10.1 Derivatives of Complex Functions

You are familiar with derivatives of functions from $\mbox{{\bf R}}$ to $\mbox{{\bf R}}$, 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 $f$ be a complex valued function with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf C}}$, let $a$ be a point such that $a\in\mbox{{\rm dom}}(f)$, and $a$ is a limit point of $\mbox{{\rm dom}}(f)$. We say $f$ is differentiable at $a$ if the limit

$$\lim_{z\to a} {{f(z)-f(a)}\over {z-a}}$$

exists. In this case, we denote this limit by $f'(a)$ and call $f'(a)$ the derivative of $f$ at $a$.

By the definition of limit, we can say that $f$ is differentiable at $a$ if $a\in\mbox{{\rm dom}}(f)$, and $a$ is a limit point of $\mbox{{\rm dom}}(f)$ and there exists a function $D_a f:\mbox{{\rm dom}}(f)\to\mbox{{\bf C}}$ such that $D_af$ is continuous at $a$, and such that

$$D_a f(z)={{f(z)-f(a)}\over {z-a}}\mbox{ for all }z\in\mbox{{\rm dom}}(f)\backslash\{a\},$$ (10.2)

and in this case $f'(a)$ is equal to $D_a f(a)$.

It is sometimes useful to rephrase condition (10.2) as follows: $f$ is differentiable at $a$ if $a\in\mbox{{\rm dom}}(f)$, $a$ is a limit point of $\mbox{{\rm dom}}(f)$, and there is a function $D_af\colon\mbox{{\rm dom}}f\to\mbox{{\bf C}}$ such that $D_af$ is continuous at $a$, and

$$f(z)=f(a)+(z-a)D_af(z)\mbox{ for all }z\in\mbox{{\rm dom}}(f).$$ (10.3)

In this case, $f'(a)=D_af(a)$.

10.4   Remark. It follows immediately from (10.3) that if $f$ is differentiable at $a$, then $f$ is continuous at $a$.

10.5   Example. Let $f\colon\mbox{{\bf C}}\to\mbox{{\bf C}}$ be given by

$$f\colon z\mapsto z^2,$$

and let $a\in\mbox{{\bf C}}$. Then for all $z\neq a$,

$${{f(z)-f(a)}\over {z-a}}={{z^2-a^2}\over {z-a}}=z+a.$$

If we define $D_af\colon\mbox{{\bf C}}\to\mbox{{\bf C}}$ by

$$D_af(z)=z+a \mbox{ for all }z\in\mbox{{\bf C}},$$

then $D_af$ is continuous at $a$, so $f$ is differentiable at $a$ and

$$f'(a)=D_af(a)=a+a=2a \mbox{ for all }a\in\mbox{{\bf C}}.$$

We could also write this calculation as

$$\lim_{z\to a}{{f(z)-f(a)}\over {z-a}}=\lim_{z\to a}{{z^2-a^2}\over {z-a}}=\lim_{z\to a}z+a=a+a=2a.$$

Hence $f$ is differentiable at $a$ and $f'(a)=2a$ for all $a\in\mbox{{\bf C}}$.

10.6   Example. Let $${v(z)={1\over z}}$$ for $z\in\mbox{{\bf C}}\backslash\{0\}$ and let $a\in\mbox{{\bf C}}\backslash\{0\}$. Then for all $z\in\mbox{{\bf C}}\backslash\{a\}$

$${{v(z)-v(a)}\over {z-a}}={{ {1\over z}-{1\over a}}\over {z-a}}={{a-z}\over {za(z-a)}}=-{1\over {za}}.$$

Let $D_av\colon\mbox{{\bf C}}\backslash\{0\}\to\mbox{{\bf C}}$ be defined by

$$D_av(z)=-{1\over {za}}\mbox{ for all } z\in\mbox{{\bf C}}\backslash\{0\}.$$

Then $D_av$ is continuous at $a$, so $v$ is differentiable at $a$, and

$$\displaystyle {v'(a)=D_av(a)=-{1\over {a^2}}}$$

for all $a\in\mbox{{\bf C}}\backslash\{0\}$.

10.7   Warning. The function $D_af$ should not be confused with $f'$. In the example above

$$D_av(z)=-{1\over {za}},\quad v'(z)={{-1}\over {z^2}}.$$

Also it is not good form to say

$$D_af(z)={{f(z)-f(a)}\over {z-a}}$$ (10.8)

without specifying the condition `` for $z\neq a$," since someone reading (10.8) would assume $D_af$ is undefined at $a$.

10.9   Example. Let $f(z)=z^*$ for all $z\in\mbox{{\bf C}}$, and let $a\in\mbox{{\bf C}}$. Let

$$D_a f(z) = {{f(z)-f(a)}\over {z-a}}={{z^*-a^*}\over {z-a}} \mbox{ for all }z \in \mbox{{\bf C}}\setminus \{a\}.$$

I claim that $${ D_a f}$$ does not have a limit at $a$, and hence $f$ is a nowhere differentiable function.

Let

$$\displaystyle { \{a_n\}_{n\geq 1}=\left\{ a+{1\over n}\right\... ...eq 1},\{b_n\}_{n\geq 1}=\left\{a+{i\over n}\right\}_{n\geq 1}}.$$

Then $\{a_n\}_{n\geq 1}$ and $\{b_n\}_{n\geq 1}$ are sequences in $\mbox{{\rm dom}}(f)\backslash\{a\}$ both of which converge to $a$. For all $n\in\mbox{{\bf Z}}_{\geq 1}$,

$$\begin{eqnarray*} D_a f(a_n)&=&{{\left(a+{1\over n}\right)^*-a^*}\over {a+{1\ove... ...}\over {a+{i\over n}-a}}={{ {-i\over n}}\over { {i\over n}}}=-1, \end{eqnarray*}$$

so $\{D_a f(a_n)\}_{n\geq 1}\to 1$ and $\{D_a f(b_n)\}_{n\geq 1}\to -1$, and hence $D_af$ does not have a limit at $a$.

10.10   Exercise. A Investigate the following functions for differentiability at an arbitrary point $a\in\mbox{{\bf C}}$. Calculate the derivatives of any differentiable functions.

a)

$f(z)=Az+B$ $\qquad A,B$ are given complex numbers.

b)

$${g(z)={1\over {(z+i)^2}}\qquad z\in\mbox{{\bf C}}\backslash\{-i\}}$$.

c)

$h(z)= \mbox{\rm Re}(z)$, i.e. $h(x+iy)=x$.

10.11   Theorem (Sum theorem for differentiable functions.)Let $f,g$
be complex functions, and suppose $f$ and $g$ are differentiable at $a\in\mbox{{\bf C}}$. Suppose $a$ is a limit point of $\mbox{{\rm dom}}(f)\cap\mbox{{\rm dom}}(g)$. Then $f+g$ is differentiable at $a$ and $(f+g)'(a)=f'(a)+g'(a)$.

Proof: Since $f,g$ are differentiable at $a$, there are functions $D_af\colon\mbox{{\rm dom}}(f)\to\mbox{{\bf C}}$, $D_ag\colon\mbox{{\rm dom}}(g)\to g$ such that $D_af$, $D_ag$ are continuous at $a$, and

$$\begin{eqnarray*} f(z)&=&f(a)+(z-a)D_af(z)\mbox{ for all } z\in\mbox{{\rm dom}}(... ...(z)&=&g(a)+(z-a)D_ag(z)\mbox{ for all } z\in\mbox{{\rm dom}}(g). \end{eqnarray*}$$

It follows that

$$(f+g)(z)=(f+g)(a)+(z-a)[D_af(z)+D_ag(z)] \mbox{ for all }z\in\mbox{{\rm dom}}(f+g)$$

and $D_af+D_ag$ is continuous at $a$.

We can let $D_a(f+g)=D_af+D_ag$ and we see $f+g$ is differentiable at $a$ and

$$(f+g)'(a)=(D_af+D_ag)(a)=D_af(a)+D_ag(a)=f'(a)+g'(a).\mbox{ $\mid\!\mid\!\mid$}$$

10.12   Theorem. Let $f$ be a complex function and let $c\in\mbox{{\bf C}}$. If $f$ is differentiable at $a$, then $cf$ is differentiable at $a$ and $(cf)'(a)=c\cdot f'(a)$.

Proof: The proof is left to you. $\mid\!\mid\!\mid$

10.13   Theorem (Chain Rule.) Let $f,g$ be complex functions, and let $a\in\mbox{{\bf C}}$. Suppose $f$ is differentiable at $a$, and $g$ is differentiable at $f(a)$, and that $a$ is a limit point of $\mbox{{\rm dom}}(g\circ f)$. Then the composition $(g\circ f)$ is differentiable at $a$, and

$$(g\circ f)'(a)=g'\left(f(a)\right)\cdot f'(a).$$

Proof: From our hypotheses, there exist functions

$$D_af\colon\mbox{{\rm dom}}(f)\to\mbox{{\bf C}},\hspace{2em} D_{f(a)}g\colon\mbox{{\rm dom}}(g)\to\mbox{{\bf C}}$$

such that $D_af$ is continuous at $a$, $D_{f(a)}g$ is continuous at $f(a)$ and

$$f(z) = f(a)+(z-a)D_af(z)\mbox{ for all } z\in\mbox{{\rm dom}}(f)$$ (10.14)

$$g(z) = g\left(f(a)\right)+\left(z-f(a)\right)D_{f(a)}g(z) \mbox{ for all }z\in\mbox{{\rm dom}}(g).$$ (10.15)

If $z\in\mbox{{\rm dom}}(g\circ f)$, then $f(z)\in\mbox{{\rm dom}}(g)$, so we can replace $z$ in (10.15) by $f(z)$ to get

$$g\left(f(z)\right)=g\left(f(a)\right)+\left(f(z)-f(a)\right)D... ...ft(f(z)\right) \mbox{ for all }z\in\mbox{{\rm dom}}(g\circ f).$$

Using (10.14) to rewrite $f(z)-f(a)$, we get

$$(g\circ f)(z)=(g\circ f)(a)+(z-a)D_af(z)(D_{f(a)}g\circ f)(z)\mbox{ for all }z\in\mbox{{\rm dom}}(g\circ f).$$

Hence we have

$$D_a(g\circ f)=D_af\cdot\left((D_{f(a)}g)\circ f\right)$$

and $D_a(g\circ f)$ is continuous at $a$. Hence $g\circ f$ is differentiable at $a$ and

$$\begin{eqnarray*} (g\circ f)'(a)&=&D_a(g\circ f)(a)=D_af(a)D_{f(a)}g\left(f(a)\r... ...\\ &=&f'(a)\cdot g'\left(f(a)\right).\mbox{ $\mid\!\mid\!\mid$} \end{eqnarray*}$$

10.16   Theorem (Reciprocal rule.) Let $f$ be a complex function, and let $a\in\mbox{{\rm dom}}(f)$. If $f$ is differentiable at $a$ and $f(a)\neq 0$, then $${{1\over f}}$$ is differentiable at $a$ and $${ \left({1\over f}\right)'(a)={{-f'(a)}\over {\left(f(a)\right)^2}}}$$.

Proof: If $${v(z)={1\over z}\mbox{ for all }z\in\mbox{{\bf C}}\backslash\{0\}}$$, we saw above that $v$ is differentiable and $${ v'(z)=-{1\over {z^2}}}$$. Let $f$ be a complex function, and let $a\in\mbox{{\bf C}}$. Suppose $f$ is differentiable at $a$, and $f(a)\neq 0$. Then $${(v\circ f)(z)={1\over {f(z)}}}$$. By the chain rule $v\circ f$ is differentiable at $a$, and

$$(v\circ f)'(a)=v'\left(f(a)\right)\cdot f'(a)=-{1\over {f(a)^2}}f'(a).\mbox{ $\mid\!\mid\!\mid$}$$

10.17   Exercise (Product rule.) A Let $f,g$ be complex functions. Suppose $f$ and $g$ are both differentiable at $a$, and that $a$ is a limit point of $\mbox{{\rm dom}}(f)\cap\mbox{{\rm dom}}(g)$. Show that $fg$ is differentiable at $a$, and that

$$(fg)'(a)=f'(a)g(a)+f(a)g'(a).$$

10.18   Exercise (Power rule.) Let $f$ be a complex function, and suppose that $f$ is differentiable at $a\in\mbox{{\bf C}}$. Show that $f^n$ is differentiable at $a$ for all $n\in\mbox{{\bf Z}}_{\geq 1}$ and

$$(f^n)'(a)=n\left(f(a)\right)^{n-1}f'(a).$$

(Use induction.)

10.19   Exercise (Power rule.) A Let $f$ be a complex function. suppose that $f$ is differentiable at $a\in\mbox{{\bf C}}$, and $f(a)\neq 0$. Show that $f^n$ is differentiable at $a$ for all $n \in \mbox{{\bf Z}}^-$, and that

$$(f^n)'(a)=n\left(f(a)\right)^{n-1}f'(a).$$

for all $n \in \mbox{{\bf Z}}^-.$

10.20   Exercise (Quotient rule.) A Let $f,g$ be complex functions and let
$a\in\mbox{{\bf C}}$. Suppose $f$ and $g$ are differentiable at $a$ and $g(a)\neq 0$, and $a$ is a limit point of $${\mbox{{\rm dom}}\left({f\over g}\right)}$$. Show that $${{f\over g}}$$ is differentiable at $a$ and

$$\left( {f\over g}\right)'(a)={{g(a)f'(a)-f(a)g'(a)}\over {g(a)^2}}.$$