Limit and continuity

1.

In order for the set $D$ to be called a region, it must satisfy the following two conditions.

$$\begin{array}{ll} 1 & z_{0} \in D, z \in N(z_{0},\delta) = \{... ...tinuous curve in} D. (\mbox{arc-like connectivity}) \end{array}$$

Note $N(z_{0},\delta) = \{z : \vert z - z_{0}\vert < \delta\}$ is said to be a $\delta$ neighborhood of a point $z_{0}$.

(a) Let $D$ be the set of the $z$ plane minus the origin O.

1. If you select a point $z_{0}$ other than the origin and let $z_{0}$ and half the distance to the origin $\vert z_{0}\vert/2$ be $\delta$, then the $\delta$ neighborhood $N(z_{0},\delta)$ is in $D$. Thus $D$ is open set.

2. Any two points in $D$ can be connected by the continuous curve contained in $D$, so arc-like connectivity is satisfied.

Therefore, $D$ is a region.

(b) Let $D = \{z : \Re{z} > 0\}$.

1. Choose $z_{0}$ other than the origin. Let $\delta$ be the half of the distance from $z_{0}$ to the imaginary axis. The $\delta$ neighborhood $N(z_{0},\delta)$ is in $D$. Thus, $D$ is an open set.

2. Any two points in $D$ can be connected by the continuous curve contained in $D$, so arc-like connectivity is satisfied.

Thus, $D$ is a region.

(c) Let $D = \{z : \Im {z} \geq 0\}$.

1. If we choose $z_{0}$ on the real axiz, then for any $\delta$ neighborhoos $N(z_{0},\delta)$ contains a point other that $D$. Thus it is not a open set. But, any point in $\{z : \Im {z} = 0\}$, $\delta$ neighborhood of its point contains a point in $D$ and not in $D$. A collection of such points is calledboundary of $D$. Thus, $D$ is a closed set.

2. Any two points in $D$ can be connected by the continuous curve contained in $D$, so arc-like connectivity is satisfied. .

Therefore, $D$ is a closed set.

(d) Let $\{z : 1 < \vert z\vert < 2\}$.

1. Choose $z_{0}$ other than the origin. Then let $\delta$ be half of the distance of the shorter of two: the distance from $z_{0}$ to $\vert z\vert = 1$ or to $\vert z\vert = 2$. Then $\delta$ neighborhood $N(z_{0},\delta)$ is in $D$. Therefore, $D$ is an open set.

2. Any two points in $D$ can be connected by the continuous curve contained in $D$, so arc-like connectivity is satisfied. .

Therefore, $D$ is a region.

2.

(a)

$$\lim_{z \to i}(z^2 + 2z) = i^2 + 2i = 2i -1$$

(b)

$$\lim_{z \to \frac{i}{2}}\frac{(2z-3)(z+i)}{(iz - 1)^2} = \frac{(i-3)(\frac{3i}{2})}{(-\frac{3}{2})^{2}} = -\frac{2}{3}(1+3i)$$

(c)

$$\lim_{z \to 1+i}\frac{z - 1 -i}{z^2 - 2z + 2} = \lim_{z \to 1+i}\frac{z - (1 +i)}{(z-(1-i))(z-(1+i))} = \frac{1}{2i} = - \frac{i}{2}$$

(d)

$$\lim_{z \to e^{i\pi/4}}\frac{z^2}{z^4 + z + 1} = \frac{e^{\pi i/2... ...1 + \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} + 1} = \frac{\sqrt{2}}{2}(1 + i)$$

3.

Note

1. The exponential function $e^{z}$ is continuous at all points on the complex plane, the logarithmic function $\log{z}$ is continuous except at the origin, and the meromorphic function is continuous except when the denominator is 0

2. The sum, difference, product, and composition of continuous functions are also continuous, the denominator is non-zero, and the quotient is also continuous.

(a) $z^2$ is a meromorphic function. Thus, it is continuous on all points.

(b) $e^{z}$ is continuous at all points..

(c) $\frac{2z}{z+ i}$ is a meromorphic function. Thus it is discontinuous at $z = -i$.

(d) $\frac{2z - 3}{z^2 + 2z + 2}$ is a meromorphic function. Thus it is discontinuous ar $z = -1 +i, -1 - i$.

(e) $\frac{z+1}{z^4 + 1}$ is a meromorphic function. Thus it is discontinuous at the point where $z^4 + 1 = 0$.

$$z^4 = -1 = e^{i \pi}$$   implies$$z_{k} = (\cos{\frac{pi + 2k\pi}{4}} + i\sin{\frac{\pi + 2k\pi}{4}}) (k = 0,1,2,3)$$

Therefore,

$$z_{0} = \frac{(1+i)}{\sqrt{2}}, z_{1} = \frac{(-1+i)}{\sqrt{2}}, z_{2} = \frac{(-1-i)}{\sqrt{2}}, z_{3} = \frac{(1-i)}{\sqrt{2}}$$

(f) $\frac{z^2 + 4}{z - 2i}$ is a meromorphic function. Thus it is discontinuous at $z = 2i$.

(g) $f(z) = \left\{\begin{array}{ll} \frac{z^2 + 4}{z - 2i} & (z \neq 2i)\\ 4i & (z = 2i) \end{array}\right.$ is meromorphic function except at $z = 2i$. Thus it is continuous except at $z = 2i$. Note that $f(z)$ is a piecewise function. Thus to check to see if it is continuous, we have to go back to the definition. In other words, if

$$\lim_{z \to 2i}f(z) = f(2i)$$

then it is continuous at $z = 2i$.

$$\lim_{z \to 2i}f(z) = \lim_{z \to 2i}\frac{z^2 + 4}{z - 2i} = \lim_{z \to 2i}\frac{(z +2i)(z - 2i)}{z - 2i} = 4i$$

or $f(2i) = 4i$. Thus it is continuous at $z = 2i$.