Complex Functions and Analyticity

2.1 Complex Functions

A complex function is a function $f : D \subseteq \mathbb{C} \to \mathbb{C}$ . We can write $f(z) = u(x, y) + iv(x, y)$ where $z = x + iy$ and $u, v$ are real-valued functions.

Example. $f(z) = z^2 = (x + iy)^2 = (x^2 - y^2) + i(2xy)$ . Here $u = x^2 - y^2$ and $v = 2xy$ .

Example. $f(z) = \bar{z} = x - iy$ . Here $u = x$ and $v = -y$ .

Example. $f(z) = |z|^2 = x^2 + y^2$ . Here $u = x^2 + y^2$ and $v = 0$ .

2.2 Limits and Continuity

The limit $\lim_{z \to z_0} f(z) = L$ means: for every $\varepsilon \gt 0$ There exists $\delta \gt 0$ Such that $0 \lt |z - z_0| \lt \delta$ implies $|f(z) - L| \lt \varepsilon$ .

Unlike the real case, $z$ can approach $z_0$ from any direction in $\mathbb{C}$ . This makes limits More restrictive.

Proposition 2.1. $\lim_{z \to z_0} f(z) = L$ if and only if $\lim_{(x,y) \to (x_0, y_0)} u(x, y) = a$ And $\lim_{(x,y) \to (x_0, y_0)} v(x, y) = b$ where $L = a + bi$ .

Definition. $f$ is continuous at $z_0$ if $\lim_{z \to z_0} f(z) = f(z_0)$ .

Solution

Problem. Show that $\lim_{z \to 0} \frac{\bar{z}}{z}$ does not exist.

Let $z = re^{i\theta}$ . Then $\frac{\bar{z}}{z} = e^{-2i\theta}$ . As $z \to 0$ along different Rays ( $\theta = 0, \pi/2, \pi/4$ Etc.), the ratio takes different values ( $1, -1, -i$ Etc.). Since the limit depends on the direction of approach, it does not exist.

Problem. Determine whether $f(z) = \frac{z^2 - 1}{z - 1}$ is continuous at $z = 1$ .

For $z \neq 1$ : $f(z) = z + 1$ . The limit as $z \to 1$ is $2$ But $f(1)$ is undefined (division by zero). If we define $f(1) = 2$ Then $f$ becomes continuous at $z = 1$ .

2.3 The Derivative

Definition. $f$ is differentiable at $z_0$ if

$f"(z_0) = \lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}$

Exists (and is independent of how $h \to 0$ in $\mathbb{C}$ ).

Remark. The requirement that the limit be the same for all directions of approach of $h$ is what Makes complex differentiability far more restrictive than real differentiability.

2.4 Analytic Functions

Definition. A function $f$ is analytic (or holomorphic) on an open set $U \subseteq \mathbb{C}$ if $f$ is differentiable at every point of $U$ . A function that is analytic On all of $\mathbb{C}$ is called entire.

Examples of entire functions: $z^n$ , $e^z$ , $\sin z$ , $\cos z$ Polynomials.

Example of a non-analytic function: $f(z) = \bar{z}$ is nowhere differentiable (except at $z = 0$ if we define it, but still not analytic there).

Solution

Problem. Show that $f(z) = |z|^2$ is differentiable only at $z = 0$ .

$f(z) = x^2 + y^2$ So $u = x^2 + y^2$ and $v = 0$ . $u_x = 2x$ , $u_y = 2y$ , $v_x = 0$ , $v_y = 0$ . The Cauchy-Riemann equations require $2x = 0$ and $2y = 0$ So $x = y = 0$ . Thus $f$ satisfies CR only at $z = 0$ .

At $z = 0$ : $f'(0) = \lim_{h \to 0} \frac{|h|^2}{h} = \lim_{h \to 0} \bar{h} = 0$ So $f$ is Differentiable at $0$ but not analytic anywhere (no neighbourhood of $0$ is analytic).

Problem. Show that $f(z) = z\bar{z} + z$ is differentiable only at $z = 0$ .

$f(z) = |z|^2 + z = (x^2 + y^2 + x) + iy$ . $u_x = 2x + 1$ , $u_y = 2y$ , $v_x = 0$ , $v_y = 1$ . CR equations: $2x + 1 = 1 \Rightarrow x = 0$ And $2y = 0 \Rightarrow y = 0$ . At $(0, 0)$ : $f'(0) = \lim_{h \to 0} \frac{h\bar{h} + h}{h} = \lim_{h \to 0} (\bar{h} + 1) = 1$ . So $f$ is differentiable at $z = 0$ only, hence nowhere analytic.

2.5 Branch Cuts and Multi-Valued Functions

Many important functions in complex analysis are inherently multi-valued. To work with them as Single-valued functions, we must restrict the domain.

Definition. A branch of a multi-valued function $f$ is a single-valued analytic function $g$ Defined on a domain $D$ such that $g(z) \in f(z)$ for all $z \in D$ .

The Complex Logarithm. We define $\log z = \ln|z| + i\arg(z)$ Which is multi-valued because $\arg(z) = \mathrm{Arg}(z) + 2\pi k$ for $k \in \mathbb{Z}$ . The principal branch is

$\mathrm{Log}\, z = \ln|z| + i\,\mathrm{Arg}(z)$

Defined on $\mathbb{C} \setminus (-\infty, 0]$ . The negative real axis is called the branch cut.

Proposition 2.2. The principal branch $\mathrm{Log}\, z$ is analytic on $\mathbb{C} \setminus (-\infty, 0]$ and $\frac{d}{dz}\,\mathrm{Log}\, z = \frac{1}{z}$ .

Complex Powers. For $z, \alpha \in \mathbb{C}$ with $z \neq 0$ :

$z^\alpha = e^{\alpha \log z}$

This is multi-valued . When $\alpha$ is rational with reduced form $p/q$ There are exactly $q$ distinct values.

Solution

Problem. Find all values of $(-1)^i$ .

$(-1)^i = e^{i \log(-1)} = e^{i(i\pi + 2\pi i k)} = e^{-\pi - 2\pi k}$ for $k \in \mathbb{Z}$ .

These are all positive real numbers: $\ldots, e^{3\pi}, e^{\pi}, e^{-\pi}, e^{-3\pi}, \ldots$ . The principal value (using the principal branch) is $e^{-\pi}$ .

Problem. Find all values of $i^{1/2}$ .

$i^{1/2} = e^{(1/2)\log i} = e^{(1/2)(i\pi/2 + 2\pi i k)} = e^{i\pi/4 + i\pi k}$ .

For $k = 0$ : $e^{i\pi/4} = \frac{\sqrt{2}}{2}(1 + i)$ . For $k = 1$ : $e^{i5\pi/4} = -\frac{\sqrt{2}}{2}(1 + i)$ . These are the two square roots of $i$ .

Problem. Find the domain of analyticity of $f(z) = \mathrm{Log}(z^2 + 1)$ .

$\mathrm{Log}\, w$ is analytic on $\mathbb{C} \setminus (-\infty, 0]$ So we need $z^2 + 1 \notin (-\infty, 0]$ .

$z^2 + 1 \leq 0$ when $z^2 \leq -1$ I.e., $z \in [-i, 0] \cup [0, i]$ (the imaginary axis Segment from $-i$ to $i$ ). Also $z^2 + 1 = 0$ at $z = \pm i$ .

Domain: $\mathbb{C} \setminus \{z : z = iy,\, y \in [-1, 1]\}$ .

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