4.3 Cauchy's integral theorem

1. Let $C_1$ be the curve connecting from a starting point $a$ to $b$. Then the curve connecting from a point $b$ to $a$ can be represented by $-C_{2}$. Now let $C = C_{1} - C_{2}$. Then the curve $C$ is a closed curve in the region $D$. Here using Cauchy's integral theorem, we have

$$= \int_{C}f(z) dz = \int_{C_{1}}f(z) dz + \int_{-C_{2}}f(z) dz$$ 0

$$= \int_{C_{1}}f(z) dz - \int_{C_{2}}f(z) dz$$

Therefore,

$$\int_{C_{1}}f(z)dz = \int_{C_{2}}f(z) dz$$

First, bridge from curve $C_{1}$ to curve $C_{2}$. Then, while turning along the curve $C_{1}$, cross the bridge and move to the curve $C_{2}$, Let $C$ be the curve that goes in the opposite direction, crosses the original bridge, returns to the curve $C_{1}$, and goes around. At this time, $C$ is a closed curve included in the region $D$, so if Cauchy's integral theorem is used,

$$= \int_{C}f(z) dz = \int_{C_{1}}f(z) dz + \int_{-C_{2}}f(z) dz$$ 0

$$= \int_{C_{1}}f(z) dz - \int_{C_{2}}f(z) dz$$

Therefore,

$$\int_{C_{1}}f(z)dz = \int_{C_{2}}f(z) dz$$

Since the curve $C$ is the circumference of a radius $r > 1$ centered at the origin, $f(z) = \frac{1}{z^2 + 1}$ is not analytic inside of this circle. Then, expand $f(z) = \frac{1}{z^2 + 1}$ by using partial fraction.

$$\frac{1}{z^2 + 1} = \frac{1}{(z+i)(z-i)} = \frac{1}{2i}[\frac{1}{z-i} - \frac{1}{z+i}]$$

Here we use the basic integral formula

$$\int_{\vert z-a\vert=r}\frac{1}{(z-a)^{n}} = \left\{\begin{array}{ll} 2\pi i & n = 1\\ 0 & n \neq i \end{array}\right.$$

$$\int_{\vert z\vert = r}\frac{1}{z^2 + 1} dz = \frac{1}{2i}[\int_{\vert z\vert=r}\frac{1}{z-1} dz - \int_{\vert z\vert=r}\frac{1}{z-1} dz$$

$$= \frac{1}{2i}[2\pi i - 2\pi i] = 0$$

Since the curve $C$ is the circumference of a radius $r > 1$ centered at the origin, $f(z) = \frac{z}{(2z + i)(z-2)}$ is not analytic inside of this circle. Then , expand $f(z) = \frac{z}{(2z + i)(z-2)}$ by using the partial fraction expansion.

$$\frac{z}{(2z + i)(z-2)} = \frac{A}{2z+i} + \frac{B}{z-2}$$

Clear the denominators,

$$z = A(z-2) + B(2z+i)$$

Now let $z = 2$. Then

$$2 = B(4 + i) \Rightarrow B = \frac{2}{4 + i}$$

Also, let $z = -\frac{i}{2}$. Then

$$-\frac{i}{2} = A(-\frac{i}{2} - 2) \Rightarrow A = \frac{i}{4 + i}$$

Thus,

$$\int_{\vert z\vert=1}\frac{z}{(2z + i)(z-2)} = \frac{i}{4+i}\int_{\vert z\vert=1}\frac{1}{2z + i} dz + \frac{2}{4+i}\int{\vert z\vert=1}\frac{1}{z-2} dz$$

Here using the basic integral formula, we have

$$\int_{\vert z-a\vert=r}\frac{1}{(z-a)^{n}} = \left\{\begin{array}{ll} 2\pi i & n = 1\\ 0 & n \neq i \end{array}\right.$$

$$\int_{\vert z\vert=1}\frac{z}{(2z + i)(z-2)} = \frac{i}{4+i}\int_{\vert z\vert=1}\frac{1}{2(z + i/2)} dz + 0$$

$$= \frac{i}{2(4+i)} \cdot 2\pi i = \frac{-\pi}{4 + i} = -\frac{4-i}{17}\pi$$

Since this curve is centered on the origin, the circumference of the upper half of a circle with radius r> 1, and the diameter on the real axis, $f(z) = \frac{1}{z^4 +1}$ is not analytic inside of this circle. Then expand $f(z) = \frac{1}{z^4 +1}$ by using the partial fraction expansion. Solutions of $z^4 = -1$ are give by

$$z_{k} = \cos{(\frac{\pi + 2k\pi}{4})} + \sin{(\frac{\pi + 2k\pi}{4}}, k =0,1,2,3$$

Thus,

$$z_{0} = \frac{1+i}{\sqrt{2}} = e^{\frac{\pi}{4}}, z_{1} = \frac{-... ...{\frac{3\pi}{4}}, z_{2} = \frac{-1-i}{\sqrt{2}}, z_{4} = \frac{1 - i}{\sqrt{2}}$$

Therefore,

$$\frac{1}{z^4 + 1} = \frac{A}{z - z_{0}} + \frac{B}{z - z_{1}} + \frac{C}{z - z_{2}} + \frac{D}{z - z_{3}}$$

Here we only need to find $A$ and $B$.

$$A = \lim_{z \to z_{0}}(z - z_{0})\frac{1}{z^4 + 1} = \frac{1}{4z_{0}^3}$$

$$= \frac{1}{4}z_{0}^{-3} = \frac{1}{4}e^{-3\pi}{4}$$

$$B = \lim_{z \to z_{1}}(z - z_{1})\frac{1}{z^4 + 1} = \frac{1}{4z_{1}^3}$$

$$= \frac{1}{4}z_{1}^{-3} = \frac{1}{4}e^{-9\pi}{4}$$

Thus

$$\int_{C}\frac{1}{z^4 + 1} dz = A \int_{C}\frac{1}{z - z_{0}} dz + B\int_{C}\frac{1}{z-z_{1}}$$

$$= \frac{1}{4}e^{-3\pi}{4}\cdot 2\pi i + \frac{1}{4}e^{-9\pi i}\cdot 2\pi i$$

$$= \frac{\pi i}{2}[e^{\frac{-3\pi i}{4}} + e^{\frac{-9 \pi i}{4}}]$$

$$= \frac{\pi i}{2}[\frac{-1-i}{\sqrt{2}} + \frac{1-i}{\sqrt{2}}] = \frac{\pi}{2}$$

3.

(a)

$$\int_{i}^{1} z^2 dz = \frac{1}{3}z^{3} \mid_{i}^{1} = \frac{1 + i}{3}$$

(b)

$$\int_{0}^{i}ze^{z} dz = [ze^{z} - e^{z}\mid_{0}^{i} = ie^{i} - e^{i}$$

(c) Recall that

$$\log{z} = \log_{e}\vert z\vert + i\arg{z}$$

$$\int_{0}^{1+i}\frac{z}{z+1} dz = \int_{0}^{1+i}(1 - \frac{1}{z+1}) dz$$

$$= [z - \log(z+1)\mid_{0}^{1+i} = 1+i - \log(2+i) = 1+i - (\log_{e}\vert 2+i\vert + i\arg(2+i))$$

$$= 1+i - \frac{1}{2}\log_{e}(5) - i \tan^{-1}{\frac{1}{2}}$$

$$= 1 - \frac{1}{2}\log_{e}{5} + i(1 - \tan^{-1}{(\frac{1}{2})})$$

(d) Recall that

$$(\sin^{-1}{z})' = \frac{1}{\sqrt{1 - z^2}}$$

$$\sin^{-1}{z} = \frac{1}{i}\log(iz + \sqrt{1 - z^2})$$

$$\int_{0}^{i}\frac{1}{\sqrt{1-z^2}} dz = [\sin^{-1}{z}\mid_{0}^{i}$$

$$= \sin^{-1}{i} - \sin^{0} = \frac{1}{i}[\log(i^2 + \sqrt{1 - i^2}) - \log{1}$$

$$= \frac{1}{i}\log(-1 + \sqrt{2}) = \frac{1}{i}[\log_{e}\vert-1 + \sqrt{2}\vert + i\arg(-1 + \sqrt{2})]$$

$$= \frac{1}{i}[\log_{e}(\sqrt{2} - 1)]$$

4. A function $u(x,y)$ is called a harmonic function if $\Delta u = u_{xx} + u_{yy} = 0$. Also, $\Delta$ is said to be Laplacian, the equation $\Delta u = 0$ is called a Laplace equation. Make sure that the holomorphic function $w = u + iv$ with $u$ in the real part satisfies Cauchy-Riemann's equation.

(a) For $u = x^2 - y^2$, we find $\Delta u$. Then

$$u_{x} = 2x, u_{xx} = 2, u_{y} = -2y, u_{yy} = -2$$

Thus, $\Delta u = 0$ and $u$ is a harmonic function. Next find a holomorphic function $w$ which has $u$ as a real part. Let $w = u + iv$. Then

$$u_{x} = v_{y} = 2x, v_{x} = -u_{y} = 2y$$

Thus,

$$v(x,y) = \int{v_{y} dy} = \int{2x dy} = 2xy + \phi(x)$$ (A.1)

Differentiate the equation A.1 by $x$,

$$v_{x} = 2y + (\phi(x))'$$ (A.2)

Now by the condition $v_{x} = 2y$, we have $\phi'(x) = 0$ which implies $\phi(x) = c$ and

$$v(x,y) = 2xy + c$$

(b) Find $\Delta u$ for $u = e^{x}\cos{y}$. Then

$$u_{x} = e^{x}\cos{y}, u_{xx} = e^{x}\cos{y}, u_{y} = -e^{x}\sin{y}, u_{yy} = -e^{x}\cos{y}$$

Thus, $\Delta u = 0$ and $u$ is a harmonic function. Next find $w$ which has $u$ as a real part. Let $w = u + iv$. Then

$$u_{x} = v_{y} = e^{x}\cos{y}, v_{x} = -u_{y} = e^{x}\sin{y}$$

which implies

$$v(x,y) = \int{v_{y} dy} = \int{e^{x}\cos{y}} dy = e^{x}\sin{y} + \phi(x)$$

$$v_{x} = e^{x}\sin{y} + (\phi(x))' = e^{x}\sin{y} \phi(x) = c ( )$$

$$v(x,y) = e^{x}\sin{y} + c$$

(c) Find $\Delta u$ for $u = \cos{x}\sinh{y}$.

$$u_{x} = -\sin{x}\sinh{y}, u_{xx} = -\cos{x}\sinh{y}, u_{y} = \cos{x}\cosh{y}, u_{yy} = \cos{x}\sinh{y}$$

Thus, $\Delta u = 0$ and $u$ is a harmonic function. Next find $w$ which has $u$ as a real part. Let $w = u + iv$. Then

$$u_{x} = v_{y} = e^{x}\cos{y}, v_{x} = -u_{y} = e^{x}\sin{y}$$

which implies

$$v(x,y) = \int{v_{y} dy} = \int{e^{x}\cos{y}} dy = e^{x}\sin{y} + \phi(x)$$

$$v_{x} = e^{x}\sin{y} + (\phi(x))' = e^{x}\sin{y} \phi(x) = c ( )$$

$$v(x,y) = e^{x}\sin{y} + c$$

(d) Find $\Delta u$ for $u = \frac{1}{2}\log_{e}(x^2 + y^2)$.

$$u_{x} = \frac{x}{x^2 + y^2}, u_{xx} = \frac{x^2 + y^2 - x(2x)}{(x^2 + y^2)^{2}} = \frac{y^2 - x^2}{(x^2 + y^2)^{2}}$$

$$u_{y} = \frac{y}{x^2 + y^2}, u_{xx} = \frac{x^2 + y^2 - y(2y)}{(x^2 + y^2)^{2}} = \frac{x^2 - y^2}{(x^2 + y^2)^{2}}$$

Thus, $\Delta u = 0$ and $u$ is a harmonic function. Next find $w$ which has $u$as a real part. Let $w = u + iv$. Then

$$u_{x} = v_{y} = \frac{x}{x^2 + y^2}, v_{x} = -u_{y} = -\frac{y}{x^2 + y^2}$$

which implies

$$v(x,y) = \int{v_{y} dy} = \int{\frac{x}{x^2 + y^2}} dy$$

$$= x(\frac{1}{x}\tan^{-1}{(\frac{y}{x})}) + \phi(x) ( \int{\frac{1}{a^2 + t^2}} dx = \frac{1}{a}\tan^{-1}{(\frac{t}{a})}$$

$$= \tan^{-1}{(\frac{y}{x})} + \phi(x)$$

Now take partial derivative with respect to $x$,

$$v_{x} = \frac{-\frac{y}{x}}{(1 + (\frac{y}{x}))^{2}} + \phi'(x)$$

$$= -\frac{y}{(x^2 + y^2)^{2}} + \phi'(x)$$

By the condition $v_{x} = -\frac{y}{(x^2 + y^2)^{2}}$. Thus $\phi'(x) = 0$ and $\phi(x) = c$. Therefore,

$$v(x,y) = \tan^{-1}{(\frac{y}{x})} + c$$