3.z沿正负实轴趋于0,f( Sin 极限不同,故z=0是本性奇点。 (4分 留数只能通过 Laurent展开计算,难求。 利用有限远与无穷远留数和为0,可求z=∞处的留数。 (4分) 作变量代换:(=1/z,f(z)=<sine, 只需求g()=f(2)2在(=0处的留数。 (4分) Res [g(0)= sin l, #x: Res [ f(oo) in l 分 ap: Res [f(O)]=sin l,I=2Tisin 1 (2分) 4.作1()√如的割线:连接z=0与z=∞的直线。上岸:argz=0(3分) 因为limf(2)=0,故:/f(2)dz=0 (2分 因为imf(x)=0,故:/f()dz=0 (2分) )=2143/=271+Rm=列 (3分) Res f(a)=T(1+i) (3分) Res(-1)=3V2n-) (3分) I (2分) 4 5.f(2)=|2l2-1-ci+(1-p)2 1=arg, 02=arg(a-1) (3分) 在上岸:61=610,62=620,x0=1/2+i0+, f(20)= 故:610-62 6 (3分) imf(2)=0z=∞为可去奇点 (3分) lim zf(z)= lim =lime(p-1)(1-2) (3分) 注意在上岸:610=2r,620=丌,610-620=丌 z→∞时,1-62=2r (3分 故 e(p-1)2, Res [f(oo)l (3分)3. z ÷K¢¶ªu 0§f(z) = sin e 1/z z 4ØÓ§z = 0 ´5Û:" (4©) 3êUÏL Laurent ÐmOŽ§J¦" |^k†Ã¡3êڏ 0§Œ¦ z = ∞ ?3ê" (4©) ŠCþ†µζ = 1/z§f(z) = ζ sin e ζ§ I¦ g(ζ) = f(z)/ζ2 3 ζ = 0 ?3ê" (4©) Res [g(0)] = sin 1§µRes [f(∞)] = − sin 1§ (4©) =µRes [f(0)] = sin 1§I = 2πisin 1 (2©) 4. Š f(z) = √ z ln z z 2 + 1 ‚µë z = 0 † z = ∞ †‚"þWµarg z = 0 (3©) Ϗ limz→∞ zf(z) = 0, µ Z CR f(z)dz = 0 (2©) Ϗ limz→0 zf(z) = 0, µ Z Cε f(z)dz = 0 (2©) I f(z)dz = 2I + 2πi Z √ x x 2 + 1 dx = 2πi[Res f(i) + Res f(−i)] (3©) Res f(i) = √ 2 8 π(1 + i) (3©) Res f(−i) = 3 √ 2 8 π(1 − i) (3©) I = √ 2 4 π 2 (2©) 5. f(z) = |z| p |z − 1| 1−p e i[pθ1+(1−p)θ2] z 2 − 1 , θ1 = arg z, θ2 = arg(z − 1) (3©) 3þWµθ1 = θ10, θ2 = θ20, z0 = 1/2 + i0 +, f(z0) = − 2 3 e i[pθ10+(1−p)θ20] = 2 3 e ipπ µθ10 − θ20 = π, θ20 = π (3©) limz→∞ f(z) = 0 z = ∞ ŒÛ: (3©) a−1 = limz→∞ zf(z) = limz→∞ z p−1 (z − 1)1−p 1 − 1 z 2 = limz→∞ e i(p−1)(θ1−θ2) (3©) 5¿3þWµθ10 = 2π, θ20 = π, θ10 − θ20 = π z → ∞ ž§θ1 − θ2 = 2π (3©) µa−1 = e i(p−1)2π , Res [f(∞)] = −a−1 = −e i2pπ (3©)