- §1.9(i) Complex Numbers
- §1.9(ii) Continuity, Point Sets, and Differentiation
- §1.9(iii) Integration
- §1.9(iv) Conformal Mapping
- §1.9(v) Infinite Sequences and Series
- §1.9(vi) Power Series
- §1.9(vii) Inversion of Limits

1.9.1 | $$z=x+\mathrm{i}y,$$ | ||

$x,y\in \mathbb{R}$, | |||

such that ${\mathrm{i}}^{2}=-1$.

1.9.2 | $\mathrm{\Re}z$ | $=x,$ | ||

$\mathrm{\Im}z$ | $=y.$ | |||

1.9.3 | $x$ | $=r\mathrm{cos}\theta ,$ | ||

$y$ | $=r\mathrm{sin}\theta ,$ | |||

where

1.9.4 | $$r={({x}^{2}+{y}^{2})}^{1/2},$$ | ||

and when $z\ne 0$,

1.9.5 | $$\theta =\omega ,\pi -\omega ,-\pi +\omega ,\text{or}-\omega ,$$ | ||

according as $z$ lies in the 1st, 2nd, 3rd, or 4th quadrants. Here

1.9.6 | $$\omega =\mathrm{arctan}\left(|y/x|\right)\in [0,\frac{1}{2}\pi ].$$ | ||

1.9.7 | $|z|$ | $=r,$ | ||

$\mathrm{ph}z$ | $=\theta +2n\pi ,$ | |||

$n\in \mathbb{Z}$. | ||||

The *principal value*
of $\mathrm{ph}z$ corresponds to $n=0$, that is, $-\pi \le \mathrm{ph}z\le \pi $. It
is single-valued on $\u2102\setminus \{0\}$, except on the interval
$(-\mathrm{\infty},0)$ where it is discontinuous and two-valued. *Unless indicated
otherwise*, these principal values are assumed throughout
the DLMF. (However, if we
require a principal value to be single-valued, then we can restrict
$$.)

1.9.8 | $|\mathrm{\Re}z|$ | $\le |z|,$ | ||

$|\mathrm{\Im}z|$ | $\le |z|,$ | |||

1.9.9 | $$z=r{\mathrm{e}}^{\mathrm{i}\theta},$$ | ||

where

1.9.10 | $${\mathrm{e}}^{\mathrm{i}\theta}=\mathrm{cos}\theta +\mathrm{i}\mathrm{sin}\theta ;$$ | ||

see §4.14.

1.9.11 | $\overline{z}$ | $=x-\mathrm{i}y,$ | ||

1.9.12 | $|\overline{z}|$ | $=|z|,$ | ||

1.9.13 | $\mathrm{ph}\overline{z}$ | $=-\mathrm{ph}z.$ | ||

If ${z}_{1}={x}_{1}+\mathrm{i}{y}_{1}$, ${z}_{2}={x}_{2}+\mathrm{i}{y}_{2}$, then

1.9.14 | $${z}_{1}\pm {z}_{2}={x}_{1}\pm {x}_{2}+\mathrm{i}({y}_{1}\pm {y}_{2}),$$ | ||

1.9.15 | $${z}_{1}{z}_{2}={x}_{1}{x}_{2}-{y}_{1}{y}_{2}+\mathrm{i}({x}_{1}{y}_{2}+{x}_{2}{y}_{1}),$$ | ||

1.9.16 | $$\frac{{z}_{1}}{{z}_{2}}=\frac{{z}_{1}{\overline{z}}_{2}}{{|{z}_{2}|}^{2}}=\frac{{x}_{1}{x}_{2}+{y}_{1}{y}_{2}+\mathrm{i}({x}_{2}{y}_{1}-{x}_{1}{y}_{2})}{{x}_{2}^{2}+{y}_{2}^{2}},$$ | ||

provided that ${z}_{2}\ne 0$. Also,

1.9.17 | $$|{z}_{1}{z}_{2}|=|{z}_{1}||{z}_{2}|,$$ | ||

1.9.18 | $$\mathrm{ph}\left({z}_{1}{z}_{2}\right)=\mathrm{ph}{z}_{1}+\mathrm{ph}{z}_{2},$$ | ||

1.9.19 | $$\left|\frac{{z}_{1}}{{z}_{2}}\right|=\frac{|{z}_{1}|}{|{z}_{2}|},$$ | ||

1.9.20 | $$\mathrm{ph}\frac{{z}_{1}}{{z}_{2}}=\mathrm{ph}{z}_{1}-\mathrm{ph}{z}_{2}.$$ | ||

Equations (1.9.18) and (1.9.20) hold for general values of the phases, but not necessarily for the principal values.

1.9.21 | $${z}^{n}=\left({x}^{n}-\left(\genfrac{}{}{0.0pt}{}{n}{2}\right){x}^{n-2}{y}^{2}+\left(\genfrac{}{}{0.0pt}{}{n}{4}\right){x}^{n-4}{y}^{4}-\mathrm{\cdots}\right)+\mathrm{i}\left(\left(\genfrac{}{}{0.0pt}{}{n}{1}\right){x}^{n-1}y-\left(\genfrac{}{}{0.0pt}{}{n}{3}\right){x}^{n-3}{y}^{3}+\mathrm{\cdots}\right),$$ | ||

$n=1,2,\mathrm{\dots}$. | |||

1.9.22 | $$\mathrm{cos}n\theta +\mathrm{i}\mathrm{sin}n\theta ={(\mathrm{cos}\theta +\mathrm{i}\mathrm{sin}\theta )}^{n},$$ | ||

$n\in \mathbb{Z}$. | |||

1.9.23 | $$\left|\left|{z}_{1}\right|-\left|{z}_{2}\right|\right|\le \left|{z}_{1}+{z}_{2}\right|\le \left|{z}_{1}\right|+\left|{z}_{2}\right|.$$ | ||

A function $f(z)$ is *continuous* at a point ${z}_{0}$ if
$\underset{z\to {z}_{0}}{lim}f(z)=f({z}_{0})$. That is, given any positive number
$\u03f5$, however small, we can find a positive number $\delta $ such that
$$ for all $z$ in the open disk $$.

A *neighborhood of a point* ${z}_{0}$ is a disk $$. An
*open set* in $\u2102$ is one in which each point has a neighborhood
that is contained in the set.

A point ${z}_{0}$ is a *limit point* (*limiting point* or
*accumulation point*) of a set of points $S$ in $\u2102$ (or
$\u2102\cup \mathrm{\infty}$) if every neighborhood of ${z}_{0}$ contains a point of $S$
distinct from ${z}_{0}$. (${z}_{0}$ may or may not belong to $S$.) As a consequence,
every neighborhood of a limit point of $S$ contains an infinite number of
points of $S$.
Also, the union of $S$ and its limit points is the *closure* of $S$.

A *domain*
$D$, say, is an open set in $\u2102$ that is *connected*,
that is, any two points can be joined by a polygonal arc (a finite chain of
straight-line segments) lying in the set. Any point whose neighborhoods always
contain members and nonmembers of $D$ is a *boundary point*
of $D$. When its boundary points are added the domain is said to be
*closed*,
but unless specified otherwise a domain is assumed to be open.

A *region* is an open domain together with none, some, or all of its
boundary points. Points of a region that are not boundary points are called
*interior points*.

A function $f(z)$ is *continuous on a region* $R$ if for each point ${z}_{0}$
in $R$ and any given number $\u03f5$ ($>0$) we can find a neighborhood of
${z}_{0}$ such that $$ for all points $z$ in the
intersection of the neighborhood with $R$.

A function $f(z)$ is *differentiable* at a point $z$ if the following
limit exists:

1.9.24 | $${f}^{\prime}(z)=\frac{df}{dz}=\underset{h\to 0}{lim}\frac{f(z+h)-f(z)}{h}.$$ | ||

Differentiability automatically implies continuity.

If ${f}^{\prime}(z)$ exists at $z=x+\mathrm{i}y$ and $f(z)=u(x,y)+\mathrm{i}v(x,y)$, then

1.9.25 | $\frac{\partial u}{\partial x}$ | $={\displaystyle \frac{\partial v}{\partial y}},$ | ||

$\frac{\partial u}{\partial y}$ | $=-{\displaystyle \frac{\partial v}{\partial x}}$ | |||

at $(x,y)$.

Conversely, if at a given point $(x,y)$ the partial derivatives $\partial u/\partial x$, $\partial u/\partial y$, $\partial v/\partial x$, and $\partial v/\partial y$ exist, are continuous, and satisfy (1.9.25), then $f(z)$ is differentiable at $z=x+\mathrm{i}y$.

A function $f(z)$ is said to be *analytic* (*holomorphic*) at
$z={z}_{0}$ if it is differentiable in a neighborhood of ${z}_{0}$.

A function $f(z)$ is *analytic in a domain*
$D$ if it is analytic at each point of $D$. A function analytic at every point
of $\u2102$ is said to be *entire*.

If $f(z)$ is analytic in an open domain $D$, then each of its derivatives ${f}^{\prime}(z)$, ${f}^{\prime \prime}(z)$, $\mathrm{\dots}$ exists and is analytic in $D$.

If $f(z)=u(x,y)+\mathrm{i}v(x,y)$ is analytic in an open domain $D$, then $u$ and
$v$ are
*harmonic* in $D$, that is,

1.9.26 | $$\frac{{\partial}^{2}u}{{\partial x}^{2}}+\frac{{\partial}^{2}u}{{\partial y}^{2}}=\frac{{\partial}^{2}v}{{\partial x}^{2}}+\frac{{\partial}^{2}v}{{\partial y}^{2}}=0,$$ | ||

or in polar form ((1.9.3)) $u$ and $v$ satisfy

1.9.27 | $$\frac{{\partial}^{2}u}{{\partial r}^{2}}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{{r}^{2}}\frac{{\partial}^{2}u}{{\partial \theta}^{2}}=0$$ | ||

at all points of $D$.

An *arc* $C$ is given by $z(t)=x(t)+\mathrm{i}y(t)$, $a\le t\le b$, where
$x$ and $y$ are continuously differentiable. If $x(t)$ and $y(t)$ are
continuous and ${x}^{\prime}(t)$ and ${y}^{\prime}(t)$ are piecewise continuous, then $z(t)$
defines a *contour*.

A contour is *simple* if it contains no multiple points, that is, for
every pair of distinct values ${t}_{1},{t}_{2}$ of $t$, $z({t}_{1})\ne z({t}_{2})$. A
*simple closed contour* is a simple contour, except that $z(a)=z(b)$.

Next,

1.9.28 | $${\int}_{C}f(z)dz={\int}_{a}^{b}f(z(t))({x}^{\prime}(t)+\mathrm{i}{y}^{\prime}(t))dt,$$ | ||

for a contour $C$ and $f(z(t))$ continuous, $a\le t\le b$. If $f(z({t}_{0}))=\mathrm{\infty}$, $a\le {t}_{0}\le b$, then the integral is defined analogously to the infinite integrals in §1.4(v). Similarly when $a=-\mathrm{\infty}$ or $b=+\mathrm{\infty}$.

Any simple closed contour $C$ divides $\u2102$ into two open domains that
have $C$ as common boundary. One of these domains is bounded and is called the
*interior domain of* $C$; the other is unbounded and is called the
*exterior domain of* $C$.

If $f(z)$ is continuous within and on a simple closed contour $C$ and analytic within $C$, then

1.9.29 | $${\int}_{C}f(z)dz=0.$$ | ||

If $f(z)$ is continuous within and on a simple closed contour $C$ and analytic within $C$, and if ${z}_{0}$ is a point within $C$, then

1.9.30 | $$f({z}_{0})=\frac{1}{2\pi \mathrm{i}}{\int}_{C}\frac{f(z)}{z-{z}_{0}}dz,$$ | ||

and

1.9.31 | $${f}^{(n)}({z}_{0})=\frac{n!}{2\pi \mathrm{i}}{\int}_{C}\frac{f(z)}{{(z-{z}_{0})}^{n+1}}dz,$$ | ||

$n=1,2,3,\mathrm{\dots}$, | |||

provided that in both cases $C$ is described in the positive rotational (anticlockwise) sense.

Any bounded entire function is a constant.

If $C$ is a closed contour, and ${z}_{0}\notin C$, then

1.9.32 | $$\frac{1}{2\pi \mathrm{i}}{\int}_{C}\frac{1}{z-{z}_{0}}dz=\mathcal{N}(C,{z}_{0}),$$ | ||

where $\mathcal{N}(C,{z}_{0})$ is an integer called the *winding number of $C$
with respect to ${z}_{\mathrm{0}}$*. If $C$ is simple and oriented in the positive
rotational sense, then $\mathcal{N}(C,{z}_{0})$ is $1$ or $0$ depending whether
${z}_{0}$ is inside or outside $C$.

For $u(z)$ harmonic,

1.9.33 | $$u(z)=\frac{1}{2\pi}{\int}_{0}^{2\pi}u(z+r{\mathrm{e}}^{\mathrm{i}\varphi})d\varphi .$$ | ||

If $h(w)$ is continuous on $|w|=R$, then with $z=r{\mathrm{e}}^{\mathrm{i}\theta}$

1.9.34 | $$u(r{\mathrm{e}}^{\mathrm{i}\theta})=\frac{1}{2\pi}{\int}_{0}^{2\pi}\frac{({R}^{2}-{r}^{2})h(R{\mathrm{e}}^{\mathrm{i}\varphi})d\varphi}{{R}^{2}-2Rr\mathrm{cos}\left(\varphi -\theta \right)+{r}^{2}}$$ | ||

is harmonic in $$. Also with $\left|w\right|=R$, $\underset{z\to w}{lim}u(z)=h(w)$ as $z\to w$ within $$.

The *extended complex plane*,
$\u2102\cup \{\mathrm{\infty}\}$, consists of the points of the complex plane
$\u2102$ together with an ideal point $\mathrm{\infty}$ called the *point at
infinity*.
A system of *open disks around infinity* is given by

1.9.35 | $${S}_{r}=\{z\mid |z|>1/r\}\cup \{\mathrm{\infty}\},$$ | ||

$$. | |||

Each ${S}_{r}$ is a *neighborhood*
of $\mathrm{\infty}$. Also,

1.9.36 | $$\mathrm{\infty}\pm z=z\pm \mathrm{\infty}=\mathrm{\infty},$$ | ||

1.9.37 | $$\mathrm{\infty}\cdot z=z\cdot \mathrm{\infty}=\mathrm{\infty},$$ | ||

$z\ne 0$, | |||

1.9.38 | $$z/\mathrm{\infty}=0,$$ | ||

1.9.39 | $$z/0=\mathrm{\infty},$$ | ||

$z\ne 0$. | |||

A function $f(z)$ is *analytic at* $\mathrm{\infty}$ if $g(z)=f(1/z)$ is analytic
at $z=0$, and we set ${f}^{\prime}(\mathrm{\infty})={g}^{\prime}(0)$.

Suppose $f(z)$ is analytic in a domain $D$ and ${C}_{1},{C}_{2}$ are two arcs in $D$
passing through ${z}_{0}$. Let ${C}_{1}^{\prime},{C}_{2}^{\prime}$ be the images of ${C}_{1}$ and ${C}_{2}$ under
the mapping $w=f(z)$. The *angle between* ${C}_{1}$ *and* ${C}_{2}$ at
${z}_{0}$ is the angle between the tangents to the two arcs at ${z}_{0}$, that is, the
difference of the signed angles that the tangents make with the positive
direction of the real axis. If ${f}^{\prime}({z}_{0})\ne 0$, then the angle between ${C}_{1}$
and ${C}_{2}$ equals the angle between ${C}_{1}^{\prime}$ and ${C}_{2}^{\prime}$ both in magnitude and
sense. We then say that the mapping $w=f(z)$ is *conformal*
(angle-preserving) at ${z}_{0}$.

The *linear transformation* $f(z)=az+b$, $a\ne 0$, has ${f}^{\prime}(z)=a$
and $w=f(z)$ maps $\u2102$ conformally onto $\u2102$.

1.9.40 | $$w=f(z)=\frac{az+b}{cz+d},$$ | ||

$ad-bc\ne 0$, $c\ne 0$. | |||

1.9.41 | $f(-d/c)$ | $=\mathrm{\infty},$ | ||

$f(\mathrm{\infty})$ | $=a/c.$ | |||

1.9.42 | $${f}^{\prime}(z)=\frac{ad-bc}{{(cz+d)}^{2}},$$ | ||

$z\ne -d/c$. | |||

1.9.43 | $${f}^{\prime}(\mathrm{\infty})=\frac{bc-ad}{{c}^{2}}.$$ | ||

1.9.44 | $$z=\frac{dw-b}{-cw+a}.$$ | ||

The transformation (1.9.40) is a one-to-one conformal mapping of $\u2102\cup \{\mathrm{\infty}\}$ onto itself.

The *cross ratio*
of ${z}_{1},{z}_{2},{z}_{3},{z}_{4}\in \u2102\cup \{\mathrm{\infty}\}$ is defined by

1.9.45 | $$\frac{({z}_{1}-{z}_{2})({z}_{3}-{z}_{4})}{({z}_{1}-{z}_{4})({z}_{3}-{z}_{2})},$$ | ||

or its limiting form, and is invariant under bilinear transformations.

Other names for the bilinear transformation are *fractional linear
transformation*, *homographic transformation*, and *Möbius
transformation*.

A sequence $\{{z}_{n}\}$ *converges*
to $z$ if $\underset{n\to \mathrm{\infty}}{lim}{z}_{n}=z$. For ${z}_{n}={x}_{n}+\mathrm{i}{y}_{n}$, the
sequence $\{{z}_{n}\}$ converges iff the sequences $\{{x}_{n}\}$ and $\{{y}_{n}\}$
separately converge. A *series* ${\sum}_{n=0}^{\mathrm{\infty}}{z}_{n}$ *converges*
if the sequence ${s}_{n}={\sum}_{k=0}^{n}{z}_{k}$ converges. The series is
*divergent* if ${s}_{n}$ does not converge. The series converges
*absolutely* if ${\sum}_{n=0}^{\mathrm{\infty}}|{z}_{n}|$ converges. A series
${\sum}_{n=0}^{\mathrm{\infty}}{z}_{n}$ converges (diverges) absolutely when
$$ ($>1$), or when
$$ ($>1$).
Absolutely convergent series are also convergent.

Let $\{{f}_{n}(z)\}$ be a sequence of functions defined on a set $S$. This
sequence *converges pointwise*
to a function $f(z)$ if

1.9.46 | $$f(z)=\underset{n\to \mathrm{\infty}}{lim}{f}_{n}(z)$$ | ||

for each $z\in S$. The sequence *converges uniformly*
on $S$, if for every $\u03f5>0$ there exists an integer $N$, independent of
$z$, such that

1.9.47 | $$ | ||

for all $z\in S$ and $n\ge N$.

A series
${\sum}_{n=0}^{\mathrm{\infty}}{f}_{n}(z)$ *converges uniformly* on $S$, if the sequence
${s}_{n}(z)={\sum}_{k=0}^{n}{f}_{k}(z)$ converges uniformly on $S$.

Suppose $\{{M}_{n}\}$ is a sequence of real numbers such that ${\sum}_{n=0}^{\mathrm{\infty}}{M}_{n}$ converges and $|{f}_{n}(z)|\le {M}_{n}$ for all $z\in S$ and all $n\ge 0$. Then the series ${\sum}_{n=0}^{\mathrm{\infty}}{f}_{n}(z)$ converges uniformly on $S$.

A doubly-infinite series ${\sum}_{n=-\mathrm{\infty}}^{\mathrm{\infty}}{f}_{n}(z)$ converges (uniformly) on $S$ iff each of the series ${\sum}_{n=0}^{\mathrm{\infty}}{f}_{n}(z)$ and ${\sum}_{n=1}^{\mathrm{\infty}}{f}_{-n}(z)$ converges (uniformly) on $S$.

For a series ${\sum}_{n=0}^{\mathrm{\infty}}{a}_{n}{(z-{z}_{0})}^{n}$ there is a number $R$, $0\le R\le \mathrm{\infty}$, such that the series converges for all $z$ in $$ and
diverges for $z$ in $|z-{z}_{0}|>R$. The circle $|z-{z}_{0}|=R$ is called the
*circle of convergence*
of the series, and $R$ is the *radius of convergence*. Inside the circle
the sum of the series is an analytic function $f(z)$. For $z$ in
$|z-{z}_{0}|\le \rho $ ($$), the convergence is absolute and
uniform. Moreover,

1.9.48 | $${a}_{n}=\frac{{f}^{(n)}({z}_{0})}{n!},$$ | ||

and

1.9.49 | $$R=\underset{n\to \mathrm{\infty}}{lim\; inf}{|{a}_{n}|}^{-1/n}.$$ | ||

For the converse of this result see §1.10(i).

When $\sum {a}_{n}{z}^{n}$ and $\sum {b}_{n}{z}^{n}$ both converge

1.9.50 | $$\sum _{n=0}^{\mathrm{\infty}}({a}_{n}\pm {b}_{n}){z}^{n}=\sum _{n=0}^{\mathrm{\infty}}{a}_{n}{z}^{n}\pm \sum _{n=0}^{\mathrm{\infty}}{b}_{n}{z}^{n},$$ | ||

and

1.9.51 | $$\left(\sum _{n=0}^{\mathrm{\infty}}{a}_{n}{z}^{n}\right)\left(\sum _{n=0}^{\mathrm{\infty}}{b}_{n}{z}^{n}\right)=\sum _{n=0}^{\mathrm{\infty}}{c}_{n}{z}^{n},$$ | ||

where

1.9.52 | $${c}_{n}=\sum _{k=0}^{n}{a}_{k}{b}_{n-k}.$$ | ||

Next, let

1.9.53 | $$f(z)={a}_{0}+{a}_{1}z+{a}_{2}{z}^{2}+\mathrm{\cdots},$$ | ||

${a}_{0}\ne 0$. | |||

Then the expansions (1.9.54), (1.9.57), and (1.9.60) hold for all sufficiently small $|z|$.

1.9.54 | $$\frac{1}{f(z)}={b}_{0}+{b}_{1}z+{b}_{2}{z}^{2}+\mathrm{\cdots},$$ | ||

where

1.9.55 | ${b}_{0}$ | $=1/{a}_{0},$ | ||

${b}_{1}$ | $=-{a}_{1}/{a}_{0}^{2},$ | |||

${b}_{2}$ | $=({a}_{1}^{2}-{a}_{0}{a}_{2})/{a}_{0}^{3},$ | |||

1.9.56 | $${b}_{n}=-({a}_{1}{b}_{n-1}+{a}_{2}{b}_{n-2}+\mathrm{\cdots}+{a}_{n}{b}_{0})/{a}_{0},$$ | ||

$n\ge 1$. | |||

With ${a}_{0}=1$,

1.9.57 | $$\mathrm{ln}f(z)={q}_{1}z+{q}_{2}{z}^{2}+{q}_{3}{z}^{3}+\mathrm{\cdots},$$ | ||

(principal value), where

1.9.58 | ${q}_{1}$ | $={a}_{1},$ | ||

${q}_{2}$ | $=(2{a}_{2}-{a}_{1}^{2})/2,$ | |||

${q}_{3}$ | $=(3{a}_{3}-3{a}_{1}{a}_{2}+{a}_{1}^{3})/3,$ | |||

and

1.9.59 | $${q}_{n}=(n{a}_{n}-(n-1){a}_{1}{q}_{n-1}-(n-2){a}_{2}{q}_{n-2}-\mathrm{\cdots}-{a}_{n-1}{q}_{1})/n,$$ | ||

$n\ge 2$. | |||

Also,

1.9.60 | $${(f(z))}^{\nu}={p}_{0}+{p}_{1}z+{p}_{2}{z}^{2}+\mathrm{\cdots},$$ | ||

(principal value), where $\nu \in \u2102$,

1.9.61 | ${p}_{0}$ | $=1,$ | ||

${p}_{1}$ | $=\nu {a}_{1},$ | |||

${p}_{2}$ | $=\nu ((\nu -1){a}_{1}^{2}+2{a}_{2})/2,$ | |||

and

1.9.62 | $${p}_{n}=((\nu -n+1){a}_{1}{p}_{n-1}+(2\nu -n+2){a}_{2}{p}_{n-2}+\mathrm{\cdots}+((n-1)\nu -1){a}_{n-1}{p}_{1}+n\nu {a}_{n})/n,$$ | ||

$n\ge 1$. | |||

For the definitions of the principal values of $\mathrm{ln}f(z)$ and ${(f(z))}^{\nu}$ see §§4.2(i) and 4.2(iv).

Lastly, a power series can be differentiated any number of times within its circle of convergence:

1.9.63 | $${f}^{(m)}(z)=\sum _{n=0}^{\mathrm{\infty}}{\left(n+1\right)}_{m}{a}_{n+m}{(z-{z}_{0})}^{n},$$ | ||

$$, $m=0,1,2,\mathrm{\dots}$. | |||

A set of complex numbers $\{{z}_{m,n}\}$ where $m$ and $n$ take all positive
integer values is called a *double sequence*. It *converges to* $z$
if for every $\u03f5>0$, there is an integer $N$ such that

1.9.64 | $$ | ||

for all $m,n\ge N$. Suppose $\{{z}_{m,n}\}$ converges to $z$ and the repeated limits

1.9.65 | $\underset{m\to \mathrm{\infty}}{lim}\left(\underset{n\to \mathrm{\infty}}{lim}{z}_{m,n}\right),$ | ||

$\underset{n\to \mathrm{\infty}}{lim}\left(\underset{m\to \mathrm{\infty}}{lim}{z}_{m,n}\right)$ | |||

exist. Then both repeated limits equal $z$.

A *double series*
is the limit of the double sequence

1.9.66 | $${z}_{p,q}=\sum _{m=0}^{p}\sum _{n=0}^{q}{\zeta}_{m,n}.$$ | ||

If the limit exists, then the double series is *convergent*; otherwise it
is *divergent*. The double series is *absolutely convergent* if it is
convergent when ${\zeta}_{m,n}$ is replaced by $|{\zeta}_{m,n}|$.

If a double series is absolutely convergent, then it is also convergent and its sum is given by either of the repeated sums

1.9.67 | $\sum _{m=0}^{\mathrm{\infty}}}\left({\displaystyle \sum _{n=0}^{\mathrm{\infty}}}{\zeta}_{m,n}\right),$ | ||

$\sum _{n=0}^{\mathrm{\infty}}}\left({\displaystyle \sum _{m=0}^{\mathrm{\infty}}}{\zeta}_{m,n}\right).$ | |||

Suppose the series ${\sum}_{n=0}^{\mathrm{\infty}}{f}_{n}(z)$, where ${f}_{n}(z)$ is continuous,
converges uniformly on every *compact set*
of a domain $D$, that is, every closed and bounded set in $D$. Then

1.9.68 | $${\int}_{C}\sum _{n=0}^{\mathrm{\infty}}{f}_{n}(z)dz=\sum _{n=0}^{\mathrm{\infty}}{\int}_{C}{f}_{n}(z)dz$$ | ||

for any finite contour $C$ in $D$.

Let $(a,b)$ be a finite or infinite interval, and ${f}_{0}(t),{f}_{1}(t),\mathrm{\dots}$ be real or complex continuous functions, $t\in (a,b)$. Suppose ${\sum}_{n=0}^{\mathrm{\infty}}{f}_{n}(t)$ converges uniformly in any compact interval in $(a,b)$, and at least one of the following two conditions is satisfied:

1.9.69 | $$ | ||

1.9.70 | $$ | ||

Then

1.9.71 | $${\int}_{a}^{b}\sum _{n=0}^{\mathrm{\infty}}{f}_{n}(t)dt=\sum _{n=0}^{\mathrm{\infty}}{\int}_{a}^{b}{f}_{n}(t)dt.$$ | ||