- §1.10(i) Taylor’s Theorem for Complex Variables
- §1.10(ii) Analytic Continuation
- §1.10(iii) Laurent Series
- §1.10(iv) Residue Theorem
- §1.10(v) Maximum-Modulus Principle
- §1.10(vi) Multivalued Functions
- §1.10(vii) Inverse Functions
- §1.10(viii) Functions Defined by Contour Integrals
- §1.10(ix) Infinite Products
- §1.10(x) Infinite Partial Fractions

Let $f(z)$ be analytic on the disk $$. Then

1.10.1 | $$f(z)=\sum _{n=0}^{\mathrm{\infty}}\frac{{f}^{(n)}({z}_{0})}{n!}{(z-{z}_{0})}^{n}.$$ | ||

The right-hand side is the *Taylor series for* $f(z)$ *at* $z={z}_{0}$,
and its radius of convergence is at least $R$.

1.10.2 | $${\mathrm{e}}^{z}=1+\frac{z}{1!}+\frac{{z}^{2}}{2!}+\mathrm{\cdots},$$ | ||

$$, | |||

1.10.3 | $$\mathrm{ln}\left(1+z\right)=z-\frac{{z}^{2}}{2}+\frac{{z}^{3}}{3}-\mathrm{\cdots},$$ | ||

$$, | |||

1.10.4 | $${(1-z)}^{-\alpha}=1+\alpha z+\frac{\alpha (\alpha +1)}{2!}{z}^{2}+\frac{\alpha (\alpha +1)(\alpha +2)}{3!}{z}^{3}+\mathrm{\cdots},$$ | ||

$$. | |||

Note that (1.10.4) is a generalization of the binomial expansion (1.2.2) with the binomial coefficients defined in (1.2.6). Again, in these examples $\mathrm{ln}\left(1+z\right)$ and ${(1-z)}^{-\alpha}$ have their principal values; see §§4.2(i) and 4.2(iv).

An analytic function $f(z)$ has a *zero of order* (or *multiplicity*)
$m$ ($\ge 1$)
at ${z}_{0}$ if the first nonzero coefficient in its Taylor series at ${z}_{0}$ is that
of ${(z-{z}_{0})}^{m}$. When $m=1$ the zero is *simple*.

Let ${f}_{1}(z)$ be analytic in a domain ${D}_{1}$. If ${f}_{2}(z)$, analytic in ${D}_{2}$,
equals ${f}_{1}(z)$ on an arc in $D={D}_{1}\cap {D}_{2}$, or on just an infinite number
of points with a limit point in $D$, then they are equal throughout $D$ and
${f}_{2}(z)$ is called an *analytic continuation* of ${f}_{1}(z)$. We write
$({f}_{1},{D}_{1})$, $({f}_{2},{D}_{2})$ to signify this continuation.

Suppose $z(t)=x(t)+\mathrm{i}y(t)$, $a\le t\le b$, is an arc and $$. Suppose the subarc $z(t)$, $t\in [{t}_{j-1},{t}_{j}]$
is contained in a domain ${D}_{j}$, $j=1,\mathrm{\dots},n$. The function ${f}_{1}(z)$ on ${D}_{1}$
is said to be *analytically continued along the path* $z(t)$,
$a\le t\le b$, if there is a chain $({f}_{1},{D}_{1})$, $({f}_{2},{D}_{2}),\mathrm{\dots},({f}_{n},{D}_{n})$.

Analytic continuation is a powerful aid in establishing transformations or functional equations for complex variables, because it enables the problem to be reduced to: (a) deriving the transformation (or functional equation) with real variables; followed by (b) finding the domain on which the transformed function is analytic.

Let $C$ be a simple closed contour consisting of a segment $\mathit{AB}$ of the real
axis and a contour in the upper half-plane joining the ends of $\mathit{AB}$. Also, let
$f(z)$ be analytic within $C$, continuous within and on $C$, and real on $\mathit{AB}$.
Then $f(z)$ can be continued analytically across $\mathit{AB}$ by *reflection*,
that is,

1.10.5 | $$f(\overline{z})=\overline{f(z)}.$$ | ||

Suppose $f(z)$ is analytic in the *annulus*
$$, $$, and $r\in ({r}_{1},{r}_{2})$.
Then

1.10.6 | $$f(z)=\sum _{n=-\mathrm{\infty}}^{\mathrm{\infty}}{a}_{n}{(z-{z}_{0})}^{n},$$ | ||

where

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

and the integration contour is described once in the positive sense. The series (1.10.6) converges uniformly and absolutely on compact sets in the annulus.

Let ${r}_{1}=0$, so that the annulus becomes the *punctured neighborhood*
$N$: $$, and assume that $f(z)$ is analytic in $N$, but not at
${z}_{0}$. Then $z={z}_{0}$ is an *isolated singularity*
of $f(z)$. This singularity is *removable*
if ${a}_{n}=0$ for all $$, and in this case the Laurent series becomes the
Taylor series. Next, ${z}_{0}$ is a *pole*
if ${a}_{n}\ne 0$ for at least one, but only finitely many, negative $n$. If
$-n$ is the first negative integer (counting from $-\mathrm{\infty}$) with
${a}_{-n}\ne 0$, then ${z}_{0}$ is a *pole of order* (or *multiplicity*)
$n$. Lastly, if ${a}_{n}\ne 0$ for infinitely many negative $n$, then ${z}_{0}$ is
an *isolated essential singularity*.

The singularities of $f(z)$ at infinity are classified in the same way as the singularities of $f(1/z)$ at $z=0$.

An isolated singularity ${z}_{0}$ is always removable when ${lim}_{z\to {z}_{0}}f(z)$ exists, for example $(\mathrm{sin}z)/z$ at $z=0$.

The coefficient ${a}_{-1}$ of ${(z-{z}_{0})}^{-1}$ in the Laurent series for $f(z)$
is called the *residue*
of $f(z)$ at ${z}_{0}$, and denoted by ${res}_{z={z}_{0}}[f(z)]$,
$\underset{z={z}_{0}}{res}[f(z)]$, or (when there
is no ambiguity) $res[f(z)]$.

A function whose only singularities, other than the point at infinity, are
poles is called a *meromorphic function*.
If the poles are infinite in number, then the point at infinity is called an
*essential singularity*:
it is the limit point of the poles.

In any neighborhood of an isolated essential singularity, however small, an analytic function assumes every value in $\u2102$ with at most one exception.

If $f(z)$ is analytic within a simple closed contour $C$, and continuous within and on $C$—except in both instances for a finite number of singularities within $C$—then

1.10.8 | $$\frac{1}{2\pi \mathrm{i}}{\int}_{C}f(z)dz=\text{sum of the residues of}f(z)\text{within}C.$$ | ||

Here and elsewhere in this subsection the path $C$ is described in the positive sense.

If the singularities within $C$ are poles and $f(z)$ is analytic and nonvanishing on $C$, then

1.10.9 | $$N-P=\frac{1}{2\pi \mathrm{i}}{\int}_{C}\frac{{f}^{\prime}(z)}{f(z)}dz=\frac{1}{2\pi}{\mathrm{\Delta}}_{C}(\mathrm{ph}f(z)),$$ | ||

where $N$ and $P$ are respectively the numbers of zeros and poles, counting multiplicity, of $f$ within $C$, and ${\mathrm{\Delta}}_{C}(\mathrm{ph}f(z))$ is the change in any continuous branch of $\mathrm{ph}\left(f(z)\right)$ as $z$ passes once around $C$ in the positive sense. For examples of applications see Olver (1997b, pp. 252–254).

In addition,

1.10.10 | $$\begin{array}{l}\frac{1}{2\pi \mathrm{i}}{\int}_{C}\frac{z{f}^{\prime}(z)}{f(z)}dz\\ \phantom{\rule{2em}{0ex}}=\text{(sum of locations of zeros)}-\text{(sum of locations of poles)},\end{array}$$ | ||

each location again being counted with multiplicity equal to that of the corresponding zero or pole.

If $f(z)$ and $g(z)$ are analytic on and inside a simple closed contour $C$, and $$ on $C$, then $f(z)$ and $f(z)+g(z)$ have the same number of zeros inside $C$.

If $f(z)$ is analytic in a domain $D$, ${z}_{0}\in D$ and $|f(z)|\le |f({z}_{0})|$ for all $z\in D$, then $f(z)$ is a constant in $D$.

Let $D$ be a bounded domain with boundary $\partial D$ and let $\overline{D}=D\cup \partial D$. If $f(z)$ is continuous on $\overline{D}$ and analytic in $D$, then $|f(z)|$ attains its maximum on $\partial D$.

If $u(z)$ is harmonic in $D$, ${z}_{0}\in D$, and $u(z)\le u({z}_{0})$ for all $z\in D$, then $u(z)$ is constant in $D$. Moreover, if $D$ is bounded and $u(z)$ is continuous on $\overline{D}$ and harmonic in $D$, then $u(z)$ is maximum at some point on $\partial D$.

In $$, if $f(z)$ is analytic, $|f(z)|\le M$, and $f(0)=0$, then

1.10.11 | $$|f(z)|\le \frac{M|z|}{R}\text{and}|{f}^{\prime}(0)|\le \frac{M}{R}.$$ | ||

Equalities hold iff $f(z)=Az$, where $A$ is a constant such that $\left|A\right|=M/R$.

Functions which have more than one value at a given point $z$ are called
*multivalued* (or *many-valued*) functions. Let $F(z)$ be a
multivalued function and $D$ be a domain. If we can assign a unique value
$f(z)$ to $F(z)$ at each point of $D$, and $f(z)$ is analytic on $D$, then
$f(z)$ is a *branch* of $F(z)$.

$F(z)=\sqrt{z}$ is two-valued for $z\ne 0$. If $D=\u2102\setminus (-\mathrm{\infty},0]$ and $z=r{\mathrm{e}}^{\mathrm{i}\theta}$, then one branch is $\sqrt{r}{\mathrm{e}}^{\mathrm{i}\theta /2}$, the other branch is $-\sqrt{r}{\mathrm{e}}^{\mathrm{i}\theta /2}$, with $$ in both cases. Similarly if $D=\u2102\setminus [0,\mathrm{\infty})$, then one branch is $\sqrt{r}{\mathrm{e}}^{\mathrm{i}\theta /2}$, the other branch is $-\sqrt{r}{\mathrm{e}}^{\mathrm{i}\theta /2}$, with $$ in both cases.

A *cut domain*
is one from which the points on finitely many nonintersecting simple contours
(§1.9(iii)) have been
removed. Each contour is called a *cut*. A *cut neighborhood* is
formed by deleting a ray emanating from the center. (Or more generally, a
simple contour that starts at the center and terminates on the boundary.)

Suppose $F(z)$ is multivalued and $a$ is a point such that there exists a
branch of $F(z)$ in a cut neighborhood of $a$, but there does not exist a
branch of $F(z)$ in any punctured neighborhood of $a$. Then $a$ is a
*branch point*
of $F(z)$. For example, $z=0$ is a branch point of $\sqrt{z}$.

Branches can be constructed in two ways:

(a) By introducing appropriate cuts from the branch points and restricting $F(z)$ to be single-valued in the cut plane (or domain).

(b) By specifying the value of $F(z)$ at a point ${z}_{0}$ (not a branch point), and requiring $F(z)$ to be continuous on any path that begins at ${z}_{0}$ and does not pass through any branch points or other singularities of $F(z)$.

If the path circles a branch point at $z=a$, $k$ times in the positive sense, and returns to ${z}_{0}$ without encircling any other branch point, then its value is denoted conventionally as $F(({z}_{0}-a){\mathrm{e}}^{2k\pi \mathrm{i}}+a)$.

Let $\alpha $ and $\beta $ be real or complex numbers that are not integers. The function $F(z)={(1-z)}^{\alpha}{(1+z)}^{\beta}$ is many-valued with branch points at $\pm 1$. Branches of $F(z)$ can be defined, for example, in the cut plane $D$ obtained from $\u2102$ by removing the real axis from $1$ to $\mathrm{\infty}$ and from $-1$ to $-\mathrm{\infty}$; see Figure 1.10.1. One such branch is obtained by assigning ${(1-z)}^{\alpha}$ and ${(1+z)}^{\beta}$ their principal values (§4.2(iv)).

Alternatively, take ${z}_{0}$ to be any point in $D$ and set $F({z}_{0})={\mathrm{e}}^{\alpha \mathrm{ln}\left(1-{z}_{0}\right)}{\mathrm{e}}^{\beta \mathrm{ln}\left(1+{z}_{0}\right)}$ where the logarithms assume their principal values. (Thus if ${z}_{0}$ is in the interval $(-1,1)$, then the logarithms are real.) Then the value of $F(z)$ at any other point is obtained by analytic continuation.

Thus if $F(z)$ is continued along a path that circles $z=1$ $m$ times in the positive sense and returns to ${z}_{0}$ without circling $z=-1$, then $F(({z}_{0}-1){\mathrm{e}}^{2m\pi \mathrm{i}}+1)={\mathrm{e}}^{\alpha \mathrm{ln}\left(1-{z}_{0}\right)}{\mathrm{e}}^{\beta \mathrm{ln}\left(1+{z}_{0}\right)}{\mathrm{e}}^{2\pi \mathrm{i}m\alpha}$. If the path also circles $z=-1$ $n$ times in the clockwise or negative sense before returning to ${z}_{0}$, then the value of $F({z}_{0})$ becomes ${\mathrm{e}}^{\alpha \mathrm{ln}\left(1-{z}_{0}\right)}{\mathrm{e}}^{\beta \mathrm{ln}\left(1+{z}_{0}\right)}{\mathrm{e}}^{2\pi \mathrm{i}m\alpha}{\mathrm{e}}^{-2\pi \mathrm{i}n\beta}$.

Suppose $f(z)$ is analytic at $z={z}_{0}$, ${f}^{\prime}({z}_{0})\ne 0$, and $f({z}_{0})={w}_{0}$. Then the equation

1.10.12 | $$f(z)=w$$ | ||

has a unique solution $z=F(w)$ analytic at $w={w}_{0}$, and

1.10.13 | $$F(w)={z}_{0}+\sum _{n=1}^{\mathrm{\infty}}{F}_{n}{(w-{w}_{0})}^{n}$$ | ||

in a neighborhood of ${w}_{0}$, where $n{F}_{n}$ is the residue of $1/{(f(z)-f({z}_{0}))}^{n}$ at $z={z}_{0}$. (In other words $n{F}_{n}$ is the coefficient of ${(z-{z}_{0})}^{-1}$ in the Laurent expansion of $1/{(f(z)-f({z}_{0}))}^{n}$ in powers of $(z-{z}_{0})$; compare §1.10(iii).)

Furthermore, if $g(z)$ is analytic at ${z}_{0}$, then

1.10.14 | $$g(F(w))=g({z}_{0})+\sum _{n=1}^{\mathrm{\infty}}{G}_{n}{(w-{w}_{0})}^{n},$$ | ||

where $n{G}_{n}$ is the residue of ${g}^{\prime}(z)/{(f(z)-f({z}_{0}))}^{n}$ at $z={z}_{0}$.

Suppose that

1.10.15 | $$f(z)=f({z}_{0})+\sum _{n=0}^{\mathrm{\infty}}{f}_{n}{(z-{z}_{0})}^{\mu +n},$$ | ||

where $\mu >0$, ${f}_{0}\ne 0$, and the series converges in a neighborhood of ${z}_{0}$. (For example, when $\mu $ is an integer $f(z)-f({z}_{0})$ has a zero of order $\mu $ at ${z}_{0}$.) Let ${w}_{0}=f({z}_{0})$. Then (1.10.12) has a solution $z=F(w)$, where

1.10.16 | $$F(w)={z}_{0}+\sum _{n=1}^{\mathrm{\infty}}{F}_{n}{(w-{w}_{0})}^{n/\mu}$$ | ||

in a neighborhood of ${w}_{0}$, $n{F}_{n}$ being the residue of $1/{(f(z)-f({z}_{0}))}^{n/\mu}$ at $z={z}_{0}$.

It should be noted that different branches of ${(w-{w}_{0})}^{1/\mu}$ used in forming ${(w-{w}_{0})}^{n/\mu}$ in (1.10.16) give rise to different solutions of (1.10.12). Also, if in addition $g(z)$ is analytic at ${z}_{0}$, then

1.10.17 | $$g(F(w))=g({z}_{0})+\sum _{n=1}^{\mathrm{\infty}}{G}_{n}{(w-{w}_{0})}^{n/\mu},$$ | ||

where $n{G}_{n}$ is the residue of ${g}^{\prime}(z)/{(f(z)-f({z}_{0}))}^{n/\mu}$ at $z={z}_{0}$.

Let $D$ be a domain and $[a,b]$ be a closed finite segment of the real axis. Assume that for each $t\in [a,b]$, $f(z,t)$ is an analytic function of $z$ in $D$, and also that $f(z,t)$ is a continuous function of both variables. Then

1.10.18 | $$F(z)={\int}_{a}^{b}f(z,t)dt$$ | ||

is analytic in $D$ and its derivatives of all orders can be found by differentiating under the sign of integration.

This result is also true when $b=\mathrm{\infty}$, or when $f(z,t)$ has a singularity at $t=b$, with the following conditions. For each $t\in [a,b)$, $f(z,t)$ is analytic in $D$; $f(z,t)$ is a continuous function of both variables when $z\in D$ and $t\in [a,b)$; the integral (1.10.18) converges at $b$, and this convergence is uniform with respect to $z$ in every compact subset $S$ of $D$.

The last condition means that given $\u03f5$ ($>0$) there exists a number ${a}_{0}\in [a,b)$ that is independent of $z$ and is such that

1.10.19 | $$ | ||

for all ${a}_{1}\in [{a}_{0},b)$ and all $z\in S$; compare §1.5(iv).

If $|f(z,t)|\le M(t)$ for $z\in S$ and ${\int}_{a}^{b}M(t)dt$ converges, then the integral (1.10.18) converges uniformly and absolutely in $S$.

Let ${p}_{k,m}={\prod}_{n=k}^{m}(1+{a}_{n})$. If for some $k\ge 1$, ${p}_{k,m}\to {p}_{k}\ne 0$ as $m\to \mathrm{\infty}$, then we say that the infinite product
${\prod}_{n=1}^{\mathrm{\infty}}(1+{a}_{n})$ *converges*. (The integer $k$ may be greater
than one to allow for a finite number of zero factors.) The convergence of the
product is *absolute* if ${\prod}_{n=1}^{\mathrm{\infty}}(1+|{a}_{n}|)$ converges. The
product ${\prod}_{n=1}^{\mathrm{\infty}}(1+{a}_{n})$, with ${a}_{n}\ne -1$ for all $n$,
converges iff ${\sum}_{n=1}^{\mathrm{\infty}}\mathrm{ln}\left(1+{a}_{n}\right)$ converges; and it converges
absolutely iff ${\sum}_{n=1}^{\mathrm{\infty}}|{a}_{n}|$ converges.

Suppose ${a}_{n}={a}_{n}(z)$, $z\in D$, a domain. The convergence of the infinite
product is *uniform*
if the sequence of partial products converges uniformly.

Suppose that ${a}_{n}(z)$ are analytic functions in $D$. If there is an $N$, independent of $z\in D$, such that

1.10.20 | $$|\mathrm{ln}\left(1+{a}_{n}(z)\right)|\le {M}_{n},$$ | ||

$n\ge N$, | |||

and

1.10.21 | $$ | ||

then the product ${\prod}_{n=1}^{\mathrm{\infty}}(1+{a}_{n}(z))$ converges uniformly to an analytic function $p(z)$ in $D$, and $p(z)=0$ only when at least one of the factors $1+{a}_{n}(z)$ is zero in $D$. This conclusion remains true if, in place of (1.10.20), $|{a}_{n}(z)|\le {M}_{n}$ for all $n$, and again $$.

If $\{{z}_{n}\}$ is a sequence such that ${\sum}_{n=1}^{\mathrm{\infty}}|{z}_{n}^{-2}|$ is convergent, then

1.10.22 | $$P(z)=\prod _{n=1}^{\mathrm{\infty}}\left(1-\frac{z}{{z}_{n}}\right){\mathrm{e}}^{z/{z}_{n}}$$ | ||

is an entire function with zeros at ${z}_{n}$.

Suppose $D$ is a domain, and

1.10.23 | $$F(z)=\prod _{n=1}^{\mathrm{\infty}}{a}_{n}(z),$$ | ||

$z\in D$, | |||

where ${a}_{n}(z)$ is analytic for all $n\ge 1$, and the convergence of the product is uniform in any compact subset of $D$. Then $F(z)$ is analytic in $D$.

If, also, ${a}_{n}(z)\ne 0$ when $n\ge 1$ and $z\in D$, then $F(z)\ne 0$ on $D$ and

1.10.24 | $$\frac{{F}^{\prime}(z)}{F(z)}=\sum _{n=1}^{\mathrm{\infty}}\frac{{a}_{n}^{\prime}(z)}{{a}_{n}(z)}.$$ | ||

If $\{{a}_{n}\}$ and $\{{z}_{n}\}$ are sequences such that ${z}_{m}\ne {z}_{n}$ ($m\ne n$) and ${\sum}_{n=1}^{\mathrm{\infty}}|{a}_{n}{z}_{n}^{-2}|$ is convergent, then

1.10.25 | $$f(z)=\sum _{n=1}^{\mathrm{\infty}}{a}_{n}\left(\frac{1}{z-{z}_{n}}+\frac{1}{{z}_{n}}\right)$$ | ||

is analytic in $\u2102$, except for simple poles at $z={z}_{n}$ of residue ${a}_{n}$.