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Introduction to differentiation

The Open University — Fri, 24 Mar 2023 12:44:08 GMT

The derivative of a real function at a point is the gradient of the tangent to the graph of at . This gradient is calculated by finding the gradient of the chord joining the point to a (nearby) point , and taking the limit as approaches (Figure 1).

Figure _unit1.1.1 Figure 1 A chord between two points on a graph

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The figure shows Cartesian axes labelled x and y. It is focused on the upper-right quadrant. A smooth curve that looks like part of the graph of a quadratic function with positive coefficient of x-squared starts in the upper-left quadrant a little above the x-axis and a little to the left of the y-axis. At this point it is nearly horizontal. As x increases the curve slopes upwards with steadily increasing gradient. A line with positive gradient also starts in the upper-left quadrant, below the curve, and slopes upwards into the upper-right quadrant. The curve and the line intersect at two points in the upper-right quadrant, both of which are marked with solid dots. The coordinates of these points are illustrated by vertical and horizontal broken line segments joining each point to the x and y axes. From the lower point of intersection, the vertical broken line segment crosses the x-axis at a point marked c. The horizontal broken line segment crosses the y-axis at a point marked f of c. From the upper point of intersection, the vertical broken line segment crosses the x-axis at a point marked x. The horizontal broken line segment crosses the y-axis at a point marked f of x. From the lower point of intersection a horizontal line segment is drawn to the right. It meets a vertical line segment that extends downwards from the upper point of intersection. They form a right-angled triangle with a segment of the original sloping line as its hypotenuse. The length of the horizontal side of this triangle is marked x minus c. The length of its vertical side is marked f of x minus f of c.

Figure 1 A chord between two points on a graph

Now, the gradient of the chord is equal to the ratio

This ratio is often called the difference quotient for at , and its limit as tends to provides a formal definition of the (real) derivative of at , denoted by . Thus

In the case of complex functions, it is difficult to think about derivatives in terms of gradients of tangents, since the graph of a complex function is not drawn in two dimensions. Instead we define the derivative of a complex function directly in terms of difference quotients, using the notion of complex limits.

Fortunately, the derivatives of many complex functions turn out to have the same form as those of the corresponding real functions. For example, the derivative of the complex sine function is the complex cosine function, and the complex exponential function is its own derivative. On the other hand, the complex modulus function fails to be differentiable at any point of , even though the real modulus function (Figure 2) is differentiable at every point of . This reflects the fact that complex differentiation imposes a much stronger condition on functions than does real differentiation. Indeed, as the course progresses, you will see that differentiable complex functions have remarkably pleasant properties. For example, if a complex function can be differentiated once throughout a region, then it can be differentiated any number of times. There is no equivalent result for real functions.

Figure _unit1.1.2 Figure 2 Graph of

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The figure shows Cartesian axes labelled x and y. It is focused on the upper half-plane. The graph of the function y equals the modulus of x is drawn and labelled. In the upper-left quadrant the graph is a line sloping down with gradient negative 1 to the origin. In the upper-right quadrant the graph is a line sloping up with gradient 1 from the origin.

Figure 2 Graph of

Section 1 will define complex differentiation and show how the definition can be used to establish whether a function is differentiable. By introducing rules for combining differentiable functions, you will see how complex polynomial and rational functions can be differentiated just as in the real case. The end of Section 1 will give a geometric interpretation of complex differentiation by introducing the idea of a complex scale factor.

Section 2 will introduce the concept of partial differentiation for real functions of two real variables, and use it to establish a relationship between complex differentiation and real differentiation. This relationship sometimes enables us to differentiate a complex function using real derivatives. Indeed, at the end of the section, this approach will be used to show that the complex exponential function is its own derivative.

Box _unit1.1.1

This OpenLearn course is an extract from the Open University course M337 Complex analysis.

1 Derivatives of complex functions

The Open University — Fri, 24 Mar 2023 12:44:08 GMT

After working through this section, you should be able to:

use the definition of derivative to show that a given function is differentiable, and to find its derivative
use the Combination Rules for differentiation to differentiate polynomial and rational functions
use various strategies to show that a given function is not differentiable at a point
interpret the derivative of a complex function at a point as a rotation and a scaling of a small disc centred at the point.

1.1 Defining differentiable functions

The Open University — Fri, 24 Mar 2023 12:44:08 GMT

As with limits and continuity, the way in which the derivative of a complex function is defined is similar to the real case. Thus a complex function is said to have a derivative at a point if the difference quotient, defined by

tends to a limit as tends to . Equivalently, it is sometimes more convenient to replace by , and examine the corresponding limit as tends to 0. The difference quotient then has the form

where is a complex number. The equivalence of these two limits can be justified by noting that if , then ‘’ is equivalent to ‘’.

Box _unit1.2.1 Definitions

Let be a complex function whose domain contains the point . Then the derivative of at is

provided that this limit exists. If it does exist, then is differentiable at . If is differentiable at every point of a set , then is differentiable on . A function is differentiable if it is differentiable on its domain.

The derivative of at is denoted by , and the function

is called the derivative of . The domain of is the set of all complex numbers at which is differentiable.

The function is sometimes called the derived function of .

Remarks

The existence of the limit
implicitly requires the domain of to contain as one of its limit points. This always holds if the domain of is a region.
The derivative is sometimes written as or .
Some other texts use the phrase complex derivative in place of derivative to draw a distinction with the standard real derivative of a function (which we will not need).

In certain cases it is easy to find the derivative of a function directly from the definition above.

Example 1

Use the definition of derivative to find the derivative of the function .

Solution

The domain of is the whole of , so let be an arbitrary point of . Then

Now is a basic continuous function, continuous at , so we see that .

Since is an arbitrary complex number, the derivative of is the function . Its domain is the whole of .

Notice the way in which the troublesome term cancels from the numerator and the denominator in the calculation of in the preceding example. This often happens when you calculate derivatives directly from the definition.

Exercise 1

Use the definition of derivative to find the derivative of

a.the constant function
b.the function .

Answer

a. is defined on the whole of , so let . Then
Since is an arbitrary complex number, is differentiable on the whole of , and its derivative is the zero function
b. is defined on the whole of , so let . Then

Since is an arbitrary complex number, is differentiable on the whole of , and its derivative is the constant function

Example 1 and Exercise 1 show that the functions , and are differentiable on the whole of . Functions that have this property are given a special name.

Box _unit1.2.2 Definition

A function is entire if it is differentiable on the whole of .

Not all functions are entire; indeed, many interesting aspects of complex analysis arise from functions that fail to be differentiable at various points of .

Exercise 2

Use the definition of derivative to find the derivative of the function . Explain why is not entire.

Answer

The domain of is the region . Since cannot exist unless is defined at , we confine our attention to . Then

Now is a basic continuous function with domain , so we see that

Since is an arbitrary non-zero complex number, the derivative of is

The function is not entire since its domain is not .

Although the function is not entire, it is differentiable on the whole of its domain . This domain is a region because it is obtained by removing the point 0 from . (The removal of a point from a region leaves a region.) As the course progresses, you will discover that regions provide an excellent setting for analysing the properties of differentiable functions. We therefore make the following definitions.

Box _unit1.2.3 Definitions

A function that is differentiable on a region is said to be analytic on . If the domain of a function is a region, and if is differentiable on its domain, then is said to be analytic. A function is analytic at a point if it is differentiable on a region containing .

It follows immediately from the definition that if a function is analytic on a region , then it is automatically analytic at each point of .

Notice that a function can have a derivative at a point without being analytic at the point. For example, in the next section we will ask you to show that the function has a derivative at 0, but at no other point. This means that there is no region on which is differentiable, and hence no point at which is analytic.

By contrast, is analytic at every point of its domain. It is an analytic function, and it is analytic on any region that does not contain 0. Three such regions are illustrated in Figure 3.

Figure _unit1.2.1 Figure 3 Three regions on which is analytic

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This figure consists of three copies of the complex plane arranged side by side, from left to right. The axes are not labelled. Each of the three diagrams has a different region, bounded by broken lines or curves, and shaded inside. The left-hand diagram shows a horseshoe-shaped region symmetrical about the vertical axis; the boundary of the shape is drawn as a broken line. The arms of the horseshoe are pointing down and they straddle the horizontal axis. The open end of the horseshoe is at the bottom, so that the origin is not inside the region. The middle diagram shows an open annulus, and both of its boundaries are drawn as broken lines. It consists of the area between two concentric circles, centred at the origin. The right-hand diagram shows a square region centred on the origin, with two vertical and two horizontal sides; its boundaries are drawn as broken lines. It is symmetrical about both axes. The origin itself is marked with a hollow dot.

Figure 3 Three regions on which is analytic

An appropriate choice of region can often simplify the analysis of complex functions.

Exercise 3

Classify each of the following statements as True or False.

a.An entire function is analytic at every point of .
b.If a function is differentiable at each point of a set, then it is analytic on that set.

Answer

a.True.
b.False. (The set must be a region.)

There is a close connection between differentiation and continuity. The function , for example, is not only differentiable, but also continuous on its domain. This is no accident for, as in real analysis, differentiability implies continuity.

Box _unit1.2.4 Theorem 1

Let be a complex function that is differentiable at . Then is continuous at .

Proof

Let be differentiable at ; then

To prove that is continuous at , we will show that as . We do this by proving the equivalent result that as .

By the Product Rule for limits of functions, we have

Hence as , so is continuous at .∎

In fact, differentiability implies more than continuity. Continuity asserts that for all close to , is close to . For differentiable functions, this ‘closeness’ has the ‘linear’ form described in the following theorem.

Box _unit1.2.5 Theorem 2 Linear Approximation Theorem

Let be a complex function that is differentiable at . Then can be approximated near by a linear polynomial. More precisely,

where is an ‘error function’ satisfying as .

Informally speaking, the statement ‘ as ’ means that ‘ tends to zero faster than does’.

Proof

We have to show that the function defined by

satisfies as .

Dividing by and letting tend to , we obtain

as required.∎

Theorem 1 and Theorem 2 are often used to investigate the properties of differentiable functions. An illustration of this occurs in the next subsection, where Theorem 1 is used in a proof of the Combination Rules for differentiation. Later in this section we use Theorem 2 to give a geometric interpretation of complex differentiation.

1.2 Combining differentiable functions

The Open University — Fri, 24 Mar 2023 12:44:08 GMT

It would be tedious if we had to use the definition of the derivative every time we needed to differentiate a function. Fortunately, once the derivatives of simple functions like and are known, we can find the derivatives of other more complicated functions by applying the following theorem.

Box _unit1.2.6 Theorem 3 Combination Rules for Differentiation

Let and be complex functions with domains and , respectively, and let be a limit point of . If and are differentiable at , then

a.Sum Rule is differentiable at , and
b.Multiple Rule is differentiable at , for , and
c.Product Rule is differentiable at , and
d.Quotient Rule is differentiable at (provided that , and

We remark that if the domains and in Theorem 3 are regions, then every point of is a limit point of and of .

In addition to these rules, there is a corollary to Theorem 3, known as the Reciprocal Rule, which is a special case of the Quotient Rule.

Box _unit1.2.7 Corollary Reciprocal Rule for Differentiation

Let be a function that is differentiable at . If , then is differentiable at , and

The proof of the Combination Rules for differentiation uses the Combination Rules for limits of functions. In the next example we illustrate the method by proving the Product Rule for differentiation. We use the Sum, Product and Multiple Rules for limits of functions, and we also use the fact that if a function is differentiable at , then it is continuous at , so .

Example 2

Prove the Product Rule for differentiation.

Solution

Let . Then

The proofs of the other Combination Rules are similar. We ask you to prove the Sum and Multiple Rules in Exercise 4, and the Quotient Rule later in Exercise 12.

Exercise 4

Prove the following rules for differentiation.

a.Sum Rule
b.Multiple Rule

Answer

a.Let . Then
b.Let , for . Then

The Combination Rules enable us to differentiate any polynomial or rational function. (Recall that a rational function is the quotient of two polynomial functions.)

For example, since the function is entire with derivative , we can use the Product Rule repeatedly to show that the function

is entire, and that its derivative is

(This result can be proved formally using the Principle of Mathematical Induction.) Next, we can use this fact, together with the Sum and Multiple Rules, to prove that any polynomial function is entire, and that its derivative is obtained by differentiating the polynomial function term by term. For example,

In general, we have the following corollary to Theorem 3.

Box _unit1.2.8 Corollary Differentiating Polynomial Functions

Let be the polynomial function

where and . Then is entire with derivative

Since a rational function is a quotient of two polynomial functions, it follows from the corollary on differentiating polynomial functions and the Quotient Rule that a rational function is differentiable at all points where its denominator is non-zero; that is, at all points of its domain.

Example 3

Find the derivative of

and specify its domain.

Solution

By the corollary on differentiating polynomial functions, the derivative of is

and the derivative of is

Provided that is non-zero, we can apply the Quotient Rule to obtain

Since is non-zero everywhere apart from and , it follows that the domain of is .

Exercise 5

Find the derivative of each of the following functions. In each case specify the domain of the derivative.

Answer

a.By the corollary on differentiating polynomial functions, we have
b.By the Quotient Rule,
Now, if and only if , so the domain of is

So, any rational function is differentiable on the whole of its domain. What is more, this domain must be a region because it is obtained by removing a finite number of points (zeros of the denominator) from .

Box _unit1.2.9 Corollary

Any rational function is analytic.

A particularly simple example of a rational function is , where is a positive integer. This can be differentiated by means of the Reciprocal Rule:

If is used to denote the negative integer , then we can write and . In this form, it is apparent that the formula for differentiating a negative integer power is the same as the formula for differentiating a positive integer power. The only difference is that for negative powers, 0 is excluded from the domain. We state these observations as a final corollary to Theorem 3.

Box _unit1.2.10 Corollary

Let . The function has derivative

The domain of is if and if .

1.3 Non-differentiability

The Open University — Fri, 24 Mar 2023 12:44:08 GMT

In Theorem 1 you saw that differentiability implies continuity. An immediate consequence of this is the following test for non-differentiability.

Box _unit1.2.11 Strategy A for non-differentiability

If is discontinuous at , then is not differentiable at .

Example 4

Show that there are no points of the negative real axis at which the function is differentiable.

Solution

The function is discontinuous at all points of the negative real axis. It follows that there are no points of the negative real axis at which is differentiable.

Exercise 6

Show that there are no points of the negative real axis at which the principal logarithm function

is differentiable.

Answer

The function Arg is discontinuous at each point of the negative real axis. It follows that Log is discontinuous at each point of the negative real axis, and hence that there are no points on it at which Log is differentiable.

The converse of Theorem 1 is not true; if a function is continuous at a point, then it does not follow that it is differentiable at the point. A particularly striking illustration of this is provided by the modulus function . This is continuous on the whole of and yet, as you will see, it fails to be differentiable at any point of .

Since is continuous, Strategy A cannot be used to show that is not differentiable at a given point . Instead we return to the definition of derivative and show that the difference quotient for fails to have a limit.

In general, if the domain of a function contains as one of its limit points, then the existence of the limit

means that for each sequence in that converges to ,

exists, and has a value that is independent of the sequence .

So, if two such sequences and can be found for which

then cannot be differentiable at .

In the next example, you will see that is not differentiable at . This result should not surprise you because the real modulus function is not differentiable at . Indeed, the proof is identical to that of the real case.

Example 5

Prove that is not differentiable at 0.

Solution

We need to find two sequences and that converge to 0 which, when substituted into the difference quotient, yield sequences with different limits. A simple choice is to pick sequences and that approach 0 along the real axis: one from the right, and one from the left, as shown in Figure 4.

Figure _unit1.2.2 Figure 4 Sequences converging to from the right and left

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This figure consists of two copies of the complex plane, one above the other. The upper copy shows on the positive real axis a sequence of points that approach zero from the right. The origin is labelled zero and marked with a solid blue dot. The point 1 is labelled, and the first three points in the sequence, starting with 1, are marked with solid black dots. An ellipsis to the left of the third point (one third) indicates that the sequence continues indefinitely getting closer to zero. A horizontal arrow above the real axis, between zero and 1, points from right to left. Above it is the label, open bracket, z sub n, close bracket. The lower copy of the complex plane shows on the negative real axis a sequence of points that approach zero from the left. The origin is labelled zero and marked with a solid blue dot. The point negative 1 is labelled, and the first three points in the sequence, starting with negative 1, are marked with solid black dots. An ellipsis to the right of the third point (negative one third) indicates that the sequence continues indefinitely getting closer to zero. A horizontal arrow above the negative real axis, between negative 1 and zero, points from left to right. Above it is the label, open bracket, z prime sub n, close bracket.

Figure 4 Sequences converging to from the right and left

There is no point in picking sequences that are more complicated than they need to be, so let , . Then

Now let , . Then

Since the two limits do not agree, the difference quotient does not have a limit as tends to 0. It follows that is not differentiable at 0.

The next exercise asks you to extend the method used in Example 5 to show that is not differentiable at any point of .

Exercise 7

Let be any non-zero complex number, and consider the circle through centred at the origin. By choosing one sequence that approaches along the circumference of the circle, and another sequence that approaches along the ray from through , prove that is not differentiable at .

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This figure consists of two copies of the complex plane, arranged side by side. The axes are not labelled. The left-hand copy shows a circle centred at the origin. An arbitrary point on the circle in the upper-right quadrant is labelled alpha and marked with a solid blue dot. A sequence of points on the circumference starts in the upper-left quadrant and approaches the point alpha at decreasing intervals. The first four points in the sequence are marked with solid black dots. An ellipsis between the fourth point and the point alpha indicates that the sequence of points continues indefinitely getting closer to alpha. Above the sequence of points is a curved arrow pointing from left to right. It is labelled: open bracket, z sub n, close bracket. The right-hand copy of the complex plane shows the same circle centred at the origin, and the same arbitrary point alpha on the circle is labelled and marked with a solid blue dot. A sequence of points outside the circle approaches the point alpha at decreasing intervals along the normal to the circle at alpha. The first three points in the sequence are marked with solid black dots. An ellipsis between the third point and the point alpha indicates that the sequence of points continues indefinitely getting closer to alpha. An arrow directed towards alpha is drawn outside the circle parallel to the sequence of points. It is labelled: open bracket, z prime sub n, close bracket.

Answer

Let , . Then tends to along the circumference of the circle, and

Now let , . Then tends to along the ray from through , and

Since for , these two limits do not agree. It follows that is not differentiable at .

The modulus function illustrates an important difference between real and complex differentiation. When the modulus function is treated as a real function, the limit of its difference quotient has to be taken along the real line. But when treated as a complex function, the limit of the difference quotient is required to exist however the limit is taken. This explains why the real modulus function is differentiable at all non-zero real points, whereas the complex modulus function fails to be differentiable at any point of . More generally, it shows that complex differentiability is a much stronger condition than real differentiability.

In Exercise 7 you were asked to prove that the modulus function fails to be differentiable by observing that its behaviour along the circumference of a circle centred at 0 is different from its behaviour along a ray. Similar observations can be applied to other functions. For example, in the next exercise you may find it helpful to notice that directions of paths parallel to the imaginary axis are reversed by the function , whereas directions of paths parallel to the real axis are left unchanged (Figure 5).

Figure _unit1.2.3 Figure 5 Images of horizontal and vertical lines under

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This figure consists of two copies of the complex plane, arranged side by side. The axes are not labelled. On the left-hand diagram, a bold vertical line, drawn in black, passes through an arbitrary point on the positive horizontal axis. It is marked with a direction arrow pointing upwards, and labelled capital gamma sub 1. A bold horizontal line, drawn in blue, passes through an arbitrary point on the positive vertical axis. It is marked with a direction arrow pointing to the right, and labelled capital gamma sub 2. The two bold lines intersect at right angles in the upper-right quadrant. Between the two copies of the complex plane is a curved horizontal arrow pointing from left to right. It is labelled f of z equals z bar. On the right-hand diagram, a bold vertical line, drawn in black, passes through the same arbitrary point on the positive horizontal axis. It is marked with a direction arrow pointing downwards, and labelled f of capital gamma sub 1. A bold horizontal line, drawn in blue, passes through a point on the negative vertical axis. (This point is the same distance below the origin as the point used on the vertical axis in the left-hand diagram was above it.) The horizontal line is marked direction arrow pointing to the right, and labelled f of capital gamma sub 2. The two bold lines intersect at right angles in the lower-right quadrant.

Figure 5 Images of horizontal and vertical lines under

Exercise 8

Show that there are no points of at which the complex conjugate function is differentiable.

Answer

Let be an arbitrary complex number. Directions of paths parallel to the imaginary axis through are reversed by , while directions of paths parallel to the real axis are not. This suggests looking at the sequences and , .

First let ; then

Now let ; then

Since these two limits do not agree, and since is arbitrary, it follows that there are no points of at which is differentiable.

For some functions , you may be able to find a sequence that converges to for which the sequence

Equation label: (equation 1)

is divergent. In such cases, there is no need to look for a second sequence.

Example 6

Show that the function is not differentiable at 0.

Solution

Strategy A cannot be used here, since is continuous at 0. Instead we look for a sequence that converges to for which the sequence 1 (above) is divergent. To make the square roots easy to handle, let , . Then

This sequence tends to infinity, and is therefore divergent. It follows that is not differentiable at 0.

The methods exemplified above for showing that a function is not differentiable at a given point can be summarised as follows.

Box _unit1.2.12 Strategy B for non-differentiability

To prove that a function is not differentiable at , apply the strategy for proving that a limit does not exist to the difference quotient

If you think that a given function is not differentiable, then you should try to apply Strategy A or Strategy B. A third strategy for proving that is not differentiable at a point will appear in Section 2.1. If, on the other hand, you think that the function is differentiable, then you should try to find the derivative.

Exercise 9

Decide whether each of the following functions is differentiable at . If it is, then find its derivative at .

Answer

a.The fact that is constant along the imaginary axis, but variable parallel to the real axis, suggests that is not differentiable at (or anywhere else, for that matter). It also suggests looking at the sequences and , .
First let ; then
Now let ; then
Since these two limits do not agree, it follows that is not differentiable at .
b. is a polynomial function, so for all . Thus .
c. is not differentiable at , since it is not continuous at .

1.4 Higher-order derivatives

The Open University — Fri, 24 Mar 2023 12:44:08 GMT

In Exercise 2 you saw that the function has derivative , a result that you can also obtain using the Reciprocal Rule. If you now apply the Reciprocal Rule to the derivative , then you obtain a function

In general, for a differentiable function , the function is called the second derivative of , and is denoted by . Continued differentiation gives the so-called higher-order derivatives of . These are denoted by , and the values , are called the higher-order derivatives of at .

Since the dashes in this notation can be rather cumbersome, we often indicate the order of the derivative by a number in brackets. Thus mean the same as , respectively. Here the brackets in are needed to avoid confusion with the fourth power of .

When we wish to discuss a derivative of general order, we will refer to the th derivative of . It is often possible to find a formula for the th derivative in terms of . For example, if , then

so the th derivative is given by

(This can be proved formally by the Principle of Mathematical Induction.)

One interesting feature about this formula is that the domain remains the same, no matter how often the function is differentiated. This is a special case of a much more general result which states that: a function that is analytic on a region has derivatives of all orders on . Here we confine our attention to first derivatives, and we continue to do this in the next subsection by giving a geometric interpretation of the first derivative.

1.5 A geometric interpretation of derivatives

The Open University — Fri, 24 Mar 2023 12:44:08 GMT

As we mentioned in this session’s introduction, the derivative of a real function is often pictured geometrically as the gradient of the graph of the function. This interpretation is useful in real analysis, but it is of little use in complex analysis, since the graph of a complex function is not two-dimensional.

Fortunately, there is another way of interpreting derivatives that works for complex functions.

If a complex function is differentiable at a point , then any point close to is mapped by to a point close to .

Indeed, by the Linear Approximation Theorem,

where as . So if , then, to a close approximation,

Multiplication of by has the effect of scaling by the factor and rotating it about 0 through the angle ; see Figure 6. We refer to as a complex scale factor, because it causes both a scaling and a rotation.

Figure _unit1.2.4 Figure 6 Scaling and rotating

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This figure consists of a diagram of the complex plane, focused on the right half-plane. The axes are not labelled. A line segment extends from the origin to a point in the lower-right quadrant. The point is marked with a solid dot and labelled z minus alpha. A second line segment extends from the origin to a point in the upper-right quadrant. The second line segment is longer than the first. Above it is the text, scale by the modulus of f prime of alpha. The point at the end of the second line segment is marked with a solid dot and labelled f prime of alpha times open bracket, z minus alpha, close bracket. A curved arrow between the two line segments points in the anticlockwise direction. Next to it is the text, rotate by Arg f prime of alpha.

Figure 6 Scaling and rotating

We can rewrite the equation above as

Equation label: (equation 2)

From this we see that is obtained by scaling and rotating the vector based at by the complex scale factor , as illustrated in Figure 7.

Figure _unit1.2.5 Figure 7 Interpreting a derivative as a complex scale factor

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This figure consists of two copies of the complex plane side by side, both focused on the upper-right quadrant. The axes are not labelled. On the left-hand diagram there are two points in the upper-right quadrant, alpha marked with a solid black dot and z marked with a solid blue dot. The point z lies to the right of the point alpha and a little below it. The vector from alpha to z is drawn as a solid blue line with an arrowhead. Between the left-hand and right-hand diagrams is a curved arrow pointing from left to right and labelled f. On the right-hand diagram, three points are marked with solid dots. The first, which is the closest to the origin, is drawn in black labelled f of alpha. The second, also in black, lies above and to the right of the first point, and is labelled f of z. The third point, drawn in blue, lies to the right and a little below the point labelled f of alpha. It is not labelled. Two vectors are also shown, both starting from the point f of alpha. The first vector, drawn in blue, runs from the point f of alpha to the unlabelled blue point. It is an identical copy of the vector from alpha to z that was drawn in the left-hand diagram. The second vector, drawn in black, starts at the point f of alpha and finishes at the point f of z. Above this second vector is the text, scale by the modulus of f prime of alpha. Between the two vectors is a curved arrow, pointing in the anticlockwise direction. Next to it is the text, rotate by Arg f prime of alpha.

Figure 7 Interpreting a derivative as a complex scale factor

Another useful way to picture how behaves geometrically is to consider the effect it has on a small disc centred at (still assuming that ). From equation 2 (above), we see that, to a close approximation, a small disc centred at is mapped to a small disc centred at . In the process, the disc is rotated through the angle , and it is scaled by the factor (see Figure 8). As usual, the rotation is anticlockwise if is positive, and clockwise if it is negative.

Figure _unit1.2.6 Figure 8 The approximate image of a disc centred at a point , where

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This figure consists of two copies of the complex plane side by side, both focused on the upper-right quadrant. The axes are not labelled. On the left-hand diagram a point in the upper-right quadrant is marked with a solid black dot and labelled alpha. A small circular disc is drawn with alpha as its centre and a solid black circle as its boundary. Inside the disc a second point, a little to the right and below the point alpha, is marked with a solid green dot. It is labelled z. The disc is shaded blue. Between the left-hand and right-hand diagrams is a curved arrow pointing from left to right. Above it is the label f. On the right-hand diagram the point f of alpha is marked with a solid black dot and labelled. A circular disc is drawn with the point f of alpha as its centre and a solid black circle as its boundary. This disc has radius approximately 50% greater than the radius of the disc surrounding the point alpha in the left-hand diagram. Inside the disc a second point, above and to the right of the point f of alpha, is marked with a solid green dot and labelled f of z. The disc is shaded blue. Outside the disc, above it and slightly to the left, is the text, scale by the modulus of f prime of alpha. Outside the disc, diametrically opposite the first text, is a curved arrow parallel to the circular boundary of the disc, which points in an anticlockwise direction. Next to it is the text, rotate by Arg f prime of alpha.

Figure 8 The approximate image of a disc centred at a point , where

The geometric interpretation of derivatives is more complicated if , and we do not discuss it here.

Example 7

Using the notion of a complex scale factor, describe what happens to points close to under the function .

Solution

To a close approximation, a small disc centred at is mapped by to a small disc centred at

In the process, the disc is scaled by the factor and rotated through the angle .

Now , so

which has modulus and principal argument .

So scales the disc by the factor and rotates it anticlockwise through the angle .

Exercise 10

Using the notion of a complex scale factor, describe what happens to points close to under the function

Answer

To a close approximation, a small disc centred at is mapped by to a small disc centred at

In the process the disc is scaled by the factor and rotated through the angle .

By the Quotient Rule,

This has modulus and principal argument .

So scales the disc by the factor and rotates it clockwise through the angle .

It is important to bear in mind that the complex scale factor interpretation of a derivative is only an approximation, and that it is unlikely to be reliable far from the point under consideration.

1.6 Further exercises

The Open University — Fri, 24 Mar 2023 12:44:08 GMT

Here are some further exercises to end this section.

Exercise 11

Use the definition of derivative to find the derivative of the function

Answer

The function is defined on the whole of . Let . Then

Since is an arbitrary complex number, is differentiable on the whole of , and the derivative is the function

Exercise 12

Prove the Quotient Rule for differentiation.

Answer

Let . Then

Using the Combination Rules for limits of functions, the continuity of , and the fact that , we can take limits to obtain

Exercise 13

Find the derivative of each of the following functions . In each case specify the domain of .

Answer

a.By the Combination Rules,
The domain of is .
b.By the Combination Rules,
Since , the domain of is .
c.By the Reciprocal Rule,
The roots of are . The domain of is therefore .
d.By the Sum Rule and the rule for differentiating integer powers,
The domain of is .

Exercise 14

Use Strategy B to show that there are no points of at which the function

is differentiable.

Answer

Consider an arbitrary complex number , where . Let , . Then , and

Now let , . Then , and

Since the two limits do not agree, it follows that fails to be differentiable at each point of .

Exercise 15

Describe the approximate geometric effect of the function

on a small disc centred at the point 2.

Answer

To a close approximation, a small disc centred at 2 is mapped by to small disc centred at . In the process, the disc is scaled by the factor and rotated through the angle .

By the Quotient Rule,

So scales the disc by the factor 4 and rotates it anticlockwise through the angle .