutio: a) We have Write e = B! =0 e e =. ad Multiplyig these two series together we obtai m=0 e = =.! B0 m!! + B m )!! +... + B ) m m =.! m )! From here the desired formula follows. b) From the formula i a) for = 0, it follows B 0 =. By iductio o, we coclude that B Q for all N. Note that B = 2. To show that B 2+ = 0 for all, it is eough to show that the fuctio f) = e + 2 is a eve fuctio. This is true, sice f ) = e 2 = The series expasio for f) is e e 2 = + f) = 2 B!. e 2 = f).
Futioetheorie, SS 204 3 Sice f ) = 2 ) B! ad f) eve, we must have B = ) B. This forces B = 0 for all odd, >. c) We have 2πiw e 2πiw = πiw 2 e 2πiw = πw cotπw) i) = πw cotπw) iπw. From Beispiel 5.5. i Sript, we have πw cotπw) = πw πw + 2πw πw) 2 π) 2 w 2 = + 2 w 2 2 = 2 w 2 w = 2 w w ) 2 2 = 2 0 0 = 2 2 w 2 = 2 ζ2)w 2 ) 2 2+2 w 2+2 O the other had, usig the power expasio for have πw cotπw) = iπw + 2πiw e 2πiw = iπw + B! 2πiw) 0 = iπw + 0 B! 2πi) w = iπw + B 0 + B 2πi) + 2 = iπw + iπw + 2πiw e 2πiw give by formula ), we B! 2πi) w B 2 2)! 2πi)2 w 2 give that B 0 = 0, B = 2, ad B 2+ = 0 for all. Comparig the coefficiets of the two power series, we obtai the desired formula.
Futioetheorie, SS 204 4 Aufgabe 3: Zeige Sie das lette Gleichheitseiche i Sat 5.22 im Sript):! Γ) = lim + )... + ) Solutio: Use the first equality i Sat 5.22 i Sript: Sice by defiitio we have Γ) = lim eγ γ = lim N = + ) e / N j log N j= Γ) = lim e N j= j e log N N = + ) e / We ca thi of this limit as the limit of a double sequece a,n as ad N. Sice it coverges, the limit is the same as the limit of the subsequece obtaied for = N. Therefore Γ) = lim e j= j e log = + ) e / = lim e j= j e log e = / log = lim e log = lim e = = + + = lim + )... + ).! Multiplyig the above by we obtai the desired formula. = +
Futioetheorie, SS 204 5 Aufgabe 4: Zeige Sie: a) + x 6 dx = 2π 3 ud b) π/ dx = 2 + x si π/ für alle gerade N \ {0}. Hiweise: Nehme Sie de Weg vo 0 ach R, da vo R ach Re 2πi/, ud da urüc ach 0. Oder beute Sie de allgemeie Sat 3.25. Solutio: We are goig to use Folgerug 3.25 i Sript to compute these itegrals. The fuctio Rx) = +x is a ratioal fuctio which does ot have ay sigularities alog the real axis. Sice 2, the order of the ero at is also 2. Therefore we are i the hypothesis of Folgerug 3.25. Cosider R) = + as a fuctio o C. Its sigularities are at the poits where + = 0. These are the poits From Folgerug 3.25 we ow ζ = e πi 2πi ) e = e 2 )πi with. + x dx = 2πi Im ζ >0 Sice ζ is a simple eroe for +, Propositio 3.2 2) gives Res ζ R) 2) Res ζ R) = ζ = ζ. 3) The ζ with Im ζ > 0 correspod to those with argumet i 0, π). This meas 0 < 2 π < π which is equivalet to 2 < < +. 2 Give that is eve, this correspods to 2 4) a) I this case = 6. Therefore, i formula 2) we oly eed to cosider those that satisfy < 3. This meas ζ, ζ 2, ad ζ 3. We have ζ = e πi/6 = cosπ/6) + i siπ/6) = 3/2 + i/2, ζ 2 = e πi/2 = i, ad ζ 3 = e 5πi/6 = 3/2 + i/2.
Futioetheorie, SS 204 6 With this dx = 2πi + x6 6 ζ 6 ζ 2 ) 6 ζ 3 = 3 πi ζ + ζ 2 + ζ 3 ) = 3 πi2i) = 2 3 π. b) I the geeral case = 2m, the roots of + = 0 that have strictly positive imagiary part are ζ,..., ζ m. From formulas 2) ad 3) it follows dx = 2πi + x m = ζ Oe way to compute m = ζ is to observe that ζ = ζ ζ 2 ). m ζ = ζ = m ζ 2 ) = ζ 2m = ζ ζ 2 = 2ζ ζ 2 = 2 = 2 2i si π/ = i ζ ζ + si π/. sice ad ζ π ) π ) = cos + i si = cos π + i si π ζ + π + ) π + ) = cos + i si = cos π i si π. Therefore π/ dx = 2 + x si π/.