4.4 Cauchy's integral formula

is inside of the curve $C$

and

is analytic inside of the region containing curve $C$

, then

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

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

(a) Since $\vert z\vert = 3$ , $z = 2$

is inside of this curve. Then by Cauchy's integral formula, we have

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

(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,

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

is inside of this circle. Then by Cauchy's integral formula,

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

$\displaystyle \int_{\vert z\vert=3}\frac{e^{3z}}{2z - \pi i} dz$	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{\vert z\vert=3}\frac{e^{3z}}{z - \frac{\pi i}{2}} dz$
	$\displaystyle =$	$\displaystyle \pi i f(\frac{\pi i}{2}) = \pi i e^{\frac{3\pi i}{2}}$
	$\displaystyle =$	$\displaystyle \pi i (\cos{(3\pi /2)} + i\sin{(3\pi/2)}) = \pi i (-i) = \pi$

(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,

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

$\displaystyle =$

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

$\displaystyle \int_{\vert z\vert=3}\frac{\cos{z}}{z^2 + 1} dz$	$\displaystyle =$	$\displaystyle \frac{1}{2i}\int_{\vert z\vert=3}{\cos{z}(\frac{1}{z-i} - \frac{1}{z+i})} dz$
	$\displaystyle =$	$\displaystyle \frac{1}{2i}[2\pi i (f(i) - f(-i))] = \pi[\cos{(i)} - \cos{(-i)}]$
	$\displaystyle =$	$\displaystyle \pi[\frac{e^{i^2} + e^{-i^2}}{2} - (\frac{e^{-i^2} + e^{i^2}}{2})] = 0$

(g) Since $\vert z\vert = 1$ , $z = 0$

is inside of this circle. Then by Cauchy's integral formula,

$\displaystyle \int_{\vert z\vert=1}\frac{e^{z}}{z^4} dz$

$\displaystyle =$

$\displaystyle \frac{2\pi i}{3!}f^{(3)}(0) = \frac{2\pi i e^{0}}{6} = \frac{\pi i}{3}$

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

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

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

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

(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,

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

$\displaystyle =$