% ----------------------------------------------------------------------

\chapter{The complex exponential}
\label{cexp}
\index{exponential, complex}
\index{complex exponential}
\index{natural exponential!complex}

The complex exponential, especially the complex natural exponential, is
ubiquitous in higher mathematics.  There seems hardly a corner of
calculus, basic or advanced, in which the complex exponential does not
strongly impress itself and frequently arise.  Because the complex
exponential emerges (at least pedagogically) out of the real
exponential---and, especially, because the complex \emph{natural}
exponential emerges out of the real natural
% bad break
ex\-po\-nen\-tial---this chapter introduces first the real natural
exponential and its inverse, the real natural logarithm; and then
proceeds to show how the two can operate on complex arguments.

This chapter develops the close relationship between the natural
exponential and chapter~\ref{trig}'s trigonometrics, showing that all
these (including the complex natural exponential) belong to a single
exponential/trigonometric family.  After developing the relationship,
the chapter works out the derivatives of the family's several functions.
Also, the chapter treats the functions' inverses and works out the
derivatives of these inverses---the derivatives of the inverses turning
out to be particularly interesting.

% ----------------------------------------------------------------------

\section{The real exponential}
\label{cexp:220}
\index{exponential, real}
\index{real exponential}
\index{natural exponential!real}
\index{natural exponential}
\index{exponential, natural}
\index{natural exponential!existence of}
\index{exponential, natural!existence of}

\index{$e$}
Consider the factor
\[
  ( 1 + \ep )^N.
\]
This is the overall factor by which a quantity grows after~$N$ iterative
rounds of multiplication\footnote{\label{cexp:230fn1}%
  For example, let a quantity~$A$ be multiplied by $(1+\ep)$, then by
  $(1+\ep)$ again, and then by $(1+\ep)$ a third time.  The product
  $(1+\ep)^3A$ results from these three rounds of multiplication.
  Overall, such a product is $(1+\ep)^3$ times the quantity~$A$ with
  which we started.  In this example, $N=3$.%
}
by $(1+\ep)$.  What happens when~$\ep$ is very
small but~$N$ is very large?  The really interesting question is, what
happens in the limit, as $\ep\rightarrow 0$ and $N\rightarrow\infty$,
while $x=\ep N$ remains a finite number?  The answer is that the factor
becomes
\bq{cexp:220:exp}
  \exp x \equiv \lim_{\ep\rightarrow 0} ( 1 + \ep )^{x/\ep}.
\eq
Equation~(\ref{cexp:220:exp}) defines the \emph{natural exponential
function}---commonly, more briefly named the \emph{exponential
function.}  Another way to write the same definition is
\bqa
  \exp x &=& e^x, \label{cexp:220:26}\\
  e &\equiv& \lim_{\ep\rightarrow 0} ( 1 + \ep )^{1/\ep}.
  \label{cexp:220:econst}
\eqa

\index{tangent line}
In whichever form we write it, the question remains as to whether the
limit actually exists; that is, whether $1<e<\infty$; whether in fact
we can put some concrete bound on~$e$.  To show that we can,%
\footnote{
  Excepting~(\ref{cexp:220:dexp}), the author would prefer to omit much
  of the rest of this section, but even at the applied level cannot
  think of a logically permissible way to do it.  It seems nonobvious
  that the limit $\lim_{\ep\rightarrow 0} ( 1 + \ep )^{1/\ep}$
  actually does exist.  The rest of this section shows why it does.
}
we observe per~(\ref{drvtv:defz}) and~(\ref{drvtv:230:apxex}) that the
derivative of the exponential function is
\bqb
  \frac{d}{dx}\exp x
    &=& \lim_{\delta\rightarrow 0}
      \frac{\exp(x+\delta/2)-\exp(x-\delta/2)}{\delta}\\
    &=& \lim_{\delta,\ep\rightarrow 0}
      \frac{
        (1+\ep)^{(x+\delta/2)/\ep} - (1+\ep)^{(x-\delta/2)/\ep}
      }{\delta} \\
    &=& \lim_{\delta,\ep\rightarrow 0}
      (1+\ep)^{x/\ep}
      \frac{
        (1+\ep)^{+\delta/2\ep} - (1+\ep)^{-\delta/2\ep}
      }{\delta} \\
    &=& \lim_{\delta,\ep\rightarrow 0}
      (1+\ep)^{x/\ep}
      \frac{
        (1+\delta/2) - (1-\delta/2)
      }{\delta} \\
    &=& \lim_{\ep\rightarrow 0}
      (1+\ep)^{x/\ep},
\eqb
which is to say that
\bq{cexp:220:dexp}
  \index{natural exponential!derivative of}
  \index{exponential, natural!derivative of}
  \index{derivative!of the natural exponential}
  \frac{d}{dx}\exp x = \exp x.
\eq
This is a curious, important result: the derivative of the exponential
function is the exponential function itself; the slope and height of the
exponential function are everywhere equal.  For the moment, however,
what interests us is that
\[
  \left. \frac{d}{dx}\exp x \right|_{x=0} = \exp 0 = \lim_{\ep\rightarrow 0}(1+\ep)^0 = 1,
\]
which says that the slope and height of the exponential function are both
unity at $x=0$, implying that the straight line which best approximates
the exponential function in that neighborhood---the \emph{tangent line,}
which just grazes the curve---is
\[
  y(x) = 1 + x.
\]

With the tangent line~$y(x)$ found, the next step toward putting a
concrete bound on~$e$ is to show that $y(x) \le \exp x$ for all
real~$x$, that the curve runs nowhere below the line.  To show this, we
observe per~(\ref{cexp:220:exp}) that the essential action of the
exponential function is to multiply repeatedly by $1+\ep$ as~$x$
increases, to divide repeatedly by $1+\ep$ as~$x$ decreases.  Since
$1+\ep > 1$, this action means for real~$x$ that
\[
  \exp x_1 \le \exp x_2 \ \ \mbox{if}\ \ x_1 \le x_2.
\]
However, a positive number remains positive no matter how many times one
multiplies or divides it by $1+\ep$, so the same action also means that
\[
  0 \le \exp x
\]
for all real~$x$.
In light of~(\ref{cexp:220:dexp}), the last two equations imply further
that
\bqb
  \left.\frac{d}{dx}\exp x\right|_{x=x_1} &\le& \left.\frac{d}{dx}\exp
  x\right|_{x=x_2}
    \ \ \ \mbox{if $x_1 \le x_2$,} \\
  0 &\le& \frac{d}{dx}\exp x.
\eqb
But we have purposely defined the tangent line $y(x) = 1+x$ such that
\vspace{-1.0ex}
\[
  \renewcommand\arraystretch{2.0}
  \br{rcccl}
    \ds\exp 0 &=& \ds y(0) &=& \ds 1, \\
    \ds\left.\frac{d}{dx}\exp x\right|_{x=0}
      &=& \ds\left.\frac{dy}{dx}\right|_{x=0} &=& \ds 1;
  \er
\]
that is, such that the line just grazes the curve of $\exp x$ at $x=0$.
Rightward, at $x > 0$, evidently the curve's slope only increases,
bending upward away from the line.  Leftward, at $x < 0$, evidently the
curve's slope only decreases, again bending upward away from the line.
Either way, the curve never crosses below the line for real~$x$.  In
symbols,
\[
  y(x) \le \exp x.
\]
Figure~\ref{cexp:220:fig1} depicts.
\begin{figure}
  \caption{The natural exponential.}
  \label{cexp:220:fig1}
  \bc
    \nc\fxa{-4.5} \nc\fxb{ 3.3}
    \nc\fya{-1.0} \nc\fyb{ 3.2}
    \nc\xta{-3.8} \nc\xtb{ 1.0}
    \nc\xxa{-4.0} \nc\xxb{ 2.5}
    \nc\xya{-0.6} \nc\xyb{ 2.5}
    \nc\xxt{0.2}
    \nc\xxk{-1.0} \nc\xxl{ 1.8}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      {
        \small
        \psset{dimen=middle}
        {
          \psset{linewidth=0.5pt}
          \psline(\xxa,0)(\xxb,0)
          \psline(0,\xya)(0,\xyb)
          \uput[r](\xxb,0){$x$}
          \uput[u](0,\xyb){$\exp x$}
          \psline(0,1.0)(\xxt,1.0)    \uput[r](\xxt,1.0){$1$}
          \psline(-1.0,0)(-1.0,-\xxt) \uput[d](-1.0,-\xxt){$-1$}
        }
        \psplot[linewidth=2.0pt,plotpoints=200]{\xta}{\xtb}{
          2.7183 x 1.0 div exp 1.0 mul
        }
        \rput(-0.5,0.5){
          \psline[linewidth=1.0pt,linestyle=dashed](\xxk,\xxk)(\xxl,\xxl)
        }
      }
    \end{pspicture}
  \ec
\end{figure}

Evaluating the last inequality at $x=-1/2$ and $x=1$, we have that
\settowidth\tla{$\ds\frac{1}{2}$}
\bqb
  \frac{1}{2} &\le& \exp\left(-\frac{1}{2}\right), \\
  \makebox[\tla][c]{$2$} &\le& \exp\left(1\right).
\eqb
But per~(\ref{cexp:220:26}) $\exp x = e^x$, so
\settowidth\tla{$\ds\frac{1}{2}$}
\bqb
  \frac{1}{2} &\le& e^{-1/2}, \\
  \makebox[\tla][c]{$2$} &\le& e^{1},
\eqb
or in other words,
\bq{cexp:220:45}
  2 \le e \le 4,
\eq
which in consideration of~(\ref{cexp:220:26}) puts the desired bound on
the exponential function.  The limit does exist.

\index{logarithmic derivative}
\index{derivative!logarithmic}
Dividing~(\ref{cexp:220:dexp}) by $\exp x$ yields the \emph{logarithmic
derivative} (\S~\ref{drvtv:240.40})
\bq{cexp:220:dexp-log}
  \index{natural exponential!logarithmic derivative of}
  \index{exponential, natural!logarithmic derivative of}
  \index{logarithmic derivative!of the natural exponential}
  \index{derivative!logarithmic of the natural exponential}
  \frac{d(\exp x)}{(\exp x)\,dx} = 1,
\eq
a form which expresses or captures the deep curiosity of the natural
exponential maybe even better than does~(\ref{cexp:220:dexp}).

By the Taylor series of Table~\ref{taylor:315:tbl}, the value
\[
  e \approx \mbox{0x2.B7E1}
\]
can readily be calculated, but the derivation of that series does not
come until chapter~\ref{taylor}.

% ----------------------------------------------------------------------

\section{The natural logarithm}
\label{cexp:225}
\index{natural logarithm}
\index{logarithm, natural}

\index{exponential!general}
\index{general exponential}
In the general exponential expression~$b^x$, one can choose any base~$b$;
for example, $b = 2$ is an interesting choice.  As we shall see in
\S~\ref{cexp:230}, however, it turns out that $b=e$, where~$e$ is the
constant introduced in~(\ref{cexp:220:econst}), is the most interesting
choice of all.  For this reason among others, the base-$e$ logarithm
is similarly interesting, such that we define for it the special
notation
\[
  \ln(\cdot) = \log_e(\cdot),
\]
and call it the \emph{natural logarithm.}  Just as for any other base~$b$,
so also for base $b=e$; thus the natural logarithm inverts the
natural exponential and vice versa:
\bq{cexp:225:10}
  \br{rclcl}
    \ds\ln \exp x &=& \ds\ln e^x &=& \ds x, \\
    \ds\exp \ln x &=& \ds e^{\ln x} &=& \ds x.
  \er
\eq
Figure~\ref{cexp:225:fig2} plots the natural logarithm.
\begin{figure}
  \caption{The natural logarithm.}
  \label{cexp:225:fig2}
  \bc
    \nc\fya{-4.0} \nc\fyb{ 2.3}
    \nc\fxa{-1.2} \nc\fxb{ 4.2}
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    \nc\xya{-3.6} \nc\xyb{ 1.5}
    \nc\xxa{-0.8} \nc\xxb{ 3.5}
    \nc\xxt{0.2}
    \nc\xxk{-1.0} \nc\xxl{ 1.8}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      {
        \small
        \psset{dimen=middle}
        {
          \psset{linewidth=0.5pt}
          \psline(\xxa,0)(\xxb,0)
          \psline(0,\xya)(0,\xyb)
          \uput[r](\xxb,0){$x$}
          \uput[u](0,\xyb){$\ln x$}
          \psline(0,-1.0)(-\xxt,-1.0) \uput[l](-\xxt,-1.0){$-1$}
          \psline(1.0,0)(1.0,\xxt) \uput[u](1.0,\xxt){$1$}
        }
        \psplot[linewidth=2.0pt,plotpoints=200]{\xta}{\xtb}{
          x 1.0 div ln 1.0 mul
        }
        \rput(0.5,-0.5){
          \psline[linewidth=1.0pt,linestyle=dashed](\xxk,\xxk)(\xxl,\xxl)
        }
      }
    \end{pspicture}
  \ec
\end{figure}

If
\[
  y = \ln x,
\]
then
\[
  x = \exp y,
\]
and per~(\ref{cexp:220:dexp}),
\[
  \frac{dx}{dy} = \exp y.
\]
But this means that
\[
  \frac{dx}{dy} = x,
\]
the inverse of which is
\[
  \frac{dy}{dx} = \frac{1}{x}.
\]
In other words,
\bq{cexp:225:dln}
  \index{natural logarithm!derivative of}
  \index{logarithm, natural!derivative of}
  \index{derivative!of the natural logarithm}
  \frac{d}{dx}\ln x = \frac{1}{x}.
\eq
Like many of the equations in these early chapters, here is another
rather significant result.%
\footnote{
  Besides the result itself, the technique which leads to the result is
  also interesting and is worth mastering.  We will use the technique
  more than once in this book.
}

One can specialize Table~\ref{alggeo:230:tbl}'s logarithmic
base-conversion identity to read
\bq{cexp:225:30}
  \log_b w = \frac{\ln w}{\ln b}.
\eq
This equation converts any logarithm to a natural logarithm.
Base $b=2$ logarithms are interesting, so we note here that
\[
  \ln 2 = -\ln\frac 1 2 \approx \mbox{0x0.B172},
\]
which chapter~\ref{taylor} and its Table~\ref{taylor:315:tbl} will show how
to calculate.

% ----------------------------------------------------------------------

\section{Fast and slow functions}
\label{cexp:228}
\index{exponential, natural!compared to~$x^a$}
\index{logarithm, natural!compared to~$x^a$}
\index{natural exponential!compared to~$x^a$}
\index{natural logarithm!compared to~$x^a$}
\index{function!slow}\index{slow function}
\index{function!fast}\index{fast function}

The exponential~$\exp x$ is a \emph{fast function.}  The logarithm~$\ln
x$ is a \emph{slow function.}  These functions grow, diverge or decay
respectively faster and slower than~$x^a$.

Such claims are proved by l'H\^opital's rule~(\ref{drvtv:260:lhopital}).
Applying the rule, we have that
{
  \settowidth{\tla}{$\ds\lim_{x\ra\infty}\frac{1}{ax^a}$}
  \nc\xx[1]{\makebox[\tla][l]{$\ds{#1}$}}
  \bq{cexp:228:50}
    \begin{split}
      \lim_{x\ra\infty}\frac{\ln x}{x^a}
        &= \xx{\lim_{x\ra\infty}\frac{-1}{ax^a}} =
        \begin{cases}
          0      &\mbox{if}\ a>0, \\
          +\infty &\mbox{if}\ a\le 0,
        \end{cases} \\
      \lim_{x\ra 0}\frac{\ln x}{x^a}
        &= \xx{\lim_{x\ra 0}\frac{-1}{ax^a}} =
        \begin{cases}
          -\infty &\mbox{if}\ a\ge 0, \\
          0      &\mbox{if}\ a<0,
        \end{cases}
    \end{split}
  \eq
}%
which reveals the logarithm to be a slow function.  Since the
$\exp(\cdot)$ and $\ln(\cdot)$ functions are mutual inverses, we can
leverage~(\ref{cexp:228:50}) to show also that
\bqb
  \lim_{x\ra\infty}\frac{\exp (\pm x)}{x^a}
  &=& \lim_{x\ra\infty}\exp\left[\ln\frac{\exp(\pm x)}{x^a}\right]
  \\&=& \lim_{x\ra\infty}\exp\left[\pm x-a\ln x\right]
  \\&=& \lim_{x\ra\infty}\exp
    \left[(x)\left(\pm 1-a\frac{\ln x}{x}\right)\right]
  \\&=& \lim_{x\ra\infty}\exp
    \left[(x)\left(\pm 1-0\right)\right]
  \\&=& \lim_{x\ra\infty}\exp
    \left[\pm x\right].
\eqb
That is,
\bq{cexp:228:55}
  \begin{split}
    \lim_{x\ra\infty}\frac{\exp (+x)}{x^a} &= \infty, \\
    \lim_{x\ra\infty}\frac{\exp (-x)}{x^a} &= 0,
  \end{split}
\eq
which reveals the exponential to be a fast function.  Exponentials grow
or decay faster than powers; logarithms diverge slower.

Such conclusions are extended to bases other than the natural base~$e$
simply by observing that $\log_b x = \ln x/\ln b$ and that $b^x = \exp(
x\ln b )$.  Thus exponentials generally are fast and logarithms
generally are slow, regardless of the base.%
\footnote{
  There are of course some degenerate edge cases like $b=0$ and $b=1$.
  The reader can detail these as the need arises.
}

It is interesting and worthwhile to contrast the sequence
\[
  \ldots,-\frac{3!}{x^4},\frac{2!}{x^3},-\frac{1!}{x^2},\frac{0!}{x^1},
  \frac{x^0}{0!},\frac{x^1}{1!},\frac{x^2}{2!},\frac{x^3}{3!},\frac{x^4}{4!},\ldots
\]
against the sequence
\[
  \ldots,-\frac{3!}{x^4},\frac{2!}{x^3},-\frac{1!}{x^2},\frac{0!}{x^1},
  \ln x,\frac{x^1}{1!},\frac{x^2}{2!},\frac{x^3}{3!},\frac{x^4}{4!},\ldots
\]
As $x \ra +\infty$, each sequence increases in magnitude going
rightward.  Also, each term in each sequence is the derivative with
respect to~$x$ of the term to its right---except left of the middle
element in the first sequence and right of the middle element in the
second.  The exception is peculiar.  What is going on here?

\index{logarithm!resemblance of to~$x^0$}
The answer is that~$x^0$ (which is just a constant) and $\ln x$
\emph{both are of zeroth order in~$x$.}  This seems strange at first
because $\ln x$ diverges as $x \ra \infty$ whereas~$x^0$ does not, but
the divergence of the former is extremely slow---so slow, in fact, that
per~(\ref{cexp:228:50})
$\lim_{x \ra \infty} (\ln x)/x^\ep = 0$
for any positive~$\ep$ no matter how small.%
\footnote{
  One does not grasp how truly slow the divergence is until one
  calculates a few concrete values.  Consider for instance how far
  out~$x$ must run to make $\ln x = \mbox{0x100}$.  It's a long, long
  way.  The natural logarithm does indeed eventually diverge to
  infinity, in the literal sense that there is no height it does
  not eventually reach, but it certainly does not hurry.  As we have
  seen, it takes practically forever just to reach~0x100.
}
Figure~\ref{cexp:225:fig2} has plotted $\ln x$ only for $x \sim 1$, but
beyond the figure's window the curve (whose slope is $1/x$) flattens
rapidly rightward, to the extent that it locally resembles the plot of a
constant value; and indeed one can write,
\[
  x^0 = \lim_{u \ra \infty} \frac{\ln (x+u)}{\ln u},
\]
which casts~$x^0$ as a logarithm shifted and scaled.  Admittedly, one
ought not strain such logic too far, because $\ln x$ is not in fact a
constant, but the point nevertheless remains that~$x^0$ and $\ln x$
often play analogous roles in mathematics.  The logarithm can in some
situations profitably be thought of as a ``diverging constant'' of
sorts.

\index{exponential!resemblance of to~$x^\infty$}
Less strange-seeming perhaps is the consequence of~(\ref{cexp:228:55})
that $\exp x$ is of infinite order in~$x$, that~$x^\infty$ and $\exp x$
play analogous roles.

It befits an applied mathematician to internalize
subjectively~(\ref{cexp:228:50}) and~(\ref{cexp:228:55}), to remember
that $\ln x$ resembles~$x^0$ and that
$\exp x$ resembles~$x^\infty$.  A qualitative sense that logarithms
are slow and exponentials, fast, helps one to grasp mentally the
essential features of many mathematical models one encounters in practice.

Now leaving aside fast and slow functions for the moment, we turn our
attention in the next section to the highly important matter of the
exponential of a complex argument.

% ----------------------------------------------------------------------

\section{Euler's formula}
\label{cexp:230}
\index{Euler's formula}
\index{Euler, Leonhard (1707--1783)}
\index{circle}

\index{unit!imaginary}
\index{imaginary unit}
The result of \S~\ref{cexp:220} leads to one of the central questions in
all of mathematics.  How can one evaluate
\[
  \exp i\theta = \lim_{\ep\rightarrow 0} ( 1 + \ep )^{i\theta/\ep},
\]
where $i^2 = -1$ is the imaginary unit introduced in
\S~\ref{alggeo:225}?

To begin, one can take advantage of~(\ref{drvtv:230:apxe}) to write
the last equation in the form
\[
  \exp i\theta = \lim_{\ep\rightarrow 0} ( 1 + i\ep )^{\theta/\ep},
\]
but from here it is not obvious where to go.  The book's development up
to the present point gives no obvious direction.  In fact it appears
that the interpretation of $\exp i\theta$ remains for us to define, if we can
find a way to define it which fits sensibly with our existing notions of
the real exponential.  So, if we don't quite know where to go with this
yet, what do we know?

\index{multiplication!repeated}
\index{magnitude}
\index{phase}
\index{Argand plane}
One thing we know is that if $\theta = \ep$, then
\[
  \exp (i\ep) = (1 + i\ep)^{\ep/\ep} = 1+i\ep.
\]
But per \S~\ref{cexp:220}, the essential operation of the exponential
function is to multiply repeatedly by some factor, the factor being
not quite exactly unity and, in this case, being $1+i\ep$.   With
such thoughts in mind, let us multiply a complex number $z=x+iy$ by
$1+i\ep$, obtaining
\[
  (1+i\ep)(x+iy) = (x-\ep y) + i(y+\ep x).
\]
The resulting change in~$z$ is\footnote{%
  % Admittedly, this footnote *seems* to be restating the obvious.
  % Yet maybe the footnote is helpful, as is, anyway.
  \index{$\Delta$}%
  As the context implies, the notation~$\Delta z$ means ``the change
  in~$z$.''  We have briefly met such notation already in
  \S~\ref{alggeo:231}.%
}
\[
  \Delta z = (1+i\ep)(x+iy) - (x+iy) = (\ep)(-y+ix),
\]
in which it is seen that
\[
  \renewcommand{\arraystretch}{1.5}
  \br{rclcl}
    \ds\left|\Delta z\right| &=& \ds(\ep)\sqrt{y^2+x^2} &=& \ds\ep\rho, \\
    \ds\arg(\Delta z) &=& \ds\arctan\frac{x}{-y} &=& \ds\phi + \frac{2\pi}{4}.
  \er
\]
The~$\Delta z$, $\rho=\left|z\right|$ and $\phi=\arg z$ are as shown in
Fig.~\ref{cexp:230:fig}.  Whether in the figure or in the equations,
\emph{the change~$\Delta z$ is evidently proportional to the magnitude
of~$z$, but at a right angle to $z$'s radial arm in the Argand plane.}
\begin{figure}
  \caption{The complex exponential and Euler's formula.}
  \label{cexp:230:fig}
  \bc
    {
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        \rput{90}(0,0){\xxa}
        \rput[r](-\xxe,-1.9){$-i2$}
        \rput[r](-\xxe,-1){$-i $}
        \rput[r](-\xxe, 1){$ i $}
        \rput[r](-\xxe, 1.9){$ i2$}
        \rput[t](-1.9,-\xxf){$-2$}
        \rput[t](-1,-\xxf){$-1$}
        \rput[t]( 1,-\xxf){$ 1$}
        \rput[t]( 1.9,-\xxf){$ 2$}
        \psline[linewidth=2.0pt]{cc-*}(0,0)(2,1)
        \rput(2,1){
          \psline[linewidth=2.0pt]{cc->}(0,0)(-0.2,0.4)
        }
        \pscircle[linewidth=0.5pt,linestyle=dashed](0,0){\xxr}
        \rput(2.25,0.90){$z$}
        \rput(2.15,1.50){$\Delta z$}
        \psarc[linewidth=0.5pt]{->}(0,0){\xxp}{0}{\xxq}
        \rput{\xxqq}(0,0){
          \uput[r](\xxp,0){
            \rput{*0}(0,0){$\phi$}
          }
        }
        \rput{\xxq}(0,0){
          \uput{\xxlsep}[u](\xxrr,0){
            \rput{*0}(0,0){$\rho$}
          }
        }
        \rput[l](\xxs,0){$\Re(z)$}
        \rput[b](0,\xxs){$i\Im(z)$}
      \end{pspicture}
    }
  \ec
\end{figure}

\index{arm, radial}
\index{radial arm}
\index{travel}
\index{motion!about a circle}
\index{motion!perpendicular to a radial arm}
\index{circle!travel about}
\index{infinitesimal incrementation}
\index{incrementation!infinitesimal}
To travel about a circle wants motion always perpendicular to the
circle's radial arm, which happens to be just the kind of motion~$\Delta z$
represents.  Referring to the figure and the last equations, we have
then that
\[
  \br{rclcl}
    \ds\Delta \rho &\equiv& \ds\left|z+\Delta z\right| - \left|z\right| &=& \ds0, \\
    \ds\Delta \phi &\equiv& \ds\arg(z+\Delta z) - \arg z = \frac{\left|\Delta z\right|}{\rho} = \frac{\ep\rho}{\rho} &=& \ds\ep,
  \er
\]
which results evidently are valid for infinitesimal $\ep \ra 0$ and,
importantly, stand independently of the value of~$\rho$.
(But does~$\rho$ not grow at least a little, as the last equations
almost seem to suggest?  The answer is no; or, if you prefer, the
answer is that $\Delta\rho \approx \{[\sqrt{1+\ep^2}]-1\}\rho
\approx \ep^2\rho/2 \approx 0$, a second-order infinitesimal
inconsequential on the scale of~$\ep\rho$, utterly
vanishing by comparison in the limit $\ep \ra 0$.  Remember
that~$\Delta z$ has a \emph{phase,} a direction in the Argand plane; and
that, as the figure depicts, this phase points at a right angle to the
phase of~$z$.  In mathematical symbols, $\arg[\Delta z] - \arg z =
2\pi/4$.  Now, if the difference between these phases were greater than
$2\pi/4$, that would mean that~$\Delta z$ pointed inward, which would
cause $|z+\Delta z|<|z|$, wouldn't it?  And if the difference were less
than $2\pi/4$,
that would mean that~$\Delta z$ pointed outward, which would cause
$|z+\Delta z|>|z|$.  So, what phase differential exactly causes
$|z+\Delta z|=|z|$?  Where indeed is the boundary between the inward and
outward domains?  Answer: $2\pi/4$.
Such are the paradoxes of calculus!)  With such results in hand, now let
us recall from earlier in the section that---as we have asserted or
defined---%
\[
  \exp i\theta = \lim_{\ep\rightarrow 0} ( 1 + i\ep )^{\theta/\ep},
\]
and that this remains so for arbitrary real~$\theta$.  Yet what does
such an equation do, mechanically, but to compute $\exp i\theta$ by multiplying~$1$ by
$1 + i\ep$ repeatedly, $\theta/\ep$ times?  The plain answer is that
such an equation does precisely this and nothing else.\footnote{See
footnote~\ref{cexp:230fn1}.}  We have
recently seen how each multiplication of the kind the equation suggests
increments the phase~$\phi$ by $\Delta\phi = \ep$ while not changing the
magnitude~$\rho$.  Since the phase~$\phi$ begins from $\arg 1 = 0$
it must become
\[
  \phi = \frac\theta\ep\ep = \theta
\]
after $\theta/\ep$ increments of~$\ep$ each, while the magnitude must
remain
\[
  \rho = 1.
\]
Reversing the sequence of the last two equations and
recalling that $\rho \equiv \left|\exp i\theta\right|$ and that $\phi
\equiv \arg(\exp i\theta)$,
\[
  \begin{split}
    \left|\exp i\theta\right| &= \ds 1, \\
    \arg(\exp i\theta) &= \theta.
  \end{split}
\]
Moreover, had we known that~$\theta$ were just $\phi \equiv \arg(\exp
i\theta)$, naturally we should have represented it by the symbol~$\phi$
from the start.  Changing $\phi \la \theta$ now, we have for real~$\phi$
that
\bqb
  \left|\exp i\phi\right| &=& \ds 1, \\
  \arg(\exp i\phi) &=& \phi,
\eqb
which equations together say neither more nor less than that
\bq{cexp:euler}
  \exp i\phi = \cos \phi + i\sin \phi = \cis \phi,
\eq
where the notation $\cis(\cdot)$ is as defined in \S~\ref{trig:280}.

\index{Euler, Leonhard (1707--1783)}
\index{exponential}
\index{complex exponent}
\index{exponent!complex}
Along with the Pythagorean theorem~(\ref{alggeo:pythag}), the
fundamental theorem of calculus~(\ref{integ:antider}), Cauchy's
integral formula~(\ref{taylor:cauchy}) and Fourier's
equation
% bad break
(\ref{fouri:eqn}), eqn.~(\ref{cexp:euler}) is one of the most famous
results in all of mathematics.  It is called \emph{Euler's formula,}%
\footnote{
  For native English speakers who do not speak German, Leonhard Euler's
  name is pronounced as ``oiler.''
}$\mbox{}^,$\footnote{
  An alternate derivation of Euler's formula~(\ref{cexp:euler})---less
  intuitive and requiring slightly more advanced mathematics, but
  briefer---constructs from Table~\ref{taylor:315:tbl} the Taylor
  series for~$\exp i\phi$, $\cos \phi$ and $i\sin \phi$, then adds the
  latter two to show them equal to the first of the three.  Such an
  alternate derivation lends little insight, perhaps, but at least it
  builds confidence that we actually knew what we were doing when we
  came up with the incredible~(\ref{cexp:euler}).
}
and it opens the exponential domain fully to complex numbers, not just
for the natural base~$e$ but for any base.  How?  Consider in light of
Fig.~\ref{cexp:230:fig} and~(\ref{cexp:euler}) that one can express any
complex number in the form
\[
  z = x+iy = \rho\exp i\phi.
\]
If a complex base~$w$ is similarly expressed in the form
\[
  w = u+iv = \sigma\exp i\psi,
\]
then it follows that
\bqb
  w^z &=& \exp[\ln w^z] \xn\\
  &=& \exp[z\ln w] \xn\\
  &=& \exp[(x+iy)(i\psi + \ln\sigma)] \xn\\
  &=& \exp[(x\ln\sigma-\psi y) + i(y\ln\sigma+\psi x)]. \xn
\eqb
Since $\exp(\alpha+\beta) = e^{\alpha+\beta} = \exp \alpha\exp \beta$,
the last equation is
\bq{cexp:230:33}
  w^z = \exp(x\ln\sigma-\psi y) \exp i(y\ln\sigma+\psi x),
\eq
where
\[
  \begin{split}
    x &= \rho \cos \phi, \\
    y &= \rho \sin \phi, \\
    \sigma &= \sqrt{u^2+v^2}, \\
    \tan\psi &= \frac{v}{u}.
  \end{split}
\]
Equation~(\ref{cexp:230:33}) serves to raise any complex number to a
complex power.

\index{Euler's formula!curious consequences of}
\index{natural logarithm!of a complex number}
Curious consequences of Euler's formula~(\ref{cexp:euler}) include
that\footnote{%
  Notes of the obvious, like $n\in\mathbb Z$, are sometimes omitted by
  this book because they clutter the page.  However,
  the note is included in this instance.%
}
\bq{cexp:230:34}
  \begin{split}
    e^{\pm i2\pi/4} &= \pm i; \\
    e^{\pm i2\pi/2} &= -1; \\
    e^{i n2\pi}  &= 1,\:\ n\in\mathbb Z.
  \end{split}
\eq
For the natural logarithm of a complex number in light of Euler's
formula, we have that
\bq{cexp:225:dlnz}
  \ln w = \ln \left(\sigma e^{i\psi}\right) = \ln\sigma + i\psi.
\eq

% ----------------------------------------------------------------------

\section[Complex exponentials and de~Moivre]{Complex exponentials
  and de~Moivre's theorem}
\label{cexp:240}

\index{exponential, complex!and de~Moivre's theorem}
\index{complex exponential!and de~Moivre's theorem}
\index{de Moivre's theorem!and the complex exponential}
\index{de Moivre, Abraham (1667--1754)}
\index{Moivre, Abraham de (1667--1754)}
\index{de Moivre's theorem}
Euler's formula~(\ref{cexp:euler}) implies that complex numbers~$z_1$
and~$z_2$ can be written as
\bq{cexp:240:05}
  \begin{split}
    z_1 &= \rho_1 e^{i\phi_1}, \\
    z_2 &= \rho_2 e^{i\phi_2}.
  \end{split}
\eq
By the basic power properties of Table~\ref{alggeo:224:t1}, then,
\bq{cexp:240:10}
  \renewcommand\arraystretch{1.5}
  \br{rclcl}
    \ds z_1z_2 &=& \ds\rho_1\rho_2 e^{i(\phi_1 + \phi_2)}
      &=& \ds\rho_1\rho_2 \exp[i(\phi_1 + \phi_2)], \\
      \ds\frac{z_1}{z_2} &=& \ds\frac{\rho_1}{\rho_2} e^{i(\phi_1 - \phi_2)}
      &=& \ds\frac{\rho_1}{\rho_2} \exp[i(\phi_1 - \phi_2)], \\
    \ds z^a &=& \ds\rho^a e^{ia\phi} &=& \ds\rho^a\exp[ia\phi].
  \er
\eq
This is de~Moivre's theorem, introduced in \S~\ref{trig:280}.

% ----------------------------------------------------------------------

\section{Complex trigonometrics}
\label{cexp:250}
\index{trigonometrics!complex}
\index{complex trigonometrics}

\index{tangent!in complex exponential form}
Applying Euler's formula~(\ref{cexp:euler}) to~$+\phi$ then to~$-\phi$,
we have that
\bqb
  \exp( +i\phi ) &=& \cos \phi + i\sin\phi, \\
  \exp( -i\phi ) &=& \cos \phi - i\sin\phi.
\eqb
Adding the two equations and solving for $\cos\phi$ yields that
\bq{cexp:250:cos}
  \index{cosine!in complex exponential form}
  \cos\phi = \frac{\exp(+i\phi)+\exp(-i\phi)}{2}.
\eq
Subtracting the second equation from the first and solving for
$\sin\phi$ yields that
\bq{cexp:250:sin}
  \index{sine!in complex exponential form}
  \sin\phi = \frac{\exp(+i\phi)-\exp(-i\phi)}{i2}.
\eq
Thus are the trigonometrics expressed in terms of complex exponentials.

\subsection{The hyperbolic functions}
\label{cexp:250.20}
\index{hyperbolic functions}
\index{hyperbolic trigonometrics}
\index{trigonometrics!hyperbolic}

\index{hyperbolic sine}
\index{hyperbolic cosine}
\index{hyperbolic tangent}
\index{Pythagorean theorem!and the hyperbolic functions}
The forms~(\ref{cexp:250:cos}) and~(\ref{cexp:250:sin}) suggest the
definition of new functions
\bqa
  \cosh\phi &\equiv& \frac{\exp(+\phi)+\exp(-\phi)}{2}, \label{cexp:250:cosh}\\
  \sinh\phi &\equiv& \frac{\exp(+\phi)-\exp(-\phi)}{2}, \label{cexp:250:sinh}\\
  \tanh\phi &\equiv& \frac{\sinh\phi}{\cosh\phi}.
\eqa
These are called the \emph{hyperbolic functions.}  Their inverses $\mopx{arccosh}$,
etc., are defined in the obvious way.  The Pythagorean theorem for
trigonometrics~(\ref{trig:226:25}) is that $\cos^2\phi + \sin^2\phi=1$,
verified by combining~(\ref{cexp:250:cos}) and~(\ref{cexp:250:sin});
and from~(\ref{cexp:250:cosh}) and~(\ref{cexp:250:sinh}) one can derive
the hyperbolic analog:
\bq{cexp:250:pythag}
  \begin{split}
    \cos^2\phi + \sin^2\phi &= 1, \\
    \cosh^2\phi - \sinh^2\phi &= 1.
  \end{split}
\eq

Although Fig.~\ref{cexp:230:fig} has visualized only real~$\phi$,
complex~$\phi$ can be considered, too.  Nothing prevents
one from taking~(\ref{cexp:250:cos}) through~(\ref{cexp:250:sinh}), as
written, \emph{to define} the trigonometrics of complex~$\phi$;
so that's what we now do.  From this it follows
that~(\ref{cexp:250:pythag}) and others must likewise hold\footnote{
  Chapter~\ref{vector} teaches that the \emph{dot product} of a unit vector
  and its own conjugate is unity---$\vu v^{*} \cdot \vu v = 1$, in the
  notation of that chapter---which tempts one to suppose incorrectly by
  analogy that $(\cos\phi)^{*}\cos\phi + (\sin\phi)^{*}\sin\phi = 1$ and
  that $(\cosh\phi)^{*}\cosh\phi - (\sinh\phi)^{*}\sinh\phi = 1$ when the
  angle~$\phi$ is complex.  However,~(\ref{cexp:250:cos})
  through~(\ref{cexp:250:sinh}) can be generally true only
  if~(\ref{cexp:250:pythag}) holds exactly as written for complex~$\phi$
  as well as for real.  Hence in fact $(\cos\phi)^{*}\cos\phi +
  (\sin\phi)^{*}\sin\phi \neq 1$ and $(\cosh\phi)^{*}\cosh\phi -
  (\sinh\phi)^{*}\sinh\phi \neq 1$.

  Fig.~\ref{trig:226:f1} is quite handy for real~$\phi$ but what if
  anything the figure means when~$\phi$ is complex is not obvious.
  The~$\phi$ of the figure cannot quite be understood to mean an actual
  direction or bearing in the east-north-west-south sense.  Therefore,
  visual analogies between geometrical vectors like~$\vu v$, on the one
  hand, and Argand-plotted complex numbers, on the other, can
  analytically fail, especially in circumstances in which~$\phi$ may be
  complex.
  (The professional mathematician might smile at this, gently prodding
  us that this is why one should rely on analysis rather than on mere
  geometrical intuition.  If so, then we would acknowledge the
  prod~\cite{Carson/Huber} without further comment in this instance.)%
}
for complex~$\phi$.

\index{cis}
The notation $\exp i(\cdot)$ or $e^{i(\cdot)}$ is sometimes felt to be too
bulky.  Although less commonly seen than the other two, the notation
\[
  \cis(\cdot) \equiv \exp i(\cdot) = \cos(\cdot) +i\sin(\cdot)
\]
is also conventionally recognized, as earlier seen in \S~\ref{trig:280}.
Also conventionally recognized are $\sin^{-1}(\cdot)$ and occasionally
$\mopx{asin}(\cdot)$ for $\arcsin(\cdot)$, and likewise for the several
other trigs.

\index{hyperbolic functions!properties of}
Replacing $z\la \phi$ in this section's several equations implies a
coherent definition for trigonometric functions of a complex variable.
Then, comparing~(\ref{cexp:250:cos}) and~(\ref{cexp:250:sin})
respectively to~(\ref{cexp:250:cosh}) and~(\ref{cexp:250:sinh}), we have
that
\bq{cexp:250:20}
  \begin{split}
    \cosh z &= \cos iz, \\
    i \sinh z &= \sin iz, \\
    i \tanh z &= \tan iz,
  \end{split}
\eq
by which one can immediately adapt the many trigonometric
properties of Tables~\ref{trig:228:table} and~\ref{trig:275:table} to
hyperbolic use.

\index{natural exponential family of functions}
\index{natural logarithmic family of functions}
\index{trigonometric family of functions}
\index{inverse trigonometric family of functions}
At this point in the development one begins to notice that the $\cos$,
$\sin$, $\exp$, $\cis$, $\cosh$ and $\sinh$ functions are each really
just different facets of the same mathematical phenomenon.  Likewise
their respective inverses: $\arccos$, $\arcsin$, $\ln$, $-i\ln$, $\mopx{arccosh}$
and $\mopx{arcsinh}$.  Conventional names for these two mutually inverse
families of functions are unknown to the author, but one might
call them the \emph{natural exponential} and \emph{natural logarithmic
families.}  Or, if the various tangent functions were included, then one
might call them the \emph{trigonometric} and \emph{inverse trigonometric
families.}

\subsection{Inverse complex trigonometrics}
\label{cexp:250.30}
\index{trigonometrics!inverse complex}
\index{complex trigonometrics!inverse}

Since one can express the several trigonometric functions in terms of
complex exponentials one would like to know, complementarily, whether
one cannot express the several inverse trigonometric functions in terms
of complex logarithms.  As it happens, one can.%
\footnote{\cite[chapter~2]{Spiegel}}

\index{arcsine}
\index{arccosine}
\index{arctangent}
\index{hyperbolic arcsine}
\index{hyperbolic arccosine}
\index{hyperbolic arctangent}
Let us consider the arccosine function, for instance.  If
per~(\ref{cexp:250:cos})
\[
  z = \cos w = \frac{e^{iw}+e^{-iw}}{2},
\]
then by successive steps
\bqb
  e^{iw} &=& 2z - e^{-iw}, \\
  \left[e^{iw}\right]^2 &=&
    \left[e^{iw}\right]\left[2z - e^{-iw}\right] =
    2z\left(e^{iw}\right) - 1, \\
  e^{iw} &=& z \pm \sqrt{ z^2 - 1 },
\eqb
the last step of which has used the quadratic
formula~(\ref{alggeo:240:quad}).  Taking the logarithm, we have that
\[
  w = \frac{1}{i}\ln\left(z \pm i\sqrt{ 1 - z^2 }\right);
\]
or, since by definition $z=\cos w$, that
\bq{cexp:250:arccos}
  \index{arccosine!in complex exponential form}
  \arccos z = \frac{1}{i}\ln\left(z \pm i\sqrt{ 1 - z^2 }\right).
\eq
Similarly,
\bq{cexp:250:arcsin}
  \index{arcsine!in complex exponential form}
  \arcsin z = \frac{1}{i}\ln\left(iz \pm \sqrt{ 1 - z^2 }\right).
\eq
The arctangent goes only a little differently:
\bqb
  z = \tan w &=& -i\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}}, \\
  ze^{iw}+ze^{-iw} &=& -ie^{iw}+ie^{-iw}, \\
  (i+z)e^{iw} &=& (i-z)e^{-iw}, \\
  e^{i2w} &=& \frac{i-z}{i+z},
\eqb
implying that
\bq{cexp:250:arctan}
  \index{arctangent!in complex exponential form}
  \arctan z = \frac{1}{i2}\ln\frac{i-z}{i+z}.
\eq
By the same means, one can work out the inverse hyperbolics to be
\bq{cexp:250:archyperbolic}
  \index{hyperbolic functions!inverse, in complex exponential form}
  \index{inverse hyperbolic functions!in complex exponential form}
  \begin{split}
    \mopx{arccosh} z &= \ln\left(z \pm \sqrt{ z^2 - 1 }\right), \\
    \mopx{arcsinh} z &= \ln\left(z \pm \sqrt{ z^2 + 1 }\right), \\
    \mopx{arctanh} z &= \frac{1}{2}\ln\frac{1+z}{1-z}.
  \end{split}
\eq

% ----------------------------------------------------------------------

\section{Summary of properties}
\label{cexp:255}

Table~\ref{cexp:tbl-prop} gathers properties of the complex exponential
from this chapter and from \S\S~\ref{alggeo:225}, \ref{trig:280}
and~\ref{drvtv:240}.
\begin{table}
  \caption{Complex exponential properties.}
  \label{cexp:tbl-prop}
  \index{complex exponential!properties of}
  \settowidth\tla{$\ds\rho_1\rho_2 e^{i(\phi_1 + \phi_2)}$}
  \bqb
    i^2 &=& -1 = (-i)^2 \\
    \frac{1}{i} &=& -i \\
    e^{i\phi} &=& \cos \phi + i\sin \phi \\
    e^{iz} &=& \cos z + i\sin z \\
    z_1z_2 &=& \makebox[\tla][l]{$\ds\rho_1\rho_2 e^{i(\phi_1 + \phi_2)}$}
      = (x_1x_2 - y_1y_2) + i(y_1x_2 + x_1y_2) \\
    \frac{z_1}{z_2} &=& \makebox[\tla][l]{$\ds\frac{\rho_1}{\rho_2} e^{i(\phi_1 - \phi_2)}$}
      = \frac{(x_1x_2+y_1y_2)+i(y_1x_2-x_1y_2)}{x_2^2+y_2^2} \\
    z^a &=& \rho^a e^{ia\phi} \\
    w^z &=& e^{x\ln\sigma-\psi y} e^{i(y\ln\sigma+\psi x)} \\
    \ln w &=& \ln\sigma + i\psi
  \eqb
  \[
    \renewcommand\arraystretch{2.0}
    \br{rclcrclcrcl}
      \ds\sin z &\!=\!& \ds\frac{e^{i z}-e^{-i z}}{i2} &\sh{1.0}&
      \ds\sin iz &\!=\!& \ds i\sinh z &\sh{1.0}&
      \ds\sinh z &\!=\!& \ds\frac{e^{ z}-e^{- z}}{2} \\
      \ds\cos z &\!=\!& \ds\frac{e^{i z}+e^{-i z}}{2} &\sh{1.0}&
      \ds\cos iz &\!=\!& \ds \cosh z &\sh{1.0}&
      \ds\cosh z &\!=\!& \ds\frac{e^{ z}+e^{- z}}{2} \\
      \ds\tan z &\!=\!& \ds\frac{\sin z}{\cos z} &\sh{1.0}&
      \ds\tan iz &\!=\!& \ds i\tanh z &\sh{1.0}&
      \ds\tanh z &\!=\!& \ds\frac{\sinh z}{\cosh z}
    \er
  \]
  \[
    \renewcommand\arraystretch{2.0}
    \br{rclcrcl}
      \arcsin z &=& \ds\frac{1}{i}\ln\left(iz \pm \sqrt{ 1 - z^2 }\right) &\sh{1.0}&
      \mopx{arcsinh} z &=& \ds\ln\left(z \pm \sqrt{ z^2 + 1 }\right) \\
      \arccos z &=& \ds\frac{1}{i}\ln\left(z \pm i\sqrt{ 1 - z^2 }\right) &\sh{1.0}&
      \mopx{arccosh} z &=& \ds\ln\left(z \pm \sqrt{ z^2 - 1 }\right) \\
      \arctan z &=& \ds\frac{1}{i2}\ln\frac{i-z}{i+z} &\sh{1.0}&
      \mopx{arctanh} z &=& \ds\frac{1}{2}\ln\frac{1+z}{1-z}
    \er
  \]
  \[
    \cos^2 z + \sin^2 z = 1 = \cosh^2 z - \sinh^2 z
  \]
  \[
    \renewcommand\arraystretch{2.0}
    \br{rclcrcl}
      \ds z &\equiv& \ds x+iy = \rho e^{i\phi} &\sh{1.0}&
      \ds \frac{d}{dz}\exp z &=& \ds \exp z \\
      \ds w &\equiv& \ds u+iv = \sigma e^{i\psi} &\sh{1.0}&
      \ds \frac{d}{dw}\ln w &=& \ds \frac{1}{w} \\
      \ds \exp z &\equiv& \ds e^{z} &\sh{1.0}&
      \ds \frac{df/dz}{f(z)} &=& \ds \frac{d}{dz}\ln f(z) \\
      \ds \cis z &\equiv& \ds \cos z + i\sin z = e^{iz} &\sh{1.0}&
      \ds \log_b w &=& \ds \frac{\ln w}{\ln b}
    \er
  \]
\end{table}

% ----------------------------------------------------------------------

\section{Derivatives of complex exponentials}
\label{cexp:320}
\index{derivative!of a complex exponential}
\index{complex exponential!derivative of}
\index{complex exponential!inverse, derivative of}
\index{inverse complex exponential!derivative of}

This section computes the derivatives of the various trigonometric and
inverse trigonometric functions.

\subsection{Derivatives of sine and cosine}
\label{cexp:320.10}
\index{sine!derivative of}
\index{cosine!derivative of}
\index{derivative!of sine and cosine}

One could compute derivatives of the sine and cosine functions
from~(\ref{cexp:250:cos}) and~(\ref{cexp:250:sin}).  To do so is left as
an exercise.  Meanwhile, however, another, more sporting way to find
the derivatives is known: one can directly examine the circle from which
the sine and cosine functions come.

Refer to Fig.~\ref{cexp:320:fig}.  Suppose that
the point~$z$ in the figure is not fixed but travels steadily
about the circle such that\footnote{%
  Observe the Greek letter~$\omega$, omega, which is not a Roman~$w$.
  Refer to appendix~\ref{greek}\@.%
}
\bq{cexp:320:30}
  z(t) = (\rho)\left[\cos(\omega t+\phi_o) + i\sin(\omega t+\phi_o)\right].
\eq
How fast then is the rate $dz/dt$, and in what Argand direction?  Answer:
\bq{cexp:320:31}
  \frac{dz}{dt} = (\rho)\left[\frac{d}{dt}\cos(\omega t+\phi_o)
    + i\frac{d}{dt}\sin(\omega t+\phi_o)\right].
\eq
\begin{figure}
  \caption{The derivatives of the sine and cosine functions.}
  \label{cexp:320:fig}
  \bc
    {
      \nc\xax{-5}
      \nc\xbx{-3}
      \nc\xcx{ 5}
      \nc\xdx{ 3}
      \begin{pspicture}(\xax,\xbx)(\xcx,\xdx)
        \psset{dimen=middle}
        \small
        %\psframe[linewidth=0.5pt](\xax,\xbx)(\xcx,\xdx)
        \nc\xx{2.5}
        \nc\xxb{1.0}
        \nc\xxc{2.0}
        \nc\xxp{0.8}
        \nc\xxq{26.565}
        \nc\xxqq{13.283}
        \nc\xxr{2.2361}
        \nc\xxrr{1.30}
        \nc\xxrrr{2.3}
        \nc\xxs{2.6}
        \nc\xxlsep{0.20}
        \nc\xxa{
          \psline[linewidth=0.5pt](-\xx,0)(\xx,0)
        }
        \xxa
        \rput{90}(0,0){\xxa}
        \psline[linewidth=2.0pt]{cc-*}(0,0)(2,1)
        \rput(2,1){
          \psline[linewidth=2.0pt]{cc->}(0,0)(-0.2,0.4)
        }
        \pscircle[linewidth=0.5pt,linestyle=dashed](0,0){\xxr}
        \rput(2.25,0.90){$z$}
        \rput(2.10,1.70){$\ds\frac{dz}{dt}$}
        \psarc[linewidth=0.5pt]{->}(0,0){\xxp}{0}{\xxq}
        \rput(1.45,0.25){$\omega t+\phi_o$}
        \rput{\xxq}(0,0){
          \uput{\xxlsep}[u](\xxrr,0){
            \rput{*0}(0,0){$\rho$}
          }
        }
        \rput[l](\xxs,0){$\Re(z)$}
        \rput[b](0,\xxs){$\Im(z)$}
      \end{pspicture}
    }
  \ec
\end{figure}%
Evidently however, considering the figure,
\bi
  \item the speed $\left|dz/dt\right|$ is also $(\rho)(d\phi/dt) = \rho\omega$;
  \item the direction is at right angles to the arm of~$\rho$, which is
    to say that $\arg(dz/dt) = \phi + 2\pi/4$.
\ei
With these observations we can write that
% \footnote{
%   As an applied mathematician, you really need to be able to see the
%   truth of this equation merely by staring hard at
%   Fig.~\ref{cexp:320:fig}.  If you cannot see it yet, this is okay; but
%   stop now and stare hard at the figure until you can.  The point
%   travels about the circle.  In what direction?  At what speed?  Such
%   reflection can lead one nowhere other than to the expression given for
%   $dz/dt$.
% }
\bqa
  \frac{dz}{dt} &=& (\rho\omega)\left[
    \cos\left(\omega t+\phi_o + \frac{2\pi}{4}\right)
    +i\sin\left(\omega t+\phi_o + \frac{2\pi}{4}\right)
  \right] \xn\\
  &=& (\rho\omega)\left[-\sin(\omega t+\phi_o)
    + i\cos(\omega t+\phi_o)\right].
  \label{cexp:320:32}
\eqa
Matching the real and imaginary parts of~(\ref{cexp:320:31}) against those
of~(\ref{cexp:320:32}), we have that
\bq{cexp:320:10}
  \begin{split}
    \frac{d}{dt}\cos (\omega t + \phi_o) &= -\omega\sin (\omega t + \phi_o), \\
    \frac{d}{dt}\sin (\omega t + \phi_o) &= +\omega\cos (\omega t + \phi_o).
  \end{split}
\eq
If $\omega = 1$ and $\phi_o = 0$, these are that
\bq{cexp:320:12}
  \begin{split}
    \frac{d}{dt}\cos t &= -\sin t, \\
    \frac{d}{dt}\sin t &= +\cos t.
  \end{split}
\eq

\subsection{Derivatives of the trigonometrics}
\label{cexp:320.20}

Equations~(\ref{cexp:220:dexp}) and~(\ref{cexp:320:12}) give the
derivatives of $\exp(\cdot)$, $\sin(\cdot)$ and $\cos(\cdot)$.  From
these, with the help of~(\ref{cexp:250:pythag}) and the derivative
chain and product rules
(\S~\ref{drvtv:250}), we can calculate
the several derivatives of Table~\ref{cexp:drv}.%
\footnote{\cite[back endpaper]{Shenk}}
\begin{table}
  \caption{Derivatives of the trigonometrics.}
  \label{cexp:drv}
  \index{trigonometric function!derivative of}
  \index{natural exponential!derivative of}
  \index{exponential, natural!derivative of}
  \index{sine!derivative of}
  \index{cosine!derivative of}
  \index{tangent!derivative of}
  \index{derivative!of a trigonometric}
  \index{derivative!of the natural exponential}
  \index{derivative!of sine, cosine and tangent}
  {
    \renewcommand\arraystretch{2.0}
    \bqb
      \br{rclcrcl}
        \ds\frac{d}{dz} \exp z &=& \ds+\exp z &\sh{1.0}&
        \ds\frac{d}{dz} \frac{1}{\exp z} &=& \ds-\frac{1}{\exp z} \\
        \ds\frac{d}{dz} \sin z &=& \ds+\cos z &\sh{1.0}&
        \ds\frac{d}{dz} \frac{1}{\sin z} &=& \ds-\frac{1}{\tan z\sin z} \\
        \ds\frac{d}{dz} \cos z &=& \ds-\sin z &\sh{1.0}&
        \ds\frac{d}{dz} \frac{1}{\cos z} &=& \ds+\frac{\tan z}{\cos z}
      \er
    \eqb
    \bqb
      \br{rclcl}
        \ds\frac{d}{dz} \tan z &=& +\left(\ds 1+\tan^2 z\right)
          &=& \ds+\frac{1}{\cos^2 z} \\
        \ds\frac{d}{dz} \frac{1}{\tan z} &=& \ds-\left(1+\frac{1}{\tan^2 z}\right)
          &=& \ds-\frac{1}{\sin^2 z}
      \er
    \eqb
    \bqb
      \br{rclcrcl}
        \ds\frac{d}{dz} \sinh z &=& \ds+\cosh z &\sh{1.0}&
        \ds\frac{d}{dz} \frac{1}{\sinh z} &=& \ds-\frac{1}{\tanh z\sinh z} \\
        \ds\frac{d}{dz} \cosh z &=& \ds+\sinh z &\sh{1.0}&
        \ds\frac{d}{dz} \frac{1}{\cosh z} &=& \ds-\frac{\tanh z}{\cosh z}
      \er
    \eqb
    \bqb
      \br{rclcl}
        \ds\frac{d}{dz} \tanh z &=& \ds 1-\tanh^2 z
          &=& \ds+\frac{1}{\cosh^2 z} \\
        \ds\frac{d}{dz} \frac{1}{\tanh z} &=& \ds 1-\frac{1}{\tanh^2 z}
          &=& \ds-\frac{1}{\sinh^2 z}
      \er
    \eqb
  }
\end{table}

\subsection{Derivatives of the inverse trigonometrics}
\label{cexp:320.30}

Observe the pair
\[
  \begin{split}
    \frac{d}{dz}\exp z &= \exp{z}, \\
    \frac{d}{dw}\ln w &= \frac{1}{w}.
  \end{split}
\]
The natural exponential $\exp z$ belongs to the trigonometric family of
functions, as does its derivative.  The natural logarithm $\ln w$, by
contrast, belongs to the inverse trigonometric family of functions; but
its derivative is simpler, not a trigonometric or inverse trigonometric
function at all.  In Table~\ref{cexp:drv}, one notices that all the
trigonometrics have trigonometric derivatives.  By analogy with the
natural logarithm, do all the inverse trigonometrics have simpler
derivatives?

It turns out that they do.  Refer to the account of the natural
logarithm's derivative in \S~\ref{cexp:225}.  Following a similar
procedure, we have by successive steps that
\bqa
  \arcsin w &=& z, \xn\\
  w &=& \sin z, \xn\\
  \frac{dw}{dz} &=& \cos z, \xn\\
  \frac{dw}{dz} &=& \pm\sqrt{1 - \sin^2 z}, \xn\\
  \frac{dw}{dz} &=& \pm\sqrt{1 - w^2}, \xn\\
  \frac{dz}{dw} &=& \frac{\pm 1}{\sqrt{1 - w^2}}, \xn\\
  \frac{d}{dw} \arcsin w &=& \frac{\pm 1}{\sqrt{1 - w^2}}.
  \label{cexp:320:20}
\eqa
Similarly,
\bqa
  \arctan w &=& z, \xn\\
  w &=& \tan z, \xn\\
  \frac{dw}{dz} &=& 1 + \tan^2 z, \xn\\
  \frac{dw}{dz} &=& 1 + w^2, \xn\\
  \frac{dz}{dw} &=& \frac{1}{1 + w^2}, \xn\\
  \frac{d}{dw} \arctan w &=& \frac{1}{1 + w^2}.
  \label{cexp:320:22}
\eqa
Derivatives of the other inverse trigonometrics are found in the same
way.  Table~\ref{cexp:drvi} summarizes.
\begin{table}
  \caption{Derivatives of the inverse trigonometrics.}
  \label{cexp:drvi}
  \index{trigonometric function!inverse, derivative of}
  \index{natural logarithm!derivative of}
  \index{logarithm, natural!derivative of}
  \index{arcsine!derivative of}
  \index{arccosine!derivative of}
  \index{arctangent!derivative of}
  \index{derivative!of an inverse trigonometric}
  \index{derivative!of the natural logarithm}
  \index{derivative!of arcsine, arccosine and arctangent}
  {
    \renewcommand\arraystretch{2.0}
    \bqb
      \br{rcl}
        \ds\frac{d}{dw} \ln w &=& \ds\frac{1}{w} \\
        \ds\frac{d}{dw} \arcsin w &=& \ds\frac{\pm 1}{\sqrt{1-w^2}} \\
        \ds\frac{d}{dw} \arccos w &=& \ds\frac{\mp 1}{\sqrt{1-w^2}} \\
        \ds\frac{d}{dw} \arctan w &=& \ds\frac{1}{1+w^2} \\
        \ds\frac{d}{dw} \mopx{arcsinh} w &=& \ds\frac{\pm 1}{\sqrt{w^2+1}} \\
        \ds\frac{d}{dw} \mopx{arccosh} w &=& \ds\frac{\pm 1}{\sqrt{w^2-1}} \\
        \ds\frac{d}{dw} \mopx{arctanh} w &=& \ds\frac{1}{1-w^2}
      \er
    \eqb
  }
\end{table}

Table~\ref{cexp:drvi} may prove useful when the integration technique of
\S~\ref{inttx:210} is applied.

% ---------------------------------------------------------------------

\section{The actuality of complex quantities}
\label{cexp:260}
\index{complex number!actuality of}
\index{number!complex, actuality of}

\index{nature!complex number in}
\index{complex number in nature}
Doing all this theoretically interesting complex mathematics, the
applied mathematician can lose sight of some questions he probably ought
to keep in mind:  Do complex quantities arise in nature?  If they do
not, then what physical systems do we mean to model with them?  Hadn't
we better avoid these complex quantities, leaving them to the
professional mathematical theorists?

\index{Heaviside, Oliver (1850--1925)}
As developed by Oliver Heaviside in 1887,%
\footnote{\cite{mathbios}}
the answer depends on your point of view.  If I have~$300$~g of grapes
and~$100$~g of grapes, then I have~$400$~g altogether.  Alternately, if I
have~$500$~g of grapes and~$-100$~g of grapes, again I have~$400$~g altogether.
(What does it mean to have~$-100$~g of grapes?  Maybe that I ate some!)
But what if I have~$200+i100$~g of grapes and~$200-i100$~g of grapes?
Answer: again,~$400$~g.

\index{wave!propagating}
\index{wave!complex}
\index{conjugate}
\index{superposition}
\index{mirror}
%\index{handwriting, reflected}
\index{grapes}
Probably you would not choose to think of~$200+i100$~g of grapes
and~$200-i100$~g of grapes, but because of~(\ref{cexp:250:cos})
and~(\ref{cexp:250:sin}), one often describes wave phenomena as linear
superpositions (sums) of countervailing complex exponentials.  Consider
for instance the propagating wave
\[
  A\cos[\omega t-kz] = \frac{A}{2}\exp[+i(\omega t-kz)] + \frac{A}{2}\exp[-i(\omega t-kz)].
\]
The benefit of splitting the real cosine into two complex parts is that,
while the magnitude of the cosine changes with time~$t$, the magnitude
of either exponential alone remains steady (see the circle in
Fig.~\ref{cexp:230:fig}).  It turns out to be much easier to analyze two
complex wave quantities of constant magnitude than to analyze one real
wave quantity of varying magnitude.  Better yet, since each complex wave
quantity is the complex conjugate of the other, the analyses thereof are
mutually conjugate, too (\S~\ref{alggeo:225.2}); so you normally needn't
actually analyze the second.  The one analysis suffices for both.%
\footnote{
  If the point is not immediately clear, an example:  Suppose that by
  the Newton-Raphson iteration (\S~\ref{drvtv:270}) you have found a
  root of the polynomial $x^3 + 2x^2 + 3x + 4$ at
  $x \approx -\mbox{0x0.2D}+i\mbox{0x1.8C}$.  Where is there another root?
  Answer: there is a conjugate root at $x \approx
  -\mbox{0x0.2D}-i\mbox{0x1.8C}$.  Because the polynomial's coefficients
  are real, one need not actually run the Newton-Raphson again to find
  the conjugate root.

  \index{voltage}
  \index{tension, electric}
  \index{potential!electric}
  \index{electric tension or potential}
  \index{capacitor}
  \index{electric capacitor}
  \index{electric current}
  \index{current, electric}
  Another example, this time with a wave: suppose that, when fed by a
  time-varying electric current of
  $(5.0\:\mbox{milliamps})\exp\{+i(60\:\mbox{sec}^{-1})2\pi t\}$, an electric
  capacitor develops a voltage---that is, develops an electric tension or
  potential---of
  $(40\:\mbox{volts})\exp\{+i[(60\:\mbox{sec}^{-1})2\pi t - 2\pi/4]\}$.
  It immediately follows, without further analysis, that the same
  capacitor, if fed by a time-varying electric current of
  $(5.0\:\mbox{milliamps})\exp\{-i(60\:\mbox{sec}^{-1})2\pi t\}$, would
  develop a voltage of
  $(40\:\mbox{volts})\exp\{-i[(60\:\mbox{sec}^{-1})2\pi t - 2\pi/4]\}$.
  The conjugate current gives rise to a conjugate voltage.

  The reason to analyze an electric circuit in such a way is that, after
  analyzing it, one can sum the two complex currents to get a real a.c.\
  current like the current an electric wall receptacle supplies.  If one
  does this, then one can likewise sum the two complex voltages to
  compute the voltage the capacitor would develop.  Indeed, this is how
  electrical engineers normally analyze a.c.\ systems (well,
  electrical engineers know some shortcuts, but this is the idea),
  because~$\exp(\cdot)$ is so much easier a function to handle
  than~$\cos(\cdot)$ or~$\sin(\cdot)$ is.
}
% The following analogy admittedly remains a little silly, but it stays
% here unless and until the author thinks of something better.
%(It's like reflecting your sister's handwriting.  To read her
%handwriting backward, you needn't ask her to try writing reverse with
%the wrong hand; you can just hold her regular script up to a mirror.
%Of course, this ignores the question of why one would want to reflect
%someone's handwriting in the first place; but anyway,
%reflecting---which is to say, conjugating---complex quantities often is
%useful.)

\index{Ockham's razor!abusing}
\index{Ockham!William of (c.~1287--1347)}
\index{Aristotle (384--322~B.C.)}
Some authors have gently denigrated the use of imaginary parts in
physical applications as a mere mathematical trick, as though the parts
were not actually there.\footnote{%
  One who gently denigrates the use can nevertheless still apply the
  trick!  They often do.%
}
Well, that is one way to treat the matter, but
it is not the way this book recommends.  Nothing in the mathematics
\emph{requires} you to regard the imaginary parts as physically
nonexistent.  You need not abuse Ockham's razor!
(Ockham's razor, ``Do not multiply objects without necessity,''%
\footnote{
  \cite[chapter~12]{Stroustrup}
}
is a sound philosophical indicator when properly used.  However,
the razor is overused in some circles, particularly in circles in which
Aristotle%
\footnote{\cite{Feser}}
is believed---mistakenly, in this writer's view---to be vaguely
outdated; or more likely in circles in which Aristotle has been
altogether forgotten.  More often than one likes to believe, the
necessity to multiply objects remains
hidden until one has ventured the multiplication, nor reveals
itself to the one who wields the razor, whose hand humility should
stay.)  It is true by Euler's formula~(\ref{cexp:euler}) that a complex
exponential $\exp i\phi$ can be decomposed into a sum of trigonometrics.
However, it is equally true by the complex trigonometric
formulas~(\ref{cexp:250:cos}) and~(\ref{cexp:250:sin}) that \emph{a
trigonometric can be decomposed into a sum of complex exponentials.}
So, if each can be decomposed into the other, then which of the two is
the true decomposition?  Answer: that depends on your point of view.
Experience seems to recommend viewing the complex exponential as the
basic element---as the element of which the trigonometrics are
composed---rather than the other way around.  From this point of view,
it is~(\ref{cexp:250:cos}) and~(\ref{cexp:250:sin}) which are the true
decomposition.  Euler's formula~(\ref{cexp:euler}) itself could be viewed in
this sense as secondary.

The complex exponential method of offsetting imaginary parts offers an
elegant yet practical mathematical means to model physical wave
phenomena.  It may find other uses, too, so go ahead: regard the imaginary
parts as actual.  Aristotle would regard them so (or so the author suspects).
%\footnote{
%  A famous English-language physics book of the twentieth century, which
%  this particular footnote will not name but which was and remains
%  otherwise an excellent book, has unsubtly insinuated among a
%  generation of scientists and engineers an impertinent contempt for
%  philosophy as such.  Oddly, the book in question, whose genial, late
%  author wore Nobel laurels, seems never to denigrate ``teleology'' or
%  even ``metaphysics'' by name, only ``philosophy''---or
%  ``philosophers''---by a curiously vague terminological inexactitude.
%  The book in question seems to delight in the burning of philosophical
%  straw men in any case.  Be that as it may, the author of the book you
%  are \emph{now} reading cordially dissents from the insinuation and,
%  for what it might be worth among philosophers, inclines toward
%  Aristotle rather than toward William of Ockham.
%}
To regard the imaginary parts as actual hurts nothing, and it helps with
the math.

