1.1 Defining differentiable functions

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 alpha element of double-struck cap c if the difference quotient, defined by

f of z minus f of alpha divided by z minus alpha comma

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

f times left parenthesis alpha plus h right parenthesis minus f of alpha divided by h comma

where h is a complex number. The equivalence of these two limits can be justified by noting that if z equals alpha plus h, then ‘z right arrow alpha’ is equivalent to ‘h right arrow zero’.

Definitions

Let f be a complex function whose domain contains the point alpha. Then the derivative of bold-italic f at bold-italic alpha is

lim over z right arrow alpha of f of z minus f of alpha divided by z minus alpha times left parenthesis or lim over h right arrow zero of f times left parenthesis alpha plus h right parenthesis minus f of alpha divided by h right parenthesis comma

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

The derivative of f at alpha is denoted by f super prime of alpha, and the function

f super prime colon z long right arrow from bar f super prime of z

is called the derivative of bold-italic f. The domain of f super prime is the set of all complex numbers at which f is differentiable.

The function f super prime is sometimes called the derived function of f.

Remarks

  1. The existence of the limit

    lim over z right arrow alpha of f of z minus f of alpha divided by z minus alpha

    implicitly requires the domain of f to contain alpha as one of its limit points. This always holds if the domain of f is a region.

  2. The derivative f super prime of z is sometimes written as d times f divided by d times z times left parenthesis z right parenthesis or d divided by d times z times left parenthesis f of z right parenthesis.

  3. Some other texts use the phrase complex derivative in place of derivative to draw a distinction with the standard real derivative of a function f colon double-struck cap r squared long right arrow double-struck cap r squared (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 f of z equals z squared.

Solution

The domain of f of z equals z squared is the whole of double-struck cap c, so let alpha be an arbitrary point of double-struck cap c. Then

multiline equation row 1 f super prime of alpha equals lim over z right arrow alpha of f of z minus f of alpha divided by z minus alpha row 2 Blank equals lim over z right arrow alpha of z squared minus alpha squared divided by z minus alpha row 3 Blank equals lim over z right arrow alpha of z plus alpha full stop

Now z long right arrow from bar z plus alpha is a basic continuous function, continuous at alpha, so we see that equation sequence part 1 f super prime of alpha equals part 2 alpha plus alpha equals part 3 two times alpha.

Since alpha is an arbitrary complex number, the derivative of f is the function f super prime of z equals two times z. Its domain is the whole of double-struck cap c.

Notice the way in which the troublesome z minus alpha term cancels from the numerator and the denominator in the calculation of f super prime of alpha 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

  1. the constant function f of z equals one

  2. the function f of z equals z.

Answer

  1. f of z equals one is defined on the whole of double-struck cap c, so let alpha element of double-struck cap c. Then

    multiline equation row 1 f super prime of alpha equals lim over z right arrow alpha of f of z minus f of alpha divided by z minus alpha row 2 Blank equals lim over z right arrow alpha of one minus one divided by z minus alpha row 3 Blank equation sequence part 1 equals part 2 lim over z right arrow alpha of zero divided by z minus alpha equals part 3 zero full stop

    Since alpha is an arbitrary complex number, f is differentiable on the whole of double-struck cap c, and its derivative is the zero function

    f super prime of z equals zero times left parenthesis z element of double-struck cap c right parenthesis full stop

  2. f of z equals z is defined on the whole of double-struck cap c, so let alpha element of double-struck cap c. Then

    multiline equation row 1 f super prime of alpha equals lim over z right arrow alpha of f of z minus f of alpha divided by z minus alpha row 2 Blank equals lim over z right arrow alpha of z minus alpha divided by z minus alpha row 3 Blank equation sequence part 1 equals part 2 lim over z right arrow alpha of one equals part 3 one full stop

Since alpha is an arbitrary complex number, f is differentiable on the whole of double-struck cap c, and its derivative is the constant function

f super prime of z equals one times left parenthesis z element of double-struck cap c right parenthesis full stop

Example 1 and Exercise 1 show that the functions f of z equals one, f of z equals z and f of z equals z squared are differentiable on the whole of double-struck cap c. Functions that have this property are given a special name.

Definition

A function is entire if it is differentiable on the whole of double-struck cap c.

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

Exercise 2

Use the definition of derivative to find the derivative of the function f of z equals one solidus z. Explain why f is not entire.

Answer

The domain of f of z equals one solidus z is the region double-struck cap c minus zero. Since f super prime of alpha cannot exist unless f is defined at alpha, we confine our attention to alpha not equals zero. Then

multiline equation row 1 f super prime of alpha equals lim over z right arrow alpha of f of z minus f of alpha divided by z minus alpha row 2 Blank equals lim over z right arrow alpha of left parenthesis one solidus z right parenthesis minus left parenthesis one solidus alpha right parenthesis divided by z minus alpha row 3 Blank equals lim over z right arrow alpha of alpha minus z divided by z times alpha times left parenthesis z minus alpha right parenthesis row 4 Blank equals lim over z right arrow alpha of negative one divided by z times alpha full stop

Now z long right arrow from bar negative one solidus left parenthesis z times alpha right parenthesis is a basic continuous function with domain double-struck cap c minus zero, so we see that

equation sequence part 1 f super prime of alpha equals part 2 lim over z right arrow alpha of negative one divided by z times alpha equals part 3 negative one divided by alpha squared full stop

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

f super prime of z equals negative one divided by z squared times left parenthesis z not equals zero right parenthesis full stop

The function f is not entire since its domain is not double-struck cap c.

Although the function f of z equals one solidus z is not entire, it is differentiable on the whole of its domain double-struck cap c minus zero. This domain is a region because it is obtained by removing the point 0 from double-struck cap c. (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.

Definitions

A function that is differentiable on a region script cap r is said to be analytic on bold-script cap r. If the domain of a function f is a region, and if f is differentiable on its domain, then f is said to be analytic. A function is analytic at a point bold-italic alpha if it is differentiable on a region containing alpha.

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

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 g of z equals absolute value of z squared has a derivative at 0, but at no other point. This means that there is no region on which g is differentiable, and hence no point at which g is analytic.

By contrast, f of z equals one solidus z 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 3 Three regions on which f of z equals one solidus z 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.

  1. An entire function is analytic at every point of double-struck cap c.

  2. If a function is differentiable at each point of a set, then it is analytic on that set.

Answer

  1. True.

  2. False. (The set must be a region.)

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

Theorem 1

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

Proof

Let f be differentiable at alpha; then

lim over z right arrow alpha of f of z minus f of alpha divided by z minus alpha equals f super prime of alpha full stop

To prove that f is continuous at alpha, we will show that f of z right arrow f of alpha as z right arrow alpha. We do this by proving the equivalent result that f of z minus f of alpha right arrow zero as z right arrow alpha.

By the Product Rule for limits of functions, we have

multiline equation row 1 lim over z right arrow alpha of f of z minus f of alpha equals lim over z right arrow alpha of f of z minus f of alpha divided by z minus alpha multiplication lim over z right arrow alpha of z minus alpha row 2 Blank equation sequence part 1 equals part 2 f super prime of alpha multiplication zero equals part 3 zero full stop

Hence f of z right arrow f of alpha as z right arrow alpha, so f is continuous at alpha.

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

Theorem 2 Linear Approximation Theorem

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

f of z equals sum with 3 summands f of alpha plus left parenthesis z minus alpha right parenthesis times f super prime of alpha plus e of z comma

where e is an ‘error function’ satisfying e of z solidus left parenthesis z minus alpha right parenthesis right arrow zero as z right arrow alpha.

Informally speaking, the statement ‘e of z solidus left parenthesis z minus alpha right parenthesis right arrow zero as z right arrow alpha’ means that ‘e of z tends to zero faster than z minus alpha does’.

Proof

We have to show that the function e defined by

e of z equals f of z minus f of alpha minus left parenthesis z minus alpha right parenthesis times f super prime of alpha

satisfies e of z solidus left parenthesis z minus alpha right parenthesis right arrow zero as z right arrow alpha.

Dividing e of z by z minus alpha and letting z tend to alpha, we obtain

multiline equation row 1 lim over z right arrow alpha of e of z divided by z minus alpha equals lim over z right arrow alpha of f of z minus f of alpha divided by z minus alpha minus f super prime of alpha row 2 Blank equation sequence part 1 equals part 2 f super prime of alpha minus f super prime of alpha equals part 3 zero comma

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.

OpenLearn - Introduction to complex analysis
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