How to check if derivative of a function with respect to a complex variable exists?

For page specific messages

For page author info

Methods to check if a function has a derivative at a point.

I- Using the definition
II- Using properties of derivative
III- Using Cauchy Riemann Equations

What will not be done here :-

The functions with branch point singularity will be discussed separately

$\newcommand{\pp}[2][]{\frac{\partial #1}{\partial #2}}$

Recall and Discuss
I-Definition of derivative of a function w.r.t. a complex variable.

Let $f(z)$ be a function of a complex variable. Consider the limit
\begin{equation}\lim_{\Delta z \to 0} \frac{f(z+\Delta z) -f(z)}{\Delta z}\end{equation}
Since $\Delta z$ is a complex number, it has a real part $h$ and an imaginary part $k$.
\begin{equation}\Delta z = h+ik\label{Derv}\end{equation}
The limit $\Delta z $ going to zero means double limit of $h\to 0$ and $k\to 0$.
The answer of Eq(1) should be independent of how limit is taken.
If the limit in Eq(1) exists, it defines the derivative $\frac{df}{dz}$.
Learn More About Limits

II-Properties of derivative

If two functions $f (z)$ and $g(z)$ are differentiable at a point, the linear combination $\alpha f (z) + \beta g(z)$ is differentiable at that point.
If two functions $f (z)$ and $g(z)$ are differentiable at a point,, the product$f (z)g(z)$ is differentiable at the same point.
If two functions $f (z)$ and $g(z)$ are differentiable at a point and if the function$g(z)$ does not vanish at that point, the function $f (z)/g(z)$ is differentiable .
If $f,g$ are two functions, the composite function $h\equiv f\circ g$ is defined by$h(z)=f(g(z))$.
If $g(z)$ is differentiable at point $z$ and if $f (w)$ is differentiable at $w = g(z)$, the composite function $(f \circ g)(z)\equiv f(g(z))$ is differentiable at $z$.

III-Cauchy Riemann Equations

If a function $f(z)$ has real and imaginary parts $u(x,y), v(x,y)$ i.e. $f(z)= u(x,y) +i v(x,y)$, we say the $f(z)$ obeys Cauchy Riemann equations if
\[\pp[u]{x}=\pp[v]{y}, \qquad \pp[u]{y} =- \pp[v]{x}\]

Question : Given a function what are the ways to check if the derivative exists at a point?

ToHide:

even

Que: How to check if the derivative exists?
Ans: Click to learn three methods.

Method I : Use definition of derivative, Check if the limit exists

Method II : Use properties of the derivative and continuity of derivatives

Method III : Verify the following:
Cauchy Riemann equations are obeyed, and
All the four partial derivatives in the CR equations exist and are continuous

Give Example of Method I : Using the definition of derivative.

Method I : Use definition of derivative to check if the derivative exists
for the following functions or not?

$f(z) = constant, c$
$g(z) = z$
$h(z)=\bar{z}$

follows from the definition.

Then the derivative of $f(z)$ and $g(z)$ exists everywhere.

Details for $f(z)$ :-
\begin{eqnarray}
\dd[f(z)]{z}
&=&\lim_{\Delta z\to 0} \frac{f(z+\Delta z)-f(z)}{\Delta z}\\
&=&\frac{c-c}{\Delta z}\\
&=&0.
\end{eqnarray}

Similarly, consider $g(z)$:-

\begin{eqnarray}
\dd[g(z)]{z}
&=& \lim_{\Delta z\to 0} \frac{g(z+\Delta z)-g(z)}{\Delta z}\\
&=&\lim \frac{(z+\Delta z)-z}{\Delta z}\\
&=&\lim \frac{\Delta z}{\Delta z}\\.
&=& 1
\end{eqnarray}

The function $h(z)$ is not differentiable any where.
Let us look at $h(z)$ more closely.

\begin{eqnarray}
\frac{h(z)}{z}
&=& \lim_{\Delta z\to 0} \frac{h(z+\Delta z)-h(z)}{\Delta z}\\
&=&\lim \frac{(\bar{z}+\overline{\Delta z}\,)-\bar{z}}{\Delta z}\\
&=&\lim \frac{\overline{\Delta z}}{\Delta z}\\.
&=& ??
\end{eqnarray}

It is not obvious what is to be done?

Let us look at the last line closely.
Write $\Delta z = h+ik$. Then
\begin{eqnarray}
\dd[h(z)]{z}
&=&\lim \frac{\overline{\Delta z}}{\Delta z}\\.
&=& \lim_{h\to 0,k \to 0} \frac{h -ik}{h+ik}
\end{eqnarray}
We need to take limit $h\to 0$ and $k\to 0$.
This is a double limit.
The answer should not depend on the order in which the limit is taken.
Also the answer should not depend on how the limit is taken.

Three ways of taking limit in the above example:
Method 1: Take limit $h\to 0$ first and $k\to 0$ next. In this case we would get
\[\lim_{k\to 0} \Big(\lim_{ h\to 0} \frac{h -ik}{h+ik}\Big) =\lim_{k\to 0} \frac{-ik}{ik}= -1\]

Method 2: Take limit $k\to 0$ first.
\[\lim_{h\to 0} \Big(\lim_{ k\to 0} \frac{h -ik}{h+ik}\Big) =\lim_{k\to 0} \frac{h}{h}= 1\]

Method 3: Set $h= m \epsilon, k=n\epsilon $ where $m, n$ are some fixed numbers. In the limit $\epsilon\to0$, $h,k$ tend to zero and we get
\[\lim_{h\to 0} \Big(\lim_{ k\to 0} \frac{h -ik}{h+ik}\Big) =\lim_{\epsilon\to 0} \frac{m-in}{m+in}= \frac{m-in}{m+in}\]
the answer depends on $m$ and $n$.
Therefore $h(z)=\bar{z}$ is not differentiable.

Que: What is the use, if any, of these simple examples?
Ans: Wait till you see more examples.

Give Examples of Method-II Using properties of derivative.

$z$ is differentiable implies that $z^2, z^3,... z^n...$ are all differentiable everywhere.
It follows from (1) that all polynomials are differentiable everywhere.
If P(z) and Q(z) are polynomials, then the function $ P(z)/Q(z) $ is differentiable everywhere except at points where Q(z) is zero.

Method-III: Give Examples of using CR equations.

EXAMPLE 1

The function $f(z)=e^z$ has real and imaginary parts $u(x,y) = e^x \cos y, v(x,y) =e^x\sin y$ These obey CR equations. The four partial derivatives

\[ \frac{du}{dx}= e^x \cos y, \frac{\partial v}{\partial y}= \cos y e^x, \quad \frac{\partial u}{\partial y}=-e^x \sin y , \frac{\partial v}{\partial x} = e^x \sin y\]

satisfy Cauchy Riemann equations and are continuous.

EXAMPLE 2

If $g(z) = \bar{z}$, then $u(x,y)=x, v(x,y)=-y$ and the CR equations are not satisfied, except at $z=0$.
Recalling chain rule, it then follows that any function that involves $\bar{z}$ wil,l in general, not be differentiable except possibly at $z=0$.

An example is derivative of $h(z)=|z|^2$ exists only at $z=0$. Check!

Can we use the result that $e^z$ is differentiable everywhere to get information about other functions?

Yes, combine this with properties of the derivative. You will get that \[\sin z, \cos z, \sinh z, \cosh z \] are differentaible everywhere.

Can we say something about other trigonometric and hyperbolic functions? For example $\tan z$?

We have $\tan z =\frac{\sin z}{\cos z}$ and the numerator and denominator are differentiable everywhere.

Therefore $\tan z$ is differentiable everywhere except at points where the denominator $\cos z$ is zero. Similarly

$cot z$ is differentiable everywhere except at points where the denominator $\sin z$ is zero?
$\text{tanh} z$ is differentiable everywhere except at points where the denominator $\text{cosh} z$ is zero?
$\text{coth} z$ is differentiable everywhere except at points where the denominator $\text{sinh} z$ is zero?

Click for a few more complicated examples

You may try your hands on the following functions.

$\exp(\sin z)$ is differentiable everywhere
WHY? because chain rule tells me
\frac{f(g(z)}{z} = \frac{f(w)}{w} \frac{g(z)}{z}\) where $w=g(z)$, is differentiable if $g(z)$
is differentiable at $z$ and $f(w)$ is differentiable at $w=g(z)$.

Similarly

$e^{e^z}$, $ \sin (e^{z^2}) $, $\sin ^2z+ 3\sin^3z - 5\sin ^7z$ are all differentiable everywhere.

Exclude node summary :

Some Important Points to Remember

If we get different answers for a limit using different ways, the limit does not exist.This is particularly useful for showing that limit defining the derivative does not exist.
The composite function $f(g(z)$ is not differentiable at $z_0$ if one, or both, of the following condition holds.
1. $g(z)$ is not differentiable at $z_0$, or
2. f(w)\) is not differentiable at $w=f(z_0)$
If CR equations are not satisfied by a function, the function is not differentiable.
On the other hand if CR equations are obeyed the function may, or may not, be differentiable.In order to confirm the existence of derivative, you must check that in addition to obeying CR equations, all the four partial derivatives in CR equation must exist and be continuous.

Example to Remember

Let $P(z), Q(z)$ be polynomials with no common factor. Then rational function $P(z)/Q(z)$ is differentiable except where the polynomial $Q(z)$ vanishes.
The exponential function $\exp(z$ is differentiable everywhere.
Trigonometric functions $\sin z,\cos z$ and hyperbolic functions $\cosh z, \sinh z$ are differentiable everywhere,
If two functions $f(z), g(z)$ are differentiable everywhere, ratio $f(z)/g(z)$ is differentiable at all points where $g(z)$ is not zero. So for example $\tan z = \sin z /\cos z$ is differentiable excepts where $\cos z=0$. Similar comments apply to $\cot z, \text{coth}z, \text{tanh}z $.
Chain rule can be used to determine where the derivative of function of function exist. So we have the rule $\frac{df(g(z))}{dz} = \frac{df(w)}{dw} \frac{d g(z)}{d z}$ where $w=g(z)$, is differentiable if $g(z)$is differentiable at $z$ and $f(w)$ is differentiable at $w=g(z)$.
The derivative of $f(g(z)$ does not exist at $z=z_0$, if at any one (or more) of the following conditions is (are) not obeyed.
- $f(z)$ is not differentiable at $z=g(z_0)$
- $g(z)$ is not differentiable at $z=z_0$
- $g(z)$ vanishes at $z_0$.
The above examples and rules will help in deciding by inspection in many cases.
Practice problems will be given separately

The content in this node is based on Lecture given in Sunday Physics Group where additional material can be found

4727:DIamond Point

How to check if derivative of a function with respect to a complex variable exists?

ToHide:

Exclude node summary :

Related node(s) :