Chapter 18

\(f(z)=z^{3}-1+3 i\) is a polynomial and so is an entire function.

(a) Choosing the line \(y=\frac{1}{3} x, 0 \leq x \leq 3\) we have \(z=x+\frac{1}{3} x i, d z=\left(1+\frac{1}{3} i\right) d x .\) Thus \\[ \int_{C} e^{z} d z=\int_{0}^{3} e^{\left(1+\frac{1}{3} i\right) x}\left(1+\frac{1}{3} i\right) d x=\left.e^{\left(1+\frac{1}{3} i\right) x}\right|_{0} ^{3}=e^{3+i}-e^{0}=\left(e^{3} \cos 1-1\right)+i e^{3} \sin 1 \\] (b) \(\int_{C} e^{z} d z=\int_{0}^{3+i} e^{z} d z=\left.e^{z}\right|_{0} ^{3+i}=e^{3+i}-e^{0}=\left(e^{3} \cos 1-1\right)+i e^{3} \sin 1\)

\(f(z)=\frac{z}{2 z+3}\) is discontinuous at \(z=-3 / 2\) but is analytic within and on the circle \(|z|=1\)

\(f(z)=\frac{\sin z}{\left(z^{2}-25\right)\left(z^{2}+9\right)}\) is discontinuous at \(z=\pm 5\) and at \(z=\pm 3 i\) but is analytic within and on the circle \(|z|=1\)

Using \(z=e^{i t},-\pi / 2 \leq t \leq \pi / 2,\) and \(d z=i e^{i t} d t, \quad \int_{C} \frac{1+z}{z} d z=-\int_{-\pi / 2}^{\pi / 2}\left(1+e^{i t}\right) d t=(2+\pi) i\).

$$\int_{-2 i}^{1}\left(3 z^{2}-4 z+5 i\right) d z=z^{3}-2 z^{2}+\left.5 i z\right|_{-2 i} ^{1}=-19-3 i$$

(a) By Theorem 18.9 with \(f(z)=\frac{z^{2}}{z+2 i}\) \\[ \oint_{C} \frac{\frac{z^{2}}{z+2 i}}{\frac{z+2 i}{z-2 i}} d z=2 \pi i\left(-\frac{4}{4 i}\right)=-2 \pi \\] (b) By Theorem 18.9 with \(f(z)=\frac{z^{2}}{z-2 i}\) \\[ \oint_{C} \frac{\frac{z^{2}}{z-2 i}}{z-(-2 i)} d z=2 \pi i\left(\frac{-4}{-4 i}\right)=2 \pi \\]

$$\int_{1-i}^{1+i} z^{3} d z=\left.\frac{1}{4} z^{4}\right|_{1-i} ^{1+i}=0$$

\(f(z)=\tan z\) is discontinuous at \(z=\pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}, \ldots\) but is analytic within and on the circle \(|z|=1\)

Using \(z=e^{i t}=\cos t+i \sin t, d z=(-\sin t+i \cos t) d t\) and \(x=\cos t\), \(\begin{aligned} \oint_{C} \operatorname{Re}(z) d z &=\int_{0}^{2 \pi} \cos t(-\sin t+i \cos t) d t=-\int_{0}^{2 \pi} \sin t \cos t d t+i \int_{0}^{2 \pi} \cos ^{2} t d t \\ &=-\frac{1}{2} \int_{0}^{2 \pi} \sin 2 t d t+\frac{1}{2} i \int_{0}^{2 \pi}(1+\cos 2 t) d t=\pi i \end{aligned}\).

\(f(z)=\frac{z^{2}-9}{\cosh z}\) is discontinuous at \(\frac{\pi}{2} i, \pm \frac{3 \pi}{2} i, \ldots\) but is analytic within and on the circle \(|z|=1\)

Using \(z+i=e^{i t}, 0 \leq t \leq 2 \pi,\) and \(d z=i e^{i t} d t\), \(\oint_{C}\left[\frac{1}{(z+i)^{3}}-\frac{5}{z+i}+8\right] d z=i \int_{0}^{2 \pi}\left[e^{-2 i t}-5+8 e^{i t}\right] d t=-10 \pi i\).

$$\int_{-i / 2}^{1-i}(2 z+1)^{2} d z=\left.\frac{1}{6}(2 z+1)^{3}\right|_{-i / 2} ^{1-i}=-\frac{7}{6}-\frac{22}{3} i$$

Using \(y=-x+1,0 \leq x \leq 1, z=x+(-x+1) i, d z=(1-i) d x\), \(\int_{C}\left(x^{2}+i y^{3}\right) d z=(1-i) \int_{1}^{0}\left[x^{2}+(1-x)^{3} i\right] d x=-\frac{7}{12}+\frac{1}{12} i\).

Using \(z=e^{i t}, \pi \leq t \leq 2 \pi, d z=i e^{i t} d t, x=\cos t=\left(e^{i t}+e^{-i t}\right) / 2, y=\sin t=\left(e^{i t}-e^{-i t}\right) / 2 i\), \(\begin{aligned} \int_{C}\left(x^{3}-i y^{3}\right) d z &=\frac{1}{8} i \int_{\pi}^{2 \pi}\left(e^{3 i t}+3 e^{i t}+3 e^{-i t}+e^{-3 i t}\right) e^{i t} d t+\frac{1}{8} i \int_{\pi}^{2 \pi}\left(e^{3 i t}-3 e^{i t}+3 e^{-i t}-e^{-3 i t}\right) e^{i t} d t \\ &=\frac{1}{8} i \int_{\pi}^{2 \pi}\left(2 e^{4 i t}+6\right) d t=\frac{3 \pi}{4} i \end{aligned}\).

$$\int_{i / 2}^{i} e^{\pi z} d z=\left.\frac{1}{\pi} e^{\pi z}\right|_{i / 2} ^{i}=-\frac{1}{\pi}-\frac{1}{\pi} i$$

\(\int_{C} e^{z} d z=\int_{C_{1}} e^{z} d z+\int_{C_{2}} e^{z} d z\) where \(C_{1}\) and \(C_{2}\) are the line segments \(y=0,0 \leq x \leq 2\) and \(y=-\pi x+2 \pi\), \(1 \leq x \leq 2,\) respectively. Now \(\int_{C_{1}} e^{z} d z=\int_{0}^{2} e^{x} d x=e^{2}-1\) \(\int_{C_{2}} e^{z} d z=(1-\pi i) \int_{2}^{1} e^{x+(-\pi x+2 \pi) i} d x=(1-\pi i) \int_{2}^{1} e^{(1-\pi i) x} d x=e^{1-\pi i}-e^{2(1-\pi i)}=-e-e^{2}\). In the second integral we have used the fact that \(e^{z}\) has period \(2 \pi i .\) Thus \(\int_{C} e^{z} d z=\left(e^{2}-1\right)+\left(-e-e^{2}\right)=-1-e\).

\(\int_{C} \sin z d z=\int_{C_{1}} \sin z d z+\int_{C_{2}} \sin z d z\) where \(C_{1}\) and \(C_{2}\) are the line segments \(y=0,0 \leq x \leq 1,\) and \(x=1\), \(0 \leq y \leq 1,\) respectively. Now \(\int_{C_{1}} \sin z d z=\int_{0}^{1} \sin x d x=1-\cos 1\) \(\int_{C_{2}} \sin z d z=i \int_{0}^{1} \sin (1+i y) d y=\cos 1-\cos (1+i)\). Thus \(\int_{C} \sin z d z=(1-\cos 1)+(\cos 1-\cos (1+i))=1-\cos (1+i)=(1-\cos 1 \cosh 1)+i \sin 1 \sinh 1=0.1663+0.9889 i\).

By Theorem 18.10 with \(f(z)=\cos 2 z, f^{\prime}(z)=-2 \sin 2 z, f^{\prime \prime}(z)=-4 \cos 2 z, f^{\prime \prime \prime}(z)=8 \sin 2 z, f^{(4)}(z)=16 \cos 2 z\) \\[ \oint_{C} \frac{\cos 2 z}{z^{5}} d z=\frac{2 \pi i}{4 !}(16 \cos 0)=\frac{4 \pi}{3} i \\]

$$\int_{\pi}^{\pi+2 i} \sin \frac{z}{2} d z=-\left.2 \cos \frac{z}{2}\right|_{\pi} ^{\pi+2 i}=-2\left[\cos \left(\frac{\pi}{2}+i\right)-\cos \frac{\pi}{2}\right]=2 i \sin \frac{\pi}{2} \sinh 1=2.3504 i$$

We have \(\quad \int_{C} \operatorname{Im}(z-i) d z=\int_{C_{1}}(y-1) d z+\int_{C_{2}}(y-1) d z\) On \(C_{1}, z=e^{i t}, 0 \leq t \leq \pi / 2, d z=i e^{i t} d t, y=\sin t=\left(e^{i t}-e^{-i t}\right) / 2 i\), \(\int_{C_{1}}=(y-1) d z=\frac{1}{2} \int_{0}^{\pi / 2}\left[e^{i t}-e^{-i t}-2 i\right] e^{i t} d t=\frac{1}{2} \int_{0}^{\pi / 2}\left[e^{2 i t}-1+2 i e^{i t}\right] d t=1-\frac{\pi}{4}-\frac{1}{2} i\).On \(C_{2}, y=x+1,-1 \leq x \leq 0, z=x+(x+1) i, d z=(1+i) d x\), \(\int_{C_{2}}(y-1) d z=(1+i) \int_{0}^{-1} x d x=\frac{1}{2}+\frac{1}{2} i\). Thus \(\int_{C} \operatorname{Im}(z-i) d z=\left(1-\frac{\pi}{4}-\frac{1}{2} i\right)+\left(\frac{1}{2}+\frac{1}{2} i\right)=\frac{3}{2}-\frac{\pi}{4}\).

$$\begin{aligned} \int_{1-2 i}^{\pi i} \cos z d z &=\left.\sin z\right|_{1-2 i} ^{\pi i}=\sin \pi i-\sin (1-2 i)=i \sinh \pi-[\sinh 1 \cosh 2-i \cos 1 \sinh 2] \\ &=-\sin 1 \cosh 2+i(\sinh \pi+\cos 1 \sinh 2)=-3.1658+13.5083 i \end{aligned}$$

Using \(x=6 \cos t, y=2 \sin t, \pi / 2 \leq t \leq 3 \pi / 2, z=6 \cos t+2 i \sin t, d z=(-6 \sin t+2 i \cos t) d t\), \(\int_{C} d z=-6 \int_{\pi / 2}^{3 \pi / 2} \sin t d t+2 i \int_{\pi / 2}^{3 \pi / 2} \cos t d t=2 i(-2)=-4 i\).

(a) By Theorem 18.9 with \(f(z)=\frac{2 z+5}{z-2}\) $$\oint_{C} \frac{\frac{2 z+5}{z-2}}{z} d z=2 \pi i\left(-\frac{5}{2}\right)=-5 \pi i$$ (b) since the circle \(|z-(-1)|=2\) encloses only \(z=0\), the value of the integral is the same as in part (a). (c) From Theorem 18.9 with \(f(z)=\frac{2 z+5}{z}\) $$\oint_{C} \frac{\frac{2 z+5}{z}}{\frac{z}{z-2}} d z=2 \pi i\left(\frac{9}{2}\right)=9 \pi i$$ (d) since the circle \(|z-(-2 i)|=1\) encloses neither \(z=0\) nor \(z=2\) it follows from the Cauchy-Goursat Theorem, Theorem \(18.4,\) that $$\oint_{C} \frac{2 z+5}{z(z-2)} d z=0$$

We have \(\oint_{C} z e^{z} d z=\int_{C_{1}} z e^{z} d z+\int_{C_{2}} z e^{z} d z+\int_{C_{3}} z e^{z} d z+\int_{C_{4}} z e^{z} d z\) \(\mathrm{On} C_{1}, y=0,0 \leq x \leq 1, z=x, d z=d x\), \(\int_{C_{1}} z e^{z} d z=\int_{0}^{1} x e^{x} d x=x e^{x}-\left.e^{x}\right|_{0} ^{1}=1\). On \(C_{2}, x=1,0 \leq y \leq 1, z=1+i y, d z=i d y\), \(\int_{C_{2}} z e^{z} d z=i \int_{0}^{1}(1+i y) e^{1+i y} d y=i e^{i+1}\). On \(C_{3}, y=1,0 \leq x \leq 1, z=x+i, d z=d x\), \(\int_{C_{3}} z e^{z} d z=\int_{1}^{0}(x+i) e^{x+i} d x=(i-1) e^{i}-i e^{1+i}\). On \(C_{4}, x=0,0 \leq y \leq 1, z=i y, d z=i d y\), \(\int_{C_{4}} z e^{z} d z=-\int_{1}^{0} y e^{i y} d y=(1-i) e^{i}-1\). Thus \(\oint_{C} z e^{z} d z=1+i e^{i+1}+(i-1) e^{i}-i e^{1+i}+(1-i) e^{i}-1=0\).

By partial fractions, \(\oint_{C} \frac{2 z+1}{z(z+1)} d z=\oint_{C} \frac{1}{z} d z+\oint_{C} \frac{1}{z+1} d z\) (a) By Theorem 18.4 and (4) of Section 18.2 $$\oint_{C} \frac{1}{z} d z+\oint_{C} \frac{1}{z+1} d z=2 \pi i+0=2 \pi i$$ (b) \(\quad\) By writing \(\oint_{C}=\oint_{C_{1}}+\oint_{C_{2}}\) where \(C_{1}\) and \(C_{2}\) are the circles \(|z|=1 / 2\) and \(|z+1|=1 / 2,\) respectively we have by Theorem 18.4 and (4) of Section 18.2 $$\begin{aligned} \oint_{C} \frac{1}{z} d z+\oint_{C} \frac{1}{z+1} d z &=\oint_{C_{1}} \frac{1}{z} d z+\oint_{C_{1}} \frac{1}{z+1} d z+\oint_{C_{2}} \frac{1}{z} d z+\oint_{C_{2}} \frac{1}{z+1} d z \\ &=2 \pi i+0+0+2 \pi i=4 \pi i \end{aligned}$$(c) \(\operatorname{since} f(z)=\frac{2 z+1}{z(z+1)}\) is analytic within and on \(C\) it follows from Theorem 18.4 that $$\oint_{C} \frac{2 z+1}{z^{2}+z} d z=0$$

$$\begin{aligned} \int_{i}^{1+\frac{\pi}{2} i} \sinh 3 z d z &=\left.\frac{1}{3} \cosh 3 z\right|_{i} ^{1+\frac{\pi}{2} i}=\frac{1}{3}\left[\cosh \left(3+\frac{3 \pi}{2} i\right)-\cosh 3 i\right] \\ &=\frac{1}{3}\left[\cosh 3 \cos \frac{3 \pi}{2}+i \sinh 3 \sin \frac{3 \pi}{2}-\cos 3\right]=-\frac{1}{3} \cos 3-\frac{1}{3} i \sinh 3=0.3300-3.3393 i \end{aligned}$$

We have \(\oint_{C} x d z=\int_{C_{1}} x d z+\int_{C_{2}} x d z+\int_{C_{3}} x d z\) On \(C_{1}, y=0,0 \leq x \leq 1, z=x, d z=d x\), \(\int_{C_{1}} x d z=\int_{0}^{1} x d x=\frac{1}{2}\). On \(C_{2}, x=1,0 \leq y \leq 1, z=1+i y, d z=i d y\), \(\int_{C_{2}} x d z=i \int_{0}^{1} d y=i\). On \(C_{3}, y=x, 0 \leq x \leq 1, z=x+i x, d z=(1+i) d x\), \(\int_{C_{3}} x d z=(1+i) \int_{1}^{0} x d x=-\frac{1}{2}-\frac{1}{2} i\). Thus \(\oint_{C} x d z=\frac{1}{2}+i-\frac{1}{2}-\frac{1}{2} i=\frac{1}{2} i\).

$$\begin{aligned} \int_{i}^{1+\frac{\pi}{2} i} \sinh 3 z d z &=\left.\frac{1}{3} \cosh 3 z\right|_{i} ^{1+\frac{\pi}{2} i}=\frac{1}{3}\left[\cosh \left(3+\frac{3 \pi}{2} i\right)-\cosh 3 i\right] \\ &=\frac{1}{3}\left[\cosh 3 \cos \frac{3 \pi}{2}+i \sinh 3 \sin \frac{3 \pi}{2}-\cos 3\right]=-\frac{1}{3} \cos 3-\frac{1}{3} i \sinh 3=0.3300-3.3393 i \end{aligned}$$

(a) By Theorem 18.10 with \(f(z)=\frac{1}{z-4}, f^{\prime}(z)=-\frac{1}{(z-4)^{2}},\) and \(f^{\prime \prime}(z)=\frac{2}{(z-4)^{3}}\) $$\oint_{C} \frac{\frac{1}{z-4}}{z^{3}} d z=\frac{2 \pi i}{2 !}\left(\frac{2}{-64}\right)=-\frac{\pi}{32} i$$ (b) By the Cauchy-Goursat Theorem, Theorem 18.4 \\[ \oint_{C} \frac{1}{z^{3}(z-4)} d z=0 \\]

We have \(\oint_{C}(2 z-1) d z=\int_{C_{1}}(2 z-1) d z+\int_{C_{2}}(2 z-1) d z+\int_{C_{3}}(2 z-1) d z\) On \(C_{1}, y=0,0 \leq x \leq 1, z=x, d z=d x\), \(\int_{C_{1}}(2 z-1) d z=\int_{0}^{1}(2 x-1) d x=0\). On \(C_{2}, x=1,0 \leq y \leq 1, z=1+i y, d z=i d y\), \(\int_{C_{2}}(2 z-1) d z=-2 \int_{0}^{1} y d y+i \int_{0}^{1} d y=-1+i\). On \(C_{3}, y=x, z=x+i x, d z=(1+i) d x\), \(\int_{C_{3}}(2 z-1) d z=(1+i) \int_{1}^{0}(2 x-1+2 i x) d x=1-i\). Thus \(\oint_{C}(2 z-1) d z=0-1+i+1-i=0\).

We have \(\oint_{C} z^{2} d z=\int_{C_{1}} z^{2} d z+\int_{C_{2}} z^{2} d z+\int_{C_{3}} z^{2} d z\) On \(C_{1} y=0,0 \leq x \leq 1, z=x, d z=d x\), \(\int_{C_{1}} z^{2} d z=\int_{0}^{1} x^{2} d x=\frac{1}{3}\). On \(C_{2}, x=1,0 \leq y \leq 1, z=1+i y, d z=i d y\), \(\int_{C_{2}} z^{2} d z=\int_{0}^{1}(1+i y)^{2} i d y=-1+\frac{2}{3} i\). On \(C_{3}, y=x, 0 \leq x \leq 1, z=x+i x, d z=(1+i) d x\), \(\int_{C_{3}} z^{2} d z=(1+i)^{3} \int_{1}^{0} x^{2} d x=\frac{2}{3}-\frac{2}{3} i\). Thus \(\oint_{C} z^{2} d z=\frac{1}{3}-1+\frac{2}{3} i+\frac{2}{3}-\frac{2}{3} i=0\).

$$\int_{-4 i}^{4 i} \frac{1}{z^{2}} d z=-\left.\frac{1}{z}\right|_{-4 i} ^{4 i}=-\left[\frac{1}{4 i}-\left(\frac{1}{-4 i}\right)\right]=\frac{1}{2} i$$

We have \(\oint_{C} \bar{z}^{2} d z=\int_{C_{1}} \bar{z}^{2} d z+\int_{C_{2}} \bar{z}^{2} d z+\int_{C_{3}} \bar{z}^{2} d z\) On \(C_{1}, y=0,0 \leq x \leq 1, z=x, d z=d x\), \(\int \bar{z}^{2} d z=\int_{0}^{1} x^{2} d x=\frac{1}{3}\). On \(C_{2}, x=1,0 \leq y \leq 1, z=1+i y, d z=i d y\), \(\int_{C_{2}} \bar{z}^{2} d z=-\int_{0}^{1}(1-i y)^{2}(-i d y)=1+\frac{2}{3} i\). On \(C_{3}, y=x, 0 \leq x \leq 1, z=x+i x, d z=(1+i) d x\), \(\int_{C_{3}} \bar{z}^{2} d z=(1-i)^{2}(1+i) \int_{1}^{0} x^{2} d x=-\frac{2}{3}+\frac{2}{3} i\). Thus \(\oint_{C} \bar{z}^{2} d z=\frac{1}{3}+1+\frac{2}{3} i-\frac{2}{3}+\frac{2}{3} i=\frac{2}{3}+\frac{4}{3} i\).

We have \(\int_{C}\left(z^{2}-z+2\right) d z=\int_{C_{1}}\left(z^{2}-z+2\right) d z+\int_{C_{2}}\left(z^{2}-z+2\right) d z\) On \(C_{1}, y=1,0 \leq x \leq 1, z=x+i, d z=d x\), \(\int_{C_{1}}\left(z^{2}-z+2\right) d z=\int_{0}^{1}\left[(x+i)^{2}-x+2-i\right] d x=\frac{5}{6}\). On \(C_{2}, x=1,0 \leq y \leq 1, z=1+i y, d z=i d y\), \(\int_{C_{2}}\left(z^{2}-z+2\right) d z=i \int_{1}^{0}\left[(1+i y)^{2}+1-i y\right] d y=\frac{1}{2}-\frac{5}{3} i\). Thus \(\int_{C}\left(z^{2}-z+2\right) d z=\frac{1}{2}-\frac{5}{3} i+\frac{5}{6}=\frac{4}{3}-\frac{5}{3} i\).

Write $$\oint_{C}\left(z^{2}+z+\operatorname{Re}(z)\right) d z=\oint_{C}\left(z^{2}+z\right) d z+\oint_{C} \operatorname{Re}(z) d z$$ By Theorem \(18.4, \oint_{C}\left(z^{2}+z\right) d z=0 .\) However, since \(\operatorname{Re}(z)=x\) is not analytic $$\oint_{C} x d z=\oint_{C_{1}} x d z+\oint_{C_{2}} x d z+\oint_{C_{3}} x d z$$ where \(C_{1}\) is \(y=0,0 \leq x \leq 1, C_{2}\) is \(x=1,0 \leq y \leq 2,\) and \(C_{3}\) is \(y=2 x, 0 \leq x \leq 1 .\) Thus, $$\oint_{C} x d z=\int_{0}^{1} x d x+i \int_{0}^{2} d y+(1+2 i) \int_{1}^{0} x d x=\frac{1}{2}+2 i-\frac{1}{2}(1+2 i)=i$$

On \(C, x=\sin t, y=\cos t, 0 \leq t \leq \pi / 2\) or \(z=i e^{-i t}, d z=e^{-i t} d t\), \(\int_{C}\left(z^{2}-z+2\right) d z=\int_{0}^{\pi / 2}\left(-e^{-2 i t}-i e^{-i t}+2\right) e^{-i t} d t=\int_{0}^{\pi / 2}\left(-e^{-3 i t}-i e^{-2 i t}+2 e^{-i t}\right) d t\) \(=-\frac{1}{3} i e^{-3 \pi i / 2}+\frac{1}{2} e^{-\pi i}+2 i e^{-\pi i / 2}+\frac{1}{3} i-\frac{1}{2}-2 i=\frac{4}{3}-\frac{5}{3} i\).

\(\operatorname{On} C,\left|\frac{e^{z}}{z^{2}+1}\right| \leq \frac{\left|e^{z}\right|}{|z|^{2}-1}=\frac{e^{5}}{24} \cdot\) Thus \(\left|\oint_{C} \frac{e^{z}}{z^{2}+1} d z\right| \leq \frac{e^{5}}{24} \cdot 10 \pi=\frac{5 \pi}{12} e^{5}\).

The length of the line segment from \(z=0\) to \(z=1+i\) is \(\sqrt{2}\). In addition, on this line segment \(\left|z^{2}+4\right| \leq|z|^{2}+4 \leq|1+i|^{2}+4=6\). Thus \(\left|\int_{C}\left(z^{2}+4\right) d z\right| \leq 6 \sqrt{2}\).

For \(f(z)=1 / z, \overline{f(z)}=1 / \bar{z},\) so on \(z=2 e^{i t}, \bar{z}=2 e^{-i t}, d z=2 i e^{i t} d t,\) and \(\oint_{C} \overline{f(z)} d z=\int_{0}^{2 \pi} \frac{1}{2 e^{-i t}} \cdot 2 i e^{i t} d t=\left.\frac{1}{2} e^{2 i t}\right|_{0} ^{2 \pi}=\frac{1}{2}\left[e^{4 \pi i}-1\right]=0\). Thus circulation \(=\operatorname{Re}\left(\oint_{C} \overline{f(z)} d z\right)=0,\) and net \(\operatorname{flux}=\operatorname{Im}\left(\oint_{C} \overline{f(z)} d z\right)=0\).

For \(f(z)=2 z, \overline{f(z)}=2 \bar{z},\) so on \(z=e^{i t}, \bar{z}=e^{-i t}, d z=i e^{i t} d t,\) and \(\oint_{C} \overline{f(z)} d z=\int_{0}^{2 \pi}\left(e^{-i t}\right)\left(i e^{i t} d t\right)=2 i \int_{0}^{2 \pi} d t=4 \pi i\). Thus circulation \(=\operatorname{Re}\left(\oint_{C} \overline{f(z)} d z\right)=0,\) and net \(\operatorname{flux}=\operatorname{Im}\left(\oint_{C} \overline{f(z)} d z\right)=4 \pi\).

For \(f(z)=1 /(\overline{z-1}), \overline{f(z)}=1 /(z-1),\) so on \(z-1=2 e^{i t}, d z=2 i e^{i t} d t,\) and \(\oint_{C} \overline{f(z)} d z=\int_{0}^{2 \pi} \frac{1}{2 e^{i t}} \cdot 2 i e^{i t} d t=i \int_{0}^{2 \pi} d t=2 \pi i\). Thus circulation \(=\operatorname{Re}\left(\oint_{C} \overline{f(z)} d z\right)=0,\) and net \(\operatorname{flux}=\operatorname{Im}\left(\oint_{C} \overline{f(z)} d z\right)=2 \pi\).