4.2 Complex integral

1. Parametrize a unit circle. Then $z(t) = e^{it}, 0 \leq t \leq 2\pi$. Thus

$$\int_{C}z\cos{z}dz = \int_{0}^{2\pi}e^{it}\cos(e^{it})ie^{it}dt \ \left(\begin{array}... ... = ie^{it}\cos(e^{it})dt\\ du = ie^{it}dt & v = sin(e^{it}) \end{array}\right.$$

$$= e^{it}\sin(e^{it})\mid_{0}^{2\pi} - \int_{0}^{2\pi}ie^{it}\sin(e^{it})dt$$

$$= \cos(e^{it})\mid_{0}^{2\pi} = \cos(e^{2\pi i}) - \cos(0) = 0$$

2.

case1. The integral path is 0 to $1$ and $1$ to $1+i$ along the sides of the square.

A straight line $c_{1}$ connecting 0 and $1$ can be parametrized by $z(t) = (0,0) + (1,0)t, 0 \leq t \leq 1$. Thus, $x(t) = t, y(t) = 0, 0 \leq t \leq 1$. Therefore, $z = x + iy = t$ and

$$\int_{c_{1}}\bar{z} dz = \int_{0}^{1}t (dt) = \frac{t^2}{2} \mid_{0}^{1} = \frac{1}{2}$$

A straight line $c_{2}$ connecting $1$ and $1+i$ can be parametrized by $z(t) = (1,0) + (0,1)t, 0 \leq t \leq 1$. Thus, $x(t) = 1, y(t) = t, 0 \leq t \leq 1$. Therefore, $z = x + iy = 1 + it$ and

$$\int_{c_{2}}\bar{z} dz = \int_{0}^{1}(1 -it) (idt) = \int_{0}^{1}(i + t)dt = [it + \frac{t^2}{2} \mid_{0}^{1} = i + \frac{1}{2}$$

Then the integral is given by

$$\int_{c}\bar{z}dz = \int_{c_{1}}\bar{z} dz + \int_{c_{2}}\bar{z} dz = 1 + i$$

case2. The integral path is 0 to $i$ and $i$ to $1+i$ along the sides of the square.

A straight line $c_{3}$ connecting 0 and $i$ can be parametrized by $z(t) = (0,0) + (0,1)t, 0 \leq t \leq 1$. Thus, $x(t) = 0, y(t) = 1, 0 \leq t \leq 1$. Therefore, $z = x + iy = it$ and

$$\int_{c}\bar{z} dz = \int_{0}^{1}(-it) (i dt) = \frac{t^2}{2} \mid_{0}^{1} = \frac{1}{2}$$

A straight line $c_{4}$ connecting $i$ and $1+i$ can be parametrized by $z(t) = (0,1) + (1,0)t, 0 \leq t \leq 1$. Thus, $x(t) = t, y(t) = 1, 0 \leq t \leq 1$. Therefore, $z = x + iy = t + i$ and

$$\int_{c}\bar{z} dz = \int_{0}^{1}(t-i) (dt) = [\frac{t^2}{2} - it \mid_{0}^{1} = \frac{1}{2} - i$$

Then the integral is given by

$$\int_{c}\bar{z}dz = \int_{c_{3}}\bar{z} dz + \int_{c_{4}}\bar{z} dz = 1 - i$$

case3. The integral path is 0 to $1+i$ along the diagonal of the square.

A straight line $c_{5}$ connecting 0 and $1+i$ can be parametrized by $z(t) = (0,0) + (1,1)t, 0 \leq t \leq 1$. Thus, $x(t) = t, y(t) = t, 0 \leq t \leq 1$. Therefore, $z = x + iy = t+ it$ and

$$\int_{c_{5}}\bar{z} dz = \int_{0}^{1}(t-it) (1+i) dt = \int_{c_{5}} 2t dt = t^{2}\mid_{0}^{1} = 1$$