Limit and Continuity

Defined in the neighborhood of $z_0$

with

, for

, if $\vert z - z_{0}\vert \to 0$ then $\vert w - w_{0}\vert \to 0$ . In other words, for any positive number $\varepsilon > 0$ , there exists $\delta$ so that $0 < \delta < A$ , and for any $z$

with $0 < \vert z - z_{0}\vert < \delta$ satisfies $\vert w - w_{0}\vert < \varepsilon$ . $f(z)$

is said to have a limit $w_{0}$ and we write $\lim_{z \to z_{0}}f(z) = w_{0}$ .

Note There are innumerable directions for $z \to z_{0}$ , but no matter which direction you approach $z_{0}$ , it will be $w \to w_{0}$ .

$For \lim_{z \to z_{0}}\vert w\vert = + \infty$ , we write $\lim_{z \to z_{0}}f(z) = \infty$ . For real-valued function, there are $+\infty$ and $-\infty$ . But by the abo ve definition, for complex functions, we only have $+\infty$ .

Theorem 3..1 Suppose that $\lim_{z \to z_{0}}f(z) = A, \lim_{z \to z_{0}}f(z) = B$ exist and the limits are finite. Then

(1) $\lim_{z \to z_{0}}\{f(z) \pm g(z)\} = A \pm B = \lim_{z \to z_{0}}f(z) \pm \lim_{z \to z_{0}}g(z)$

(2) $\lim_{z \to z_{0}}\{f(z)\cdot g(z)\} = AB$

(3) $\lim_{z \to z_{0}}\{\frac{f(z)}{g(z)}\} = \frac{A}{B} (B \neq 0)$

For $w = f(z)$ defined in the region $D$ , a point $z_0$ in $D$ , $\lim_{z \to z_{0}}f(z) = f(z_{0})$ holds, In other words, for any positive number $\varepsilon > 0$ , there exists $\delta > 0$ and for any $z$ with $\vert z - z_{0}\vert < \delta$ , $\vert f(z) - f(z_{0})\vert < \varepsilon$ holds. Then $f(z)$ is said to be continuous at $z_{0}$ .

When $f(z)$ is continuous at each point in the region $D$ , $f(z)$ is said to be continuous on $D$ .

Theorem 3..2

(1) If are continuous at $z_{0}$ , $f(z) \pm g(z), f(z)\cdot g(z), f(z)/g(z) (however g(z_{0}) \neq 0)$ are continuous at $z_{0}$ .

(2) Suppose that is continuous at $w = w_{0}$ and is continuous at $z_{0}$ . If $w_{0} = f(z_{0})$ , then the composite function is continuous at $z_{0}$ .

Example 3..1 Find $\lim_{z \to 0}f(z)$ for the following . (1) $f(z) = \frac{bar{z}}{z}$

(2) $f(z) = \frac{1}{z}$

(3) $f(z) = \frac{z^2}{z}$

Solution (1) Since $f(z) = \frac{bar{z}}{z} = \frac{x - iy}{x + iy}$ , for $z$ approaches 0 along the straight line $y = mx$ , the value of $f(z) = \frac{1 - im}{1 + im}$ depends on the value of $m$ . Thus, $\lim_{z \to 0}f(z)$ does not exist. (2) $\lim_{z \to 0}\vert f(z)\vert = \lim_{z \to 0}\frac{1}{\vert z\vert} = + \infty$ . Thus, $\lim_{z \to 0}\frac{1}{z} = \infty$

(3) For $z \neq 0$ , $f(z) = \frac{z^2}{z} = z$ . Thus, $\lim_{z \to 0}f(z) = \lim_{z \to 0}z = 0$ .

Exercise3.1

1. Find out if the next set of points is a region

(a): set of points on plane excluding 0
(b): $\{z : \Re{z} > 0\}$
(c): $\{z : \Im {z} \geq 0\}$
(d): $\{z : 1 < \vert z\vert < 2\}$

2. Find the limit of the followings.

(a): $\displaystyle{\lim_{z \to i}(z^2 + 2z)}$
(b): $\displaystyle{\lim_{z \to \frac{i}{2}}\frac{(2z-3)(z+i)}{(iz - 1)^2}}$
(c): $\displaystyle{\lim_{z \to 1+i}\frac{z - 1 -i}{z^2 - 2z + 2}}$
(d): $\displaystyle{\lim_{z \to e^{i\pi/4}}\frac{z^2}{z^4 + z + 1}}$

3. Find the point where the following function is not continuous.

(a)
(b)
(c): $\frac{2z}{z+ i}$
(d): $\frac{2z - 3}{z^2 + 2z + 2}$
(e): $\frac{z+1}{z^4 + 1}$
(f): $\frac{z^2 + 4}{z - 2i}$
(g): $f(z) = \left\{\begin{array}{ll} \frac{z^2 + 4}{z - 2i} & (z \neq 2i)\\ 4i & (z = 2i) \end{array}\right.$