6.3 參考解答

  1. (a) f(z)=z2z2+1f(z)=\frac{z}{2z^2+1} analytic everywhere except at z=±i2z=\pm \frac{i}{\sqrt{2}}.

    (2z2+1=0⇒z2=−12⇒z=±i2)(2z^2+1=0 \Rightarrow z^2=-\frac{1}{2} \Rightarrow z=\pm \frac{i}{\sqrt{2}})

    ∫C1+(0)z2z2+1 dz=∫C1+(0)14z−i2 dz+∫C1+(0)14z+i2 dz\int_{C_1^+(0)} \frac{z}{2z^2+1}\space dz=\int_{C_1^+(0)} \frac{\frac{1}{4}}{z-\frac{i}{\sqrt{2}}}\space dz+\int_{C_1^+(0)} \frac{\frac{1}{4}}{z+\frac{i}{\sqrt{2}}}\space dz

    Both ±i2\pm \frac{i}{\sqrt{2}} lie inside C1+(0)C_1^+(0).

    ∴∫C1+(0)14z−i2 dz+∫C1+(0)14z+i2 dz=14(2πi)+14(2πi)=πi\therefore \int_{C_1^+(0)} \frac{\frac{1}{4}}{z-\frac{i}{\sqrt{2}}}\space dz+\int_{C_1^+(0)} \frac{\frac{1}{4}}{z+\frac{i}{\sqrt{2}}}\space dz=\frac{1}{4}(2\pi i)+\frac{1}{4}(2\pi i)=\pi i

3. 4z2−4z+5=0⇒z=4±16−808=4±−648=4±8i8=12±i4z^2-4z+5=0 \Rightarrow z=\frac{4\pm \sqrt{16-80}}{8}=\frac{4\pm \sqrt{-64}}{8}=\frac{4\pm 8i}{8}=\frac{1}{2}\pm i

Both 12±i\frac{1}{2}\pm i lie outside C1+(0)C_1^+(0), the function (4z2−4z+5)−1(4z^2-4z+5)^{-1} is analytic inside C1(0)C_1(0),

so ∫C1+(0)(4z2−4z+5)−1 dz=0\int_{C_1^+(0)} (4z^2-4z+5)^{-1}\space dz=0.

  1. (a) ∫C1z2−z dz=∫C1z(z−1) dz=∫C−1z dz+∫C1z−1 dz\int_C \frac{1}{z^2-z}\space dz=\int_C \frac{1}{z(z-1)}\space dz=\int_C \frac{-1}{z}\space dz+\int_C \frac{1}{z-1}\space dz

    =2πi+2πi=2\pi i+2\pi i

    =4πi=4\pi i

  1. (a) ∫C2z−1z2−z dz=∫C2z−1z(z−1) dz=∫C1z dz+∫C1z−1 dz\int_C \frac{2z-1}{z^2-z}\space dz=\int_C \frac{2z-1}{z(z-1)}\space dz=\int_C \frac{1}{z}\space dz+\int_C \frac{1}{z-1}\space dz

    =∫02π1z dz+∫02π1z−1 dz=\int_{0}^{2\pi} \frac{1}{z}\space dz+\int_{0}^{2\pi} \frac{1}{z-1}\space dz

    =2πi+2πi=2\pi i+2\pi i

    =4πi=4\pi i