Basic concepts of the theory of functions of a complex variable. Theory of functions of a complex variable

17.06.2019 Windows 7, XP

where
are real numbers, and is a special character called imaginary unit . For the imaginary unit, by definition, it is assumed that
.

(4.1) – algebraic form complex number, and
called real part complex number, and
-imaginary part .

Number
called complex conjugate to the number
.

Let two complex numbers be given
,
.

1. sum
complex numbers and called a complex number

2. difference
complex numbers and called a complex number

3. work
complex numbers and called a complex number

4. Private from dividing a complex number to a complex number
called a complex number

Remark 4.1. That is, operations on complex numbers are introduced according to the usual rules of arithmetic operations on literal expressions in algebra.

Example 4.1. Complex numbers are given. Find

Decision. 1) .

4) Multiplying the numerator and denominator by the complex conjugate of the denominator, we get

trigonometric form complex number:

where
is the modulus of a complex number,
is the argument of a complex number. Corner defined ambiguously, up to a term
:

,
.

- the main value of the argument, determined by the condition

, (or
).

indicative form complex number:

Root
th degree of number It has different values, which are found by the formula

where
.

Points corresponding to values
, are vertices of a regular
a square inscribed in a circle of radius
centered at the origin.

Example 4.2. Find all root values
.

Decision. Imagine a complex number
in trigonometric form:

, where
.

Then
. Therefore, by formula (4.2)
has four meanings:

,
.

Assuming
, we find

,
,

, .

Here we have converted the values of the argument to its main value.

Sets on the complex plane

Complex number
depicted on a plane
dot
with coordinates
. Module
and argument
correspond to the polar coordinates of the point
.

It is useful to remember that inequality
defines a circle centered at a point radius . Inequality
defines a half-plane located to the right of the straight line
, and the inequality
- a half-plane located above a straight line
. In addition, the system of inequalities
sets the angle between rays
and
outgoing from the origin of coordinates.

Example 4.3. Draw the area defined by the inequalities:
.

Decision. The first inequality corresponds to a ring centered at a point
and two radii 1 and 2, circles are not included in the area (Fig. 4.1).

The second inequality corresponds to the angle between the rays
(bisector of the 4th coordinate angle) and
(positive axis direction
). The rays themselves do not enter the region (Fig. 4.2).

The desired area is the intersection of the two obtained areas (Fig. 4.3)

4.2. Functions of a complex variable

Let a single-valued function
defined and continuous in the domain
, a is a piecewise-smooth closed or non-closed oriented curve lying in
. Let, as usual,
,, where
,
- real functions of variables and .

Calculating the integral of a function
complex variable reduces to the calculation of ordinary curvilinear integrals, namely

If the function
is analytic in a simply connected domain
containing points and , then the Newton-Leibniz formula holds:

where
- some antiderivative for the function
, that is
in area
.

In integrals of functions of a complex variable, one can change the variable, and integration by parts is similar to how it is done when calculating integrals of functions of a real variable.

Note also that if the integration path is a part of a straight line starting from the point , or part of a circle centered at a point , then it is useful to change the variable of the form
. In the first case
, a - real integration variable; in the second case
, a is the real integration variable.

Example 4.4. Calculate
along a parabola
from the point
to the point
(Figure 4.4).

Decision. Let us rewrite the integrand in the form

Then
,
. We apply formula (4.3):

Because
, then
,
. That's why

Example 4.5. Calculate Integral
, where - arc of a circle
,
(Fig. 4.5) .

Decision. Suppose
, then
,
,
. We get:

Function
, single-valued and analytic in the ring
, decomposes in this ring into Laurent series

In formula (4.5) the series
called main part Laurent series, and the series
called right part Laurent row.

Definition 4.1. Dot calledisolated singular point functions
if there is a neighborhood of this point where the function
is analytic everywhere except for the point itself .

Function
in the vicinity of the point can be expanded in a Laurent series. In this case, three different cases are possible when the Laurent series:

1) does not contain terms with negative degrees of difference
, that is

(the Laurent series does not contain the main part). In this case called removable singular point functions
;

2) contains a finite number of terms with negative degrees of difference
, that is

and
. In this case, the point called pole of order functions
;

3) contains an infinite number of terms with negative powers:

In this case, the point called essential point functions
.

When determining the nature of an isolated singular point, it is not necessary to look for a Laurent series expansion. You can use various properties of isolated keypoints.

1) is a removable singular point of the function
if there is a finite limit of the function
at the point :

2) is a pole of the function
, if

3) is an essential singular point of the function
, if at
the function has no limit, neither finite nor infinite.

Definition 4.2. Dot calledzero
order (or multiplicities ) functions
if the following conditions are met:

…,

.

Remark 4.2. Dot then and only then is zero
order functions
when, in some neighborhood of this point, the equality

where is the function
is analytic at the point and

4) dot is the pole of the order (
) functions
if this point is zero of order for function
.

5) let - isolated singular point of a function
, where
- functions analytic at a point . And let the dot is the order zero functions
and order zero functions
.

At
dot is the pole of the order
functions
.

At
dot is a removable singular point of the function
.

Example 4.6. Find isolated points and determine their type for the function
.

Decision. Functions
and
- analytical in the entire complex plane. Hence, the singular points of the function
are the zeros of the denominator, that is, the points where
. There are infinitely many such points. First, this is the point
, as well as points satisfying the equation
. From here
and
.

Consider a point
. At this point we get:

,
,

,
.

The order of zero is
.

,
,

,
.

So the point
is a second-order pole (
).

. Then

,
.

The order of the zero numerator is
.

,
,
.

The order of zero denominator is
. Therefore, the points
at
are first-order poles ( simple poles ).

Theorem 4.1. (Cauchy residue theorem ). If the function
is analytic on the boundary areas
and everywhere inside the region, except for a finite number of singular points
, then

When calculating integrals, it is worth carefully finding all the singular points of the function
, then draw a contour and special points, and after that select only those points that are inside the integration contour. Making the right choice without a picture is often difficult.

Method of calculating the deduction
depends on the type of singular point. Therefore, before calculating the residue, you need to determine the type of the singular point.

1) function residue at a point is equal to the coefficient of minus the first power in the Laurent expansion
in the vicinity of the point :

This statement is true for all types of isolated points, and therefore in this case it is not necessary to determine the type of a singular point.

2) the residue at the removable singular point is equal to zero.

3) if is a simple pole (pole of the first order), and the function
can be represented as
, where
,
(note that in this case
), then the residue at the point equals

In particular, if
, then
.

4) if is a simple pole, then

5) if - pole
th order function
, then

Example 4.7. Calculate Integral
.

Decision. Find singular points of the integrand
. Function
has two singular points
and
Only a point falls inside the contour
(Fig. 4.6). Dot
is a second-order pole, since
is zero of multiplicity 2 for the function
.

Then by formula (4.7) we find the residue at this point:

By virtue of Theorem 4.1 we find

Functions of a complex variable.
Differentiation of functions of a complex variable.

This article opens a series of lessons in which I will consider typical problems related to the theory of functions of a complex variable. To successfully master the examples, you must have basic knowledge of complex numbers. In order to consolidate and repeat the material, it is enough to visit the page. You will also need skills to find second order partial derivatives. Here they are, these partial derivatives ... even now I was a little surprised how often they occur ...

The topic that we are starting to analyze is not particularly difficult, and in the functions of a complex variable, in principle, everything is clear and accessible. The main thing is to adhere to the basic rule, which is derived by me empirically. Read on!

The concept of a function of a complex variable

First, let's refresh our knowledge about the school function of one variable:

Function of one variable is a rule according to which each value of the independent variable (from the domain of definition) corresponds to one and only one value of the function . Naturally, "x" and "y" are real numbers.

In the complex case, the functional dependence is given similarly:

Single-valued function of a complex variable is the rule that everyone integrated the value of the independent variable (from the domain) corresponds to one and only one comprehensive function value. In theory, multivalued and some other types of functions are also considered, but for simplicity, I will focus on one definition.

What is the function of a complex variable?

The main difference is that numbers are complex. I'm not being ironic. From such questions they often fall into a stupor, at the end of the article I will tell a cool story. In the classroom Complex numbers for dummies we considered a complex number in the form . Since now the letter "Z" has become variable, then we will denote it as follows: , while "x" and "y" can take different valid values. Roughly speaking, the function of a complex variable depends on the variables and , which take "usual" values. The following point logically follows from this fact:

The function of a complex variable can be written as:
, where and are two functions of two valid variables.

The function is called real part functions .
The function is called imaginary part functions .

That is, the function of a complex variable depends on two real functions and . To finally clarify everything, let's look at practical examples:

Example 1

Decision: The independent variable "z", as you remember, is written as , therefore:

(1) Substituted into the original function.

(2) For the first term, the reduced multiplication formula was used. In the term, the brackets were opened.

(3) Carefully squared, not forgetting that

(4) Rearrangement of terms: first rewrite terms , in which there is no imaginary unit(first group), then terms, where there is (second group). It should be noted that it is not necessary to shuffle the terms, and this step can be skipped (in fact, by performing it orally).

(5) The second group is taken out of brackets.

As a result, our function turned out to be represented in the form

Answer:
is the real part of the function .
is the imaginary part of the function .

What are these functions? The most ordinary functions of two variables, from which one can find such popular partial derivatives. Without mercy - we will find. But a little later.

Briefly, the algorithm of the solved problem can be written as follows: we substitute into the original function, carry out simplifications and divide all terms into two groups - without an imaginary unit (real part) and with an imaginary unit (imaginary part).

Example 2

Find the real and imaginary part of a function

This is a do-it-yourself example. Before you throw yourself into battle on the complex plane with checkers naked, let me give you the most important advice on the topic:

BE CAREFUL! You need to be careful, of course, everywhere, but in complex numbers you should be careful more than ever! Remember that, carefully expand the brackets, don't lose anything. According to my observations, the most common mistake is the loss of sign. Do not hurry!

Full solution and answer at the end of the lesson.

Now cube. Using the abbreviated multiplication formula, we derive:
.

Formulas are very convenient to use in practice, as they greatly speed up the solution process.

Differentiation of functions of a complex variable.

I have two news: good and bad. I'll start with a good one. For a function of a complex variable, the rules of differentiation and the table of derivatives of elementary functions are valid. Thus, the derivative is taken in exactly the same way as in the case of a function of a real variable.

The bad news is that for many functions of a complex variable, there is no derivative at all, and you have to figure out is differentiable one function or another. And “figuring out” how your heart feels is associated with additional troubles.

Consider a function of a complex variable . For this function to be differentiable, it is necessary and sufficient that:

1) For there to be partial derivatives of the first order. Forget about these notations right away, since in the theory of the function of a complex variable, another version of the notation is traditionally used: .

2) To carry out the so-called Cauchy-Riemann conditions:

Only in this case will the derivative exist!

Example 3

Decision decomposed into three successive stages:

1) Find the real and imaginary parts of the function. This task was analyzed in previous examples, so I will write it down without comment:

Since , then:

Thus:

is the imaginary part of the function .

I will dwell on one more technical point: in what order write terms in real and imaginary parts? Yes, basically it doesn't matter. For example, the real part can be written like this: , and imaginary - like this: .

2) Let us check the fulfillment of the Cauchy-Riemann conditions. There are two of them.

Let's start by checking the condition. We find partial derivatives:

Thus, the condition is fulfilled.

Undoubtedly, the good news is that partial derivatives are almost always very simple.

We check the fulfillment of the second condition:

It turned out the same thing, but with opposite signs, that is, the condition is also fulfilled.

The Cauchy-Riemann conditions are satisfied, therefore, the function is differentiable.

3) Find the derivative of the function. The derivative is also very simple and is found according to the usual rules:

The imaginary unit in differentiation is considered a constant.

Answer: - real part is the imaginary part.
The Cauchy-Riemann conditions are met, .

There are two more ways to find the derivative, they are of course used less often, but the information will be useful for understanding the second lesson - How to find the function of a complex variable?

The derivative can be found using the formula:

In this case:

Thus

It is necessary to solve the inverse problem - in the resulting expression, you need to isolate . In order to do this, it is necessary in terms and to take out of brackets:

The reverse action, as many have noticed, is somewhat more difficult to perform, for verification it is always better to take the expression and on the draft or verbally open the brackets back, making sure that it will turn out exactly

Mirror formula for finding the derivative:

In this case: , that's why:

Example 4

Determine the real and imaginary parts of a function . Check the fulfillment of the Cauchy-Riemann conditions. If the Cauchy-Riemann conditions are met, find the derivative of the function.

A short solution and an approximate sample of finishing at the end of the lesson.

Are the Cauchy-Riemann conditions always satisfied? Theoretically, they are more often not fulfilled than they are. But in practical examples, I don’t remember a case where they were not executed =) Thus, if your partial derivatives “did not converge”, then with a very high probability we can say that you made a mistake somewhere.

Let's complicate our functions:

Example 5

Determine the real and imaginary parts of a function . Check the fulfillment of the Cauchy-Riemann conditions. Calculate

Decision: The solution algorithm is completely preserved, but at the end a new fad is added: finding the derivative at a point. For the cube, the required formula has already been derived:

Let's define the real and imaginary parts of this function:

Attention and again attention!

Since , then:

Thus:
is the real part of the function ;
is the imaginary part of the function .

Checking the second condition:

It turned out the same thing, but with opposite signs, that is, the condition is also fulfilled.

The Cauchy-Riemann conditions are satisfied, therefore, the function is differentiable:

Calculate the value of the derivative at the required point:

Answer:, , the Cauchy-Riemann conditions are satisfied,

Functions with cubes are common, so an example to consolidate:

Example 6

Determine the real and imaginary parts of a function . Check the fulfillment of the Cauchy-Riemann conditions. Calculate .

Decision and sample finishing at the end of the lesson.

In the theory of complex analysis, other functions of a complex argument are also defined: exponential, sine, cosine, etc. These functions have unusual and even bizarre properties - and it's really interesting! I really want to tell you, but here, it just so happened, not a reference book or a textbook, but a solution, so I will consider the same task with some common functions.

First about the so-called Euler formulas:

For anyone valid numbers, the following formulas are valid:

You can also copy it into your notebook as a reference.

Strictly speaking, there is only one formula, but usually, for convenience, they also write a special case with a minus in the indicator. The parameter does not have to be a single letter, it can be a complex expression, a function, it is only important that they take only valid values. Actually, we will see it right now:

Example 7

Find derivative.

Decision: The general line of the party remains unshakable - it is necessary to single out the real and imaginary parts of the function. I will give a detailed solution, and comment on each step below:

Since , then:

(1) Substitute for "z".

(2) After substitution, it is necessary to separate the real and imaginary parts first in exponent exhibitors. To do this, open the brackets.

(3) We group the imaginary part of the indicator, putting the imaginary unit out of brackets.

(4) Use school action with powers.

(5) For the multiplier, we use the Euler formula , while .

(6) We open the brackets, as a result:

is the real part of the function ;
is the imaginary part of the function .

Further actions are standard, let's check the fulfillment of the Cauchy-Riemann conditions:

Example 9

Determine the real and imaginary parts of a function . Check the fulfillment of the Cauchy-Riemann conditions. So be it, we will not find the derivative.

Decision: The solution algorithm is very similar to the previous two examples, but there are very important points, so I will again comment on the initial stage step by step:

Since , then:

1) We substitute instead of "z".

(2) First, select the real and imaginary parts inside the sinus. For this purpose, open the brackets.

(3) We use the formula , while .

(4) Use parity of hyperbolic cosine: and hyperbolic sine oddness: . Hyperbolics, although not of this world, but in many ways resemble similar trigonometric functions.

Eventually:
is the real part of the function ;
is the imaginary part of the function .

Attention! The minus sign refers to the imaginary part, and in no case should we lose it! For a visual illustration, the result obtained above can be rewritten as follows:

Let's check the fulfillment of the Cauchy-Riemann conditions:

The Cauchy-Riemann conditions are met.

Answer:, , the Cauchy-Riemann conditions are satisfied.

With cosine, ladies and gentlemen, we understand on our own:

Example 10

Determine the real and imaginary parts of the function. Check the fulfillment of the Cauchy-Riemann conditions.

I deliberately picked up more complicated examples, because everyone can handle something like peeled peanuts. At the same time, train your attention! Nutcracker at the end of the lesson.

Well, in conclusion, I will consider another interesting example when the complex argument is in the denominator. We met a couple of times in practice, let's analyze something simple. Oh, I'm getting old...

Example 11

Determine the real and imaginary parts of the function. Check the fulfillment of the Cauchy-Riemann conditions.

Decision: Again, it is necessary to separate the real and imaginary parts of the function.
If , then

The question arises, what to do when "Z" is in the denominator?

Everything is simple - the standard will help method of multiplying the numerator and denominator by the conjugate expression, it has already been used in the examples of the lesson Complex numbers for dummies. Let's remember the school formula. In the denominator we already have , so the conjugate expression will be . Thus, you need to multiply the numerator and denominator by:

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Saint Petersburg State

Electrotechnical University "LETI"

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Theory of functions of a complex variable

Guidelines

to practical exercises

in higher mathematics

Saint Petersburg

Publishing house of Saint-Petersburg Electrotechnical University "LETI"

UDC 512.64(07)

TFKP: Guidelines for solving problems / comp.: V.G. Dyumin, A.M. Kotochigov, N.N. Sosnovsky. St. Petersburg: Publishing house of St.

Approved

editorial and publishing council of the university

as guidelines

The functions of the complex variable ,, in the general case differ from the mappings of the real plane
in itself only form of record. An important and extremely useful object is the class of a function of a complex variable,

having the same derivative as functions of one variable. It is known that functions of several variables can have partial and directional derivatives, but, as a rule, derivatives in different directions do not coincide, and it is not possible to speak of a derivative at a point. However, for functions of a complex variable, it is possible to describe the conditions under which they admit differentiation. The study of the properties of differentiable functions of a complex variable is the content of guidelines. The instructions are oriented towards demonstrating how the properties of such functions can be used to solve various problems. Successful mastering of the material presented is impossible without elementary skills in computing with complex numbers and familiarity with the simplest geometric objects defined in terms of inequalities relating the real and imaginary parts of a complex number, as well as its modulus and argument. A summary of all the information needed for this can be found in the guidelines.

The standard apparatus of mathematical analysis: limits, derivatives, integrals, series is widely used in the text of guidelines. Where these concepts have their own specifics, in comparison with functions of one variable, the corresponding explanations are given, but in most cases it is enough to separate the real and imaginary parts and apply the standard apparatus of real analysis to them.

1. Elementary functions of a complex variable

It is natural to start discussing the conditions for differentiability of functions of a complex variable by clarifying which elementary functions have this property. From the obvious relation

Differentiability of any polynomial follows. And, since the power series can be differentiated term by term inside the circle of its convergence,

then any function is differentiable at points around which it can be expanded in a Taylor series. This is a sufficient condition, but, as will soon become clear, it is also a necessary one. It is convenient to support the study of functions of one variable by derivative by controlling the behavior of the graph of the function. For functions of a complex variable, this is not possible. The points of the graph lie in a space of dimension 4, .

Nevertheless, some graphic representation of the function can be obtained by considering the images of sufficiently simple sets of the complex plane
arising under the influence of a given function. For example, consider, from this point of view, a few simple functions.

Linear function

This simple function is very important, since any differentiable function is locally similar to a linear one. Consider the action of the function with maximum detail

here
-- complex number modulus and is his argument. Thus, the linear function performs stretch, rotation and shear. Therefore, a linear mapping maps any set to a similar set. In particular, under the influence of a linear mapping, lines turn into lines, and circles into circles.

Function

This function is next in complexity to the linear one. It is difficult to expect that it will take any line to a line, and a circle to a circle, simple examples show that this does not happen, nevertheless, it can be shown that this function takes the set of all lines and circles into itself. To verify this, it is convenient to pass to the real (coordinate) description of the mapping

The proof requires a description of the inverse mapping

Consider the equation if
, then we get the general equation of a straight line. If
, then

Therefore, at
the equation of an arbitrary circle is obtained.

Note that if
and
, then the circle passes through the origin. If
and
, then you get a straight line passing through the origin.

Under the action of inversion, the considered equation will be rewritten in the form

, (
)

or . It can be seen that this is also an equation describing either circles or straight lines. The fact that in the equation the coefficients and
swapped means that during inversion, lines passing through 0 will turn into circles, and circles passing through 0 will turn into lines.

Power functions

The main difference between these functions and those considered earlier is that they are not one-to-one (
). We can say that the function
maps the complex plane to two instances of the same plane. A careful consideration of this topic requires the use of the cumbersome apparatus of Riemann surfaces and is beyond the scope of the questions considered here. It is important to understand that the complex plane can be divided into sectors, each of which is one-to-one mapped onto the complex plane. This is the breakdown for the function
looks like this, For example, the upper half-plane is one-to-one mapped onto the complex plane by the function
. Geometry distortions for such images are more difficult to describe than in the case of inversion. As an exercise, you can trace what the grid of rectangular coordinates of the upper half-plane goes to when displayed

It can be seen that the grid of rectangular coordinates transforms into a family of parabolas forming a system of curvilinear coordinates in the plane
. The partition of the plane described above is such that the function
displays each of sectors on the whole plane. The description of forward and backward mapping looks like this

So the function
It has various inverse functions,

given in different sectors of the plane

In such cases, the mapping is said to be multisheeted.

Zhukovsky function

The function has its own name, since it formed the basis of the theory of the aircraft wing, created by Zhukovsky (a description of this design can be found in the book). The function has a number of interesting properties, let's focus on one of them - find out on which sets this function acts one-to-one. Consider the equality

, where
.

Therefore, the Zhukovsky function is one-to-one in any domain in which, for any and their product is not equal to unity. These are, for example, the open unit circle
and the complement of the closed unit circle
.

Consider the action of the Zhukovsky function on the circle , then

Separating the real and imaginary parts, we obtain the parametric equation of the ellipse

,
.

If
, then these ellipses fill the entire plane. Similarly, it is verified that the images of segments are hyperbolas

Exponential function

The function can be expanded in a power series, which converges absolutely in the entire complex plane, therefore it is differentiable everywhere. Let us describe the sets on which the function is one-to-one. Obvious equality
shows that the plane can be divided into a family of strips, each of which is one-to-one mapped by the function onto the entire complex plane. This partition is essential in order to understand how the inverse function works, or rather, inverse functions. On each of the strips, the inverse map is naturally defined