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Introduction to complex analysis
Introduction to complex analysis

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2.1 The Cauchy–Riemann theorems

Here we will explore the relationship between complex differentiation and real differentiation. To do this, we introduce the notion of a partial derivative and use it to derive the Cauchy–Riemann equations (pronounced ‘coh-she ree-man’). These equations are conditions that any differentiable complex function must satisfy, so they can be used to test whether a given complex function is differentiable. In particular, we use them to investigate the differentiability of the complex exponential function. The technique is to split the exponential function

exp of x plus i times y equals e super x times left parenthesis cosine of y plus i times sine of y right parenthesis

into its real and imaginary parts:

u of x comma y equals e super x times cosine of y and v of x comma y equals e super x times sine of y comma

each of which is a real-valued function of the real variables x and y. The derivative of exp is then calculated by using the derivatives of the real trigonometric and exponential functions, which we assume to be known.

Before we deal with the exponential function, however, let us first consider the simpler function f of z equals z cubed. By writing z equals x plus i times y, we see that

equation sequence part 1 f times left parenthesis x plus i times y right parenthesis equals part 2 left parenthesis x plus i times y right parenthesis cubed equals part 3 left parenthesis x cubed minus three times x times y squared right parenthesis plus i times left parenthesis three times x squared times y minus y cubed right parenthesis full stop

Let us define

u of x comma y equals x cubed minus three times x times y squared and v of x comma y equals three times x squared times y minus y cubed full stop

Then u and v are the real and imaginary parts of f, respectively; that is, u equals Re of f and v equals Im of f. For the moment we will concentrate on the real part u; part of its graph (given by the equation s equals u of x comma y) is shown in Figure 9. Since u is a function of two real variables, its graph is a surface. The height of the surface above the open x comma y close-plane represents the value of the function at the point open x comma y close. For instance, the point cap p on the surface has coordinates left parenthesis two comma one comma two right parenthesis because equation sequence part 1 u of two comma one equals part 2 two cubed minus three multiplication two multiplication one squared equals part 3 two.

Described image
Figure 9 Graph of u of x comma y equals x cubed minus three times x times y squared

Let us now explore the concept of the gradient of the surface at a point such as cap p. We will find that the answer depends on the ‘direction’ from which we approach the point. To make this more precise, consider Figure 10, in which the vertical plane with equation y equals one is shown intersecting the surface in a curve that passes through cap p. By substituting y equals one into u of x comma y equals x cubed minus three times x times y squared, we see that the curve has equation x long right arrow from bar x cubed minus three times x, so we can calculate its gradient at cap p; this is the gradient of the surface in the x-direction at cap p.

Described image
Figure 10 Intersection of the graph of u of x comma y equals x cubed minus three times x times y squared with the vertical plane y equals one

More generally, whenever we intersect the surface with a vertical plane with equation y equals constant, we obtain a curve on the surface with equation x long right arrow from bar x cubed minus three times x times y squared (where y is considered to be fixed). We can find the gradient at any point left parenthesis a comma b comma u of a comma b right parenthesis on this curve by differentiating with respect to x and then substituting x equals a and y equals b. The resulting expression is called the partial derivative of u with respect to x at left parenthesis a comma b right parenthesis, and it is denoted by

prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis full stop

A curly normal partial differential is used rather than a straight d to emphasise that this is a partial derivative, for which we differentiate with respect to one variable and keep the other variable fixed. In our particular case, differentiating u of x comma y equals x cubed minus three times x times y squared with respect to x (and keeping y fixed) gives

prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals three times x squared minus three times y squared comma

and substituting x equals two and y equals one gives

prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis two comma one right parenthesis equals nine full stop

Hence the gradient of the surface in the x-direction at the point cap p is 9. This is a positive value because near the point cap p, u increases as x increases (with y equals one), as you can see from Figure 10.

Figure 11 shows the vertical plane with equation x equals two intersecting the surface in a different curve that passes through cap p.

Described image
Figure 11 Intersection of the graph of u of x comma y equals x cubed minus three times x times y squared with the vertical plane x equals two

Reasoning similarly to before, we see that intersecting the surface with a vertical plane with equation x equals constant gives a curve on the surface, and we can obtain the gradient at a point left parenthesis a comma b comma u of a comma b right parenthesis on this curve by differentiating u of x comma y with respect to y while keeping x fixed (and then substituting x equals a and y equals b). The resulting expression is called the partial derivative of u with respect to y at left parenthesis a comma b right parenthesis, and it is denoted by

prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis full stop

Differentiating u of x comma y equals x cubed minus three times x times y squared with respect to y (and keeping x fixed) gives

prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals negative six times x times y comma so prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis two comma one right parenthesis equals negative 12 semicolon

this is the gradient of the surface in the y-direction at the point cap p. It is a negative value this time, because when x and y are positive, u decreases as y increases (keeping x fixed), as you can see from Figure 11.

You will need to work with partial derivatives a good deal here, so let us state the definitions formally.

Definitions

Let u colon cap a long right arrow double-struck cap r be a function with domain cap a a subset of double-struck cap r squared that contains the point left parenthesis a comma b right parenthesis.

  • The partial derivative of bold-italic u with respect to bold-italic x at bold left parenthesis bold-italic a bold comma bold-italic b bold right parenthesis, denoted prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis, is the derivative of the function x long right arrow from bar u of x comma b at x equals a, provided that this derivative exists.

  • The partial derivative of bold-italic u with respect to bold-italic y at bold left parenthesis bold-italic a bold comma bold-italic b bold right parenthesis, denoted prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis, is the derivative of the function y long right arrow from bar u of a comma y at y equals b, provided that this derivative exists.

Partial derivatives are real derivatives, not complex derivatives.

The next exercise asks you to work out the partial derivatives of the imaginary part of the complex function f of z equals z cubed.

Exercise 16

  • a.Calculate the partial derivatives of v of x comma y equals three times x squared times y minus y cubed.

  • b.Evaluate these partial derivatives at left parenthesis two comma one right parenthesis.

Answer

  • a.Differentiating v of x comma y equals three times x squared times y minus y cubed with respect to x while keeping y fixed, we obtain

    prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals six times x times y full stop

    Differentiating v with respect to y while keeping x fixed, we obtain

    prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals three times x squared minus three times y squared full stop
  • b.So, at left parenthesis x comma y right parenthesis equals left parenthesis two comma one right parenthesis the partial derivatives have the values

    prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis two comma one right parenthesis equals 12 and prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis two comma one right parenthesis equals nine full stop

Let us collect together the partial derivatives of the real and imaginary parts u and v of the function f of z equals z cubed:

multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis equals three times a squared minus three times b squared comma prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis equals six times a times b comma row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis equals negative six times a times b comma prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis equals three times a squared minus three times b squared full stop

As you can see, we have

prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis equals prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis and prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis equals negative prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis full stop

This pair of equations is called the Cauchy–Riemann equations, and they hold true for the real and imaginary parts of any differentiable complex function, as the following important theorem testifies.

Theorem 4 Cauchy–Riemann Theorem

Let f times left parenthesis x plus i times y right parenthesis equals u of x comma y plus i times v of x comma y be defined on a region script cap r containing a plus i times b.

If f is differentiable at a plus i times b, then prefix partial differential of of u divided by prefix partial differential of of x, prefix partial differential of of u divided by prefix partial differential of of y, prefix partial differential of of v divided by prefix partial differential of of x, prefix partial differential of of v divided by prefix partial differential of of y exist at left parenthesis a comma b right parenthesis and satisfy the Cauchy–Riemann equations

prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis equals prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis and prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis equals negative prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis full stop

Proof

Let alpha equals a plus i times b. Suppose that left parenthesis z sub n right parenthesis is any sequence in script cap r minus alpha that converges to alpha. Let us write z sub n equals x sub n plus i times y sub n. According to the definition of a derivative, we have
equation sequence part 1 f super prime of alpha equals part 2 lim over z right arrow alpha of f of z minus f of alpha divided by z minus alpha equals part 3 lim over n right arrow normal infinity of f of z sub n minus f of alpha divided by z sub n minus alpha full stop

Observe that, by expressing f in terms of its real and imaginary parts, we can write

f of z sub n minus f of alpha divided by z sub n minus alpha equals left parenthesis u of x sub n comma y sub n minus u of a comma b divided by left parenthesis x sub n minus a right parenthesis plus i times left parenthesis y sub n minus b right parenthesis right parenthesis plus i of v of x sub n comma y sub n minus v of a comma b divided by left parenthesis x sub n minus a right parenthesis plus i times left parenthesis y sub n minus b right parenthesis full stop
Equation label: (equation 3)

We proceed by choosing two different types of sequences left parenthesis z sub n right parenthesis, and observing the behaviour of the expressions in large brackets in equation 3 (above) in each case.

For our first choice, let us begin by defining left parenthesis x sub n right parenthesis to be any sequence in double-struck cap r minus a that converges to a. Let z sub n equals x sub n plus i times b, so the sequence left parenthesis z sub n right parenthesis converges to alpha equals a plus i times b. By removing a finite number of terms from left parenthesis x sub n right parenthesis, if need be, we can assume that each point z sub n belongs to the open set script cap r minus alpha. Substituting z sub n equals x sub n plus i times b into equation 3 gives

f of z sub n minus f of alpha divided by z sub n minus alpha equals left parenthesis u of x sub n comma b minus u of a comma b divided by x sub n minus a right parenthesis plus i of v of x sub n comma b minus v of a comma b divided by x sub n minus a full stop

We know that the expression on the left-hand side converges (to f super prime of alpha), so its real and imaginary parts (indicated by the bracketed expressions on the right-hand side) converge too. Since left parenthesis x sub n right parenthesis was chosen to be any sequence in double-struck cap r minus a that converges to a, we see from the definition of partial derivatives that prefix partial differential of of u divided by prefix partial differential of of x and prefix partial differential of of v divided by prefix partial differential of of x exist at left parenthesis a comma b right parenthesis and

u of x sub n comma b minus u of a comma b divided by x sub n minus a right arrow prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis and v of x sub n comma b minus v of a comma b divided by x sub n minus a right arrow prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis full stop

In summary, we have

f super prime of alpha equals prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis plus i times prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis full stop
Equation label: (equation 4)

Next let left parenthesis y sub n right parenthesis be any sequence in double-struck cap r minus b that converges to b, and define z sub n equals a plus i times y sub n, so z sub n right arrow alpha. Again, by omitting a finite number of terms from left parenthesis y sub n right parenthesis, if need be, we can assume that z sub n element of script cap r minus alpha for all n. Substituting z sub n equals a plus i times y sub n into equation 3 gives

multiline equation row 1 f of z sub n minus f of alpha divided by z sub n minus alpha equals left parenthesis u of a comma y sub n minus u of a comma b divided by i times left parenthesis y sub n minus b right parenthesis right parenthesis plus i of v of a comma y sub n minus v of a comma b divided by i times left parenthesis y sub n minus b right parenthesis row 2 Blank equals left parenthesis v of a comma y sub n minus v of a comma b divided by y sub n minus b right parenthesis minus i of u of a comma y sub n minus u of a comma b divided by y sub n minus b full stop

Reasoning as before, we see that prefix partial differential of of u divided by prefix partial differential of of y and prefix partial differential of of v divided by prefix partial differential of of y exist at left parenthesis a comma b right parenthesis and

f super prime of alpha equals prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis minus i times prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis full stop
Equation label: (equation 5)
Comparing equation 4 and equation 5 (both above), and equating real and imaginary parts, we obtain the Cauchy–Riemann equations, as required.

Origin of the Cauchy–Riemann equations

The Cauchy–Riemann equations are named after the mathematicians Augustin-Louis Cauchy and Bernhard Riemann (1826–1866), who were among the first to recognise the importance of these equations in complex analysis.

The Cauchy–Riemann equations first appeared in the work of another mathematician, however: the Frenchman Jean le Rond d’Alembert (1717–1783), who is perhaps best remembered for his work in classical mechanics. Indeed, the Cauchy–Riemann equations were written down by d’Alembert in an essay on fluid dynamics in 1752 to describe the velocity components of a two-dimensional irrotational fluid flow.

Described image
Jean le Rond d’Alembert (1717–1783)

The Cauchy–Riemann Theorem gives us another strategy for proving the non-differentiability of a complex function. (Two other strategies were described earlier in Section 1.3.) If a complex function is differentiable, then it must satisfy the Cauchy–Riemann equations. So if those equations do not hold, then the function cannot be differentiable.

Strategy C for non-differentiability

Let f times left parenthesis x plus i times y right parenthesis equals u of x comma y plus i times v of x comma y. If either

prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis not equals prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis or prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis a comma b right parenthesis not equals negative prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis a comma b right parenthesis comma

then f is not differentiable at a plus i times b.

To illustrate this strategy, consider the function

f times left parenthesis x plus i times y right parenthesis equals left parenthesis x squared plus y squared right parenthesis plus i times left parenthesis two times x plus four times y right parenthesis full stop

The real part u and imaginary part v of this function are given by

u of x comma y equals x squared plus y squared and v of x comma y equals two times x plus four times y full stop

Hence

Described image

As you can see, the partial derivatives have been grouped into two pairs according to the Cauchy–Riemann equations.

prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis and prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals negative prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis full stop

In this case, these equations are two times x equals four and two equals negative two times y, which are satisfied only when x equals two and y equals negative one; that is, they are satisfied only when z equals two minus i. If z not equals two minus i, then the Cauchy–Riemann equations fail, so Strategy C tells us that f is not differentiable at z.

Notice that the Cauchy–Riemann Theorem and Strategy C do not tell us whether f is differentiable at the point two minus i at which the Cauchy–Riemann equations are satisfied. To deal with points of this type we need another theorem, which we will come to shortly. First, however, try the following exercise, to practise applying Strategy C.

Exercise 17

Show that each of the following functions fails to be differentiable at all points of double-struck cap c.

  • a.f times left parenthesis x plus i times y right parenthesis equals e super x minus i times e super y

  • b.f of z equals z macron

Answer

  • a.Writing f in the form

    f times left parenthesis x plus i times y right parenthesis equals u of x comma y plus i times v of x comma y comma

    we obtain

    u of x comma y equals e super x and v of x comma y equals negative e super y full stop

    Hence

    prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals e super x and prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals negative e super y full stop

    Since e super x is always positive, whereas negative e super y is always negative, the first of the Cauchy–Riemann equations fails to hold for each left parenthesis x comma y right parenthesis. It follows that f fails to be differentiable at all points of double-struck cap c.

  • b.Writing equation sequence part 1 f of z equals part 2 z macron equals part 3 x minus i times y in the form

    f times left parenthesis x plus i times y right parenthesis equals u of x comma y plus i times v of x comma y comma

    we obtain

    u of x comma y equals x and v of x comma y equals negative y full stop

    Hence

    prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis x comma y right parenthesis equals one and prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis x comma y right parenthesis equals negative one full stop

    It follows that the first of the Cauchy–Riemann equations fails to hold for each left parenthesis x comma y right parenthesis, so f fails to be differentiable at all points of double-struck cap c.

We have seen that if the Cauchy–Riemann equations are not satisfied, then the function is not differentiable. Let us now describe an example to show that even if the Cauchy–Riemann equations are satisfied, then the function may still not be differentiable.

Consider the function f times left parenthesis x plus i times y right parenthesis equals u of x comma y plus i times v of x comma y, where v of x comma y equals zero for all x and y, and

u of x comma y equals case statement case 1column 1 comma of minmin left curly bracket right curly bracket comma x comma y comma comma comma x comma greater than greater than y zero comma case 2column 1 comma zero comma full stop otherwise full stop

The graph of u is shown in Figure 12.

Described image
Figure 12 Graph of u

Since u and v take the value zero at all points on the x- and y-axes, we see that all the partial derivatives vanish at left parenthesis zero comma zero right parenthesis; that is,

multiline equation row 1 prefix partial differential of of u divided by prefix partial differential of of x times left parenthesis zero comma zero right parenthesis equals zero comma prefix partial differential of of v divided by prefix partial differential of of x times left parenthesis zero comma zero right parenthesis equals zero comma row 2 prefix partial differential of of u divided by prefix partial differential of of y times left parenthesis zero comma zero right parenthesis equals zero comma prefix partial differential of of v divided by prefix partial differential of of y times left parenthesis zero comma zero right parenthesis equals zero full stop

However, even though the Cauchy–Riemann equations are satisfied at the origin, f is not differentiable there. To see this, observe that if z sub n equals one solidus n, n equals one comma two comma ellipsis, then

multirelation f of z sub n minus f of zero divided by z sub n minus zero equals u of one solidus n comma zero minus zero divided by one solidus n minus zero equals zero right arrow zero comma

whereas if z sub n equals one solidus n plus i solidus n, n equals one comma two comma ellipsis, then

multirelation f of z sub n minus f of zero divided by z sub n minus zero equals u of one solidus n comma one solidus n minus zero divided by one solidus n plus i solidus n minus zero equals one solidus n divided by one solidus n plus i solidus n equals one divided by one plus i right arrow one divided by one plus i full stop

The two limits zero and one solidus left parenthesis one plus i right parenthesis differ, so f is not differentiable at zero.

This example demonstrates that the differentiability of a complex function does not follow from the Cauchy–Riemann equations alone. However, if certain extra conditions are satisfied, then f is differentiable, as the following theorem reveals.

Theorem 5 Cauchy–Riemann Converse Theorem

Let f times left parenthesis x plus i times y right parenthesis equals u of x comma y plus