4.4 Cauchy's integral formula

2. Cauchy's integral formula

If $z=a$ is inside of the curve $C$ and $f(z)$ is analytic inside of the region containing curve $C$, then

$$\int_{c}\frac{f(z)}{z-a} dz = 2\pi i f(a)$$

Cauchy's integral theorem

If $\frac{f(z)}{z-a}$ is analytic inside of curve $C$, then

$$\int_{c}\frac{f(z)}{z-a} dz = 0$$

(a) Since $\vert z\vert = 3$, $z = 2$ is inside of this curve. Then by Cauchy's integral formula, we have

$$\int_{\vert z\vert=3}\frac{e^{z}}{z - 2} dz = 2\pi i f(2) = 2\pi i e^{2}$$

(b) Since $\vert z\vert = 1$, $z = 2$ is not inside of this circle. Then the integrand is analytic. Thus by Cauchy's integral theorem,

$$\int_{\vert z\vert=1}\frac{e^{z}}{z - 2} dz = 0$$

(c) Since $\vert z\vert = 3$, $z = 0$ is inside of this circle. Then by Cauchy's integral formula,

$$\int_{\vert z\vert=3}\frac{\sin^{3}{z}}{z} dz = 2\pi i f(0) = 2\pi i \sin^{3}{0} = 0$$

(d) Since $\vert z\vert = 3$, $z=\frac{\pi i}{2}$ is inside of this circle. Then by Cauchy's integral formula,

$$\int_{\vert z\vert=3}\frac{e^{3z}}{2z - \pi i} dz = \frac{1}{2}\int_{\vert z\vert=3}\frac{e^{3z}}{z - \frac{\pi i}{2}} dz$$

$$= \pi i f(\frac{\pi i}{2}) = \pi i e^{\frac{3\pi i}{2}}$$

$$= \pi i (\cos{(3\pi /2)} + i\sin{(3\pi/2)}) = \pi i (-i) = \pi$$

(e) Since $\vert z\vert = 1$, $z=\frac{\pi i}{2}$ is not inside of this circle. Then the integrand is analytic. Thus by Cauchy's integral theorem,

$$\int_{\vert z\vert=1}\frac{e^{3z}}{2z - \pi i} dz =$$ 0

(f) Since $\vert z\vert = 3$, $z=\pm i$ is inside of this circle. Then by Cauchy's integral formula,

$$\int_{\vert z\vert=3}\frac{\cos{z}}{z^2 + 1} dz = \frac{1}{2i}\int_{\vert z\vert=3}{\cos{z}(\frac{1}{z-i} - \frac{1}{z+i})} dz$$

$$= \frac{1}{2i}[2\pi i (f(i) - f(-i))] = \pi[\cos{(i)} - \cos{(-i)}]$$

$$= \pi[\frac{e^{i^2} + e^{-i^2}}{2} - (\frac{e^{-i^2} + e^{i^2}}{2})] = 0$$

(g) Since $\vert z\vert = 1$, $z = 0$ is inside of this circle. Then by Cauchy's integral formula,

$$\int_{\vert z\vert=1}\frac{e^{z}}{z^4} dz = \frac{2\pi i}{3!}f^{(3)}(0) = \frac{2\pi i e^{0}}{6} = \frac{\pi i}{3}$$

(h) Since $\vert z\vert = 3$, $z=\frac{\pi}{2}$ is inside of thie circle. Then by Cauchy's integral formula,

$$\int_{\vert z\vert=3}\frac{\sin{z}}{(2z - \pi)^{3}} dz = \frac{1}{8}\int_{\vert z\vert=3}\frac{\sin{z}}{(z - \pi/2)^{3}} dz$$

$$= \frac{1}{8}\frac{2\pi i}{2!}f''(\frac{\pi}{2})$$

Note that $f(z) = \sin{z}$, $f'(z) = \cos{z}, f''(z) = -\sin{z}$. Then $f''(\frac{\pi}{2}) = -1$. Thus

$$\int_{\vert z\vert=3}\frac{\sin{z}}{(2z - \pi)^{3}} dz = -\frac{1}{8}\frac{2\pi i}{2!} = -\frac{\pi i}{8}$$

(i) Since $\vert z\vert = 1$, $z=\frac{\pi}{2}$ is not inside of thie circle. Then the integrand is analytic. Thus by Cauchy's integral theorem,

$$\int_{\vert z\vert=1}\frac{e^{3z}}{2z - \pi i} dz =$$ 0