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You are familiar with derivatives of functions from
to
, and with the
motivation of the definition of derivative as the slope of the tangent to a curve.
For complex functions, the geometrical motivation is missing, but the definition is
formally the same as the definition for derivatives of real functions.
10.1
Definition (Derivative.)
Let
![$f$](img13.gif)
be a complex valued function with
![$\mbox{{\rm dom}}(f)\subset\mbox{{\bf C}}$](img1013.gif)
, let
![$a$](img590.gif)
be a point
such that
![$a\in\mbox{{\rm dom}}(f)$](img1014.gif)
, and
![$a$](img590.gif)
is a limit point of
![$\mbox{{\rm dom}}(f)$](img623.gif)
. We say
![$f$](img13.gif)
is
differentiable at ![$a$](img590.gif)
if the limit
exists. In this case, we denote this limit by
![$f'(a)$](img1016.gif)
and call
![$f'(a)$](img1016.gif)
the
derivative of
at ![$a$](img590.gif)
.
By the definition of limit, we can say that
is differentiable at
if
, and
is a limit point of
and there exists
a function
such
that
is continuous at
, and
such that
![\begin{displaymath}
D_a f(z)={{f(z)-f(a)}\over {z-a}}\mbox{ for
all }z\in\mbox{{\rm dom}}(f)\backslash\{a\},
\end{displaymath}](img1019.gif) |
(10.2) |
and in this case
is equal to
.
It is sometimes useful to rephrase condition (10.2) as follows:
is
differentiable at
if
,
is a limit point of
,
and there is a function
such that
is continuous at
, and
![\begin{displaymath}
f(z)=f(a)+(z-a)D_af(z)\mbox{ for all }z\in\mbox{{\rm dom}}(f).
\end{displaymath}](img1022.gif) |
(10.3) |
In this case,
.
10.4
Remark.
It follows immediately from (
10.3) that if
![$f$](img13.gif)
is differentiable at
![$a$](img590.gif)
,
then
![$f$](img13.gif)
is continuous at
![$a$](img590.gif)
.
10.5
Example.
Let
![$f\colon\mbox{{\bf C}}\to\mbox{{\bf C}}$](img683.gif)
be given by
and let
![$a\in\mbox{{\bf C}}$](img86.gif)
. Then for all
![$z\neq a$](img1025.gif)
,
If we define
![$D_af\colon\mbox{{\bf C}}\to\mbox{{\bf C}}$](img1027.gif)
by
then
![$D_af$](img1018.gif)
is continuous at
![$a$](img590.gif)
, so
![$f$](img13.gif)
is differentiable at
![$a$](img590.gif)
and
We could also write
this calculation as
Hence
![$f$](img13.gif)
is differentiable at
![$a$](img590.gif)
and
![$f'(a)=2a$](img1031.gif)
for all
![$a\in\mbox{{\bf C}}$](img86.gif)
.
10.6
Example.
Let
![$\displaystyle {v(z)={1\over z}}$](img1032.gif)
for
![$z\in\mbox{{\bf C}}\backslash\{0\}$](img890.gif)
and let
![$a\in\mbox{{\bf C}}\backslash\{0\}$](img1033.gif)
. Then for all
Let
![$D_av\colon\mbox{{\bf C}}\backslash\{0\}\to\mbox{{\bf C}}$](img1036.gif)
be defined by
Then
![$D_av$](img1038.gif)
is continuous at
![$a$](img590.gif)
, so
![$v$](img1039.gif)
is differentiable at
![$a$](img590.gif)
, and
for all
![$a\in\mbox{{\bf C}}\backslash\{0\}$](img1033.gif)
.
10.7
Warning.
The function
![$D_af$](img1018.gif)
should not be confused with
![$f'$](img1041.gif)
. In the example above
Also it is not good form to say
![\begin{displaymath}
D_af(z)={{f(z)-f(a)}\over {z-a}}
\end{displaymath}](img1043.gif) |
(10.8) |
without specifying the condition `` for
![$z\neq a$](img1025.gif)
," since someone reading
(
10.8) would assume
![$D_af$](img1018.gif)
is undefined at
![$a$](img590.gif)
.
10.9
Example.
Let
![$f(z)=z^*$](img1044.gif)
for all
![$z\in\mbox{{\bf C}}$](img66.gif)
, and let
![$a\in\mbox{{\bf C}}$](img86.gif)
. Let
I claim that
![$\displaystyle { D_a f}$](img1046.gif)
does not have a limit at
![$a$](img590.gif)
,
and hence
![$f$](img13.gif)
is a
nowhere differentiable function.
Let
Then
![$\{a_n\}_{n\geq 1}$](img1048.gif)
and
![$\{b_n\}_{n\geq 1}$](img1049.gif)
are sequences in
![$\mbox{{\rm dom}}(f)\backslash\{a\}$](img773.gif)
both of which converge
to
![$a$](img590.gif)
. For all
![$n\in\mbox{{\bf Z}}_{\geq 1}$](img117.gif)
,
so
![$\{D_a f(a_n)\}_{n\geq 1}\to 1$](img1051.gif)
and
![$\{D_a f(b_n)\}_{n\geq 1}\to -1$](img1052.gif)
, and hence
![$D_af$](img1018.gif)
does not have a limit at
![$a$](img590.gif)
.
10.11
Theorem (Sum theorem for differentiable functions.)Let
be complex
functions, and suppose
and
are differentiable at
. Suppose
is a limit point of
. Then
is
differentiable at
and
.
Proof: Since
are differentiable at
, there are functions
,
such that
,
are continuous at
, and
It follows that
and
is continuous at
.
We can let
and we see
is differentiable at
and
10.12
Theorem.
Let
be a complex function and let
. If
is differentiable at
,
then
is differentiable at
and
.
Proof: The proof is left to you.
Proof: From our hypotheses, there exist functions
such that
is continuous at
,
is continuous at
and
If
, then
, so we can replace
in
(10.15) by
to get
Using (10.14) to rewrite
, we get
Hence we have
and
is continuous at
. Hence
is differentiable
at
and
10.16
Theorem (Reciprocal rule.)
Let
be a complex function, and let
. If
is differentiable at
and
, then
is differentiable at
and
.
Proof: If
, we saw above that
is
differentiable and
. Let
be a complex function, and
let
. Suppose
is differentiable at
, and
. Then
. By the chain rule
is differentiable
at
, and
10.17
Exercise (Product rule.)
A
Let
![$f,g$](img109.gif)
be complex
functions. Suppose
![$f$](img13.gif)
and
![$g$](img111.gif)
are both
differentiable at
![$a$](img590.gif)
, and that
![$a$](img590.gif)
is a limit point of
![$\mbox{{\rm dom}}(f)\cap\mbox{{\rm dom}}(g)$](img844.gif)
.
Show that
![$fg$](img195.gif)
is differentiable at
![$a$](img590.gif)
, and that
10.18
Exercise (Power rule.)
Let
![$f$](img13.gif)
be a complex function, and suppose that
![$f$](img13.gif)
is differentiable at
![$a\in\mbox{{\bf C}}$](img86.gif)
.
Show that
![$f^n$](img1093.gif)
is differentiable at
![$a$](img590.gif)
for all
![$n\in\mbox{{\bf Z}}_{\geq 1}$](img117.gif)
and
(Use induction.)
10.19
Exercise (Power rule.)
A
Let
![$f$](img13.gif)
be a complex function.
suppose that
![$f$](img13.gif)
is differentiable at
![$a\in\mbox{{\bf C}}$](img86.gif)
, and
![$f(a)\neq 0$](img1085.gif)
.
Show that
![$f^n$](img1093.gif)
is differentiable at
![$a$](img590.gif)
for all
![$n \in \mbox{{\bf Z}}^-$](img1095.gif)
, and
that
for all
![$n \in \mbox{{\bf Z}}^-.$](img1096.gif)
10.20
Exercise (Quotient rule.)
A
Let
![$f,g$](img109.gif)
be
complex functions and let
![$a\in\mbox{{\bf C}}$](img86.gif)
. Suppose
![$f$](img13.gif)
and
![$g$](img111.gif)
are differentiable at
![$a$](img590.gif)
and
![$g(a)\neq 0$](img1097.gif)
, and
![$a$](img590.gif)
is a limit point of
![$\displaystyle {\mbox{{\rm dom}}\left({f\over g}\right)}$](img850.gif)
. Show that
![$\displaystyle {{f\over g}}$](img290.gif)
is
differentiable at
![$a$](img590.gif)
and
Next: 10.2 Differentiable Functions on
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