6.2 參考解答

  1. (a) C1:z1(t)=2eitC_1: z_1(t)=2e^{it}, 0≤t≤π20 \leq t \leq \frac{\pi}{2}

    C2:z2(t)=−t+i(2−t)C_2: z_2(t)=-t+i(2-t), 0≤t≤20 \leq t \leq 2

7. (e) f(z)=z+1f(z)=z+1, C1+(0):z=eitC_1^+(0): z=e^{it}

∫Cf(z) dz=∫0π2(eit+1)⋅ieit dt\int_C f(z)\space dz=\int_{0}^{\frac{\pi}{2}} (e^{it}+1)\cdot ie^{it}\space dt

=∫0π2iei2t+ieit dt=\int_{0}^{\frac{\pi}{2}} ie^{i2t}+ie^{it}\space dt 

=12ei2t+ieit ∣0π2=\frac{1}{2}e^{i2t}+ie^{it}\space |_{0}^{\frac{\pi}{2}}

=(12eiπ+eiπ2)−(12e0+e0)=(\frac{1}{2}e^{i\pi}+e^{i\frac{\pi}{2}})-(\frac{1}{2}e^0+e^0)

=(−12+i)−(12+1)=(-\frac{1}{2}+i)-(\frac{1}{2}+1)

=−12+i−12−1=-\frac{1}{2}+i-\frac{1}{2}-1

=i−2=i-2

  1. (a) ∫CR+(z0)1z−z0 dz,z=z0+Reit\int_{C_R^+(z_0)} \frac{1}{z-z_0}\space dz, z=z_0+Re^{it}

    =∫02π1z0+reit−z0(iReit) dt=\int_{0}^{2\pi} \frac{1}{z_0+re^{it}-z_0}(iRe^{it})\space dt

    =∫02πi dt=\int_{0}^{2\pi} i\space dt

    =it ∣02π=it \space |_{0}^{2\pi}

    =2πi=2\pi i

12. z2=(t+it2)2=t2+i2t3−t4z^2=(t+it^2)^2=t^2+i2t^3-t^4, z’(t)=1+i2tz’(t)=1+i2t

∣z2∣=∣(t2−t4)+i2t3∣=(t2−t4)2+(2t3)2=t4−2t6+t8+4t6|z^2|=|(t^2-t^4)+i2t^3|=\sqrt{(t^2-t^4)^2+(2t^3)^2}=\sqrt{t^4-2t^6+t^8+4t^6}

=t4+2t6+t8=(t2+t4)2=t2+t4=\sqrt{t^4+2t^6+t^8}=\sqrt{(t^2+t^4)^2}=t^2+t^4

∴∫C∣z2∣ dz=∫01∣(z(t))2∣⋅(z’(t)) dt\therefore \int_C |z^2|\space dz=\int_{0}^{1} |(z(t))^2| \cdot (z’(t))\space dt

=∫01(t2+t4)⋅(1+i2t) dt=\int_{0}^{1} (t^2+t^4) \cdot (1+i2t) \space dt

=∫01t2+i2t3+t4+i2t5 dt=\int_{0}^{1} t^2+i2t^3+t^4+i2t^5 \space dt

=∫01(t2+t4)+i(2t3+2t5) dt=\int_{0}^{1} (t^2+t^4)+i(2t^3+2t^5) \space dt

=(13t3+15t5)+i(12t4+13t6) ∣01=(\frac{1}{3}t^3+\frac{1}{5}t^5)+i(\frac{1}{2}t^4+\frac{1}{3}t^6) \space |_{0}^{1}

=(13+15)+i(12+13)−0=(\frac{1}{3}+\frac{1}{5})+i(\frac{1}{2}+\frac{1}{3})-0

=815+i56=\frac{8}{15}+i\frac{5}{6}