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 about complex numbers. In order to consolidate and repeat the material, just visit the page. You will also need the skills to find second order partial derivatives. Here they are, these partial derivatives... even now I was a little surprised at how often they occur...

The topic that we are beginning to examine does not present any particular difficulties, 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 I derived experimentally. Read on!

Concept of a function of a complex variable

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

Single variable function 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” - real numbers.

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

Single-valued function of a complex variable- this is the rule according to which everyone comprehensive the value of the independent variable (from the domain of definition) corresponds to one and only one comprehensive function value. The theory also considers multi-valued and some other types of functions, but for simplicity I will focus on one definition.

What is the difference between a complex variable function?

The main difference: complex numbers. I'm not being ironic. Such questions often leave people in a stupor; at the end of the article I’ll tell you a funny story. At the lesson 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 meanings. Roughly speaking, the function of a complex variable depends on the variables and , which take on “ordinary” 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

Solution: The independent variable “zet”, as you remember, is written in the form , therefore:

(1) We substituted .

(2) For the first term, the abbreviated multiplication formula was used. In the term, the parentheses have been opened.

(3) Carefully squared, not forgetting that

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

(5) For the second group we take it out of brackets.

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

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

What kind of functions did these turn out to be? The most ordinary functions of two variables from which you can find such popular partial derivatives. Without mercy, we will find it. But a little later.

Briefly, the algorithm for 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 the function

This is an example for independent decision. Before you rush into battle on the complex plane with your checkers drawn, let me give you the most important advice on this topic:

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

Complete solution and the answer at the end of the lesson.

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

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

Differentiation of functions of a complex variable.

I have two news: good and bad. I'll start with the 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 complex variable functions there is no derivative at all, and you have to figure out is it differentiable one function or another. And “figuring out” how your heart feels is associated with additional problems.

Let's consider the function of a complex variable. In order for this function to be differentiable it is necessary and sufficient:

1) So that first-order partial derivatives exist. Forget about these notations right away, since in the theory of functions of a complex variable a different notation is traditionally used: .

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

Only in this case will the derivative exist!

Example 3

Solution is divided into three successive stages:

1) Let's find the real and imaginary parts of the function. This task was discussed in previous examples, so I’ll write it down without comment:

Since then:

Thus:

– imaginary part of the function.

I'll stop at one more technical point: in what order write the terms in the real and imaginary parts? Yes, in principle, it doesn’t matter. For example, the real part can be written like this: , and the imaginary one – 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 satisfied.

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

We check the fulfillment of the second condition:

The result is the same, but with opposite signs, that is, the condition is also fulfilled.

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

3) Let's find the derivative of the function. The derivative is also very simple and is found according to the usual rules:

The imaginary unit is considered a constant during differentiation.

Answer: – real part, – imaginary part.
The Cauchy-Riemann conditions are satisfied, .

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

The derivative can be found using the formula:

IN in this case:

Thus

We have to solve the inverse problem - in the resulting expression we need to isolate . In order to do this, it is necessary in the terms and outside the brackets:

The reverse action, as many have noticed, is somewhat more difficult to perform; to check, it is always better to take the expression on a draft or orally open the brackets back, making sure that the result is 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.

Quick Solution and an approximate sample of the final design at the end of the lesson.

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

Solution: The solution algorithm is completely preserved, but at the end a new point will be added: finding the derivative at a point. For cube required formula already output:

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

Attention and attention again!

Since then:


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

Checking the second condition:

The result is the same, but with opposite signs, that is, the condition is also fulfilled.

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

Let's calculate the value of the derivative at the required point:

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

Functions with cubes are common, so here’s an example to reinforce:

Example 6

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

Solution and example of finishing at the end of the lesson.

In theory comprehensive analysis Other functions of a complex argument are also defined: exponent, sine, cosine, etc. These functions have unusual and even bizarre properties - and this is really interesting! I really want to tell you, but here, as it happens, is not a reference book or textbook, but a solution book, so I will consider the same problem with some common functions.

First about the so-called Euler's formulas:

For anyone valid numbers, the following formulas are valid:

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

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

Example 7

Find the derivative.

Solution: The general line of the party remains unshakable - it is necessary to distinguish the real and imaginary parts of the function. I'll bring you detailed solution, and below I will comment on each step:

Since then:

(1) Substitute “z” instead.

(2) After substitution, you need to select the real and imaginary parts first in the indicator exhibitors. To do this, open the brackets.

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

(4) We use the school action with degrees.

(5) For the multiplier we use Euler’s formula, and .

(6) Open the brackets, resulting in:

– real part of the function;
– 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 won’t find the derivative.

Solution: The solution algorithm is very similar to the previous two examples, but there are very important points, That's why First stage I will comment again step by step:

Since then:

1) Substitute “z” instead.

(2) First, we select the real and imaginary parts inside the sinus. For these purposes, we open the brackets.

(3) We use the formula, and .

(4) Use parity of hyperbolic cosine: And oddity of hyperbolic sine: . Hyperbolics, although out of this world, are in many ways reminiscent of similar trigonometric functions.

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

Attention! The minus sign refers to the imaginary part, and under no circumstances should we lose it! For a clear 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 satisfied.

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

Ladies and gentlemen, let’s figure it out on our own:

Example 10

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

I deliberately chose more difficult examples, because everyone seems to be able to cope with something, like shelled peanuts. At the same time, you will train your attention! Nut cracker at the end of the lesson.

Well, in conclusion, I’ll consider one more interesting example, when the complex argument is in the denominator. It’s happened a couple of times in practice, let’s look at something simple. Eh, I'm getting old...

Example 11

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

Solution: Again it is necessary to distinguish 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 one 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. We already have in the denominator, which means the conjugate expression will be . Thus, you need to multiply the numerator and denominator by:

Federal Agency for Education

___________________________________

St. Petersburg State

Electrotechnical University "LETI"

_______________________________________

Theory of functions of a complex variable

Guidelines

to practical classes

in higher mathematics

Saint Petersburg

Publishing house SPbSETU "LETI"

UDC 512.64(07)

TFKP: Methodological instructions for solving problems / compiled by: V.G. Dyumin, A.M. Kotochigov, N.N. Sosnovsky. St. Petersburg: Publishing house of St. Petersburg State Electrotechnical University "LETI", 2010. 32 p.

Approved

Editorial and Publishing Council of the University

as methodological instructions

© SPbSETU "LETI", 2010

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

having the same derivative as functions of one variable. It is known that functions of several variables can have partial derivatives and directional derivatives, but, as a rule, the derivatives in different directions do not coincide, and it is not possible to talk about the derivative at a point. However, for functions of a complex variable it is possible to describe the conditions under which they allow differentiation. The study of the properties of differentiable functions of a complex variable is the content of methodological instructions. The instructions are aimed at demonstrating how the properties of such functions can be used to solve a variety of problems. Successful mastery of the material presented is impossible without basic skills in calculations with complex numbers and familiarity with the simplest geometric objects, defined in terms of inequalities connecting the real and imaginary parts complex number, as well as its module and argument. A summary of all the information necessary 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 the guidelines. Where these concepts have their own specifics, in comparison with functions of one variable, appropriate 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 begin a discussion of the conditions for differentiability of functions of a complex variable by finding out which elementary functions have this property. From the obvious relation

It follows that any polynomial is differentiable. And, since power series can be differentiated term by term within its circle of convergence,

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

However, some graphical representation of the function can be obtained by considering the images of fairly simple sets in the complex plane
, arising under the influence of a given function. For example, let's consider several simple functions from this point of view.

Linear function

This simple function is very important, since any differentiable function is locally similar to a linear one. Let's consider the action of the function in maximum detail

Here
-- modulus of a complex number And -- his argument. Thus, the linear function performs stretching, rotation and translation. Therefore, a linear mapping takes any set to a similar set. In particular, under the influence of a linear mapping, straight lines turn into straight lines, and circles into circles.

Function

This function is the next most complex after linear. It is difficult to expect that it will transform any line into a straight line, and a circle into a circle; simple examples show that this does not happen, however, it can be shown that this function transforms the set of all lines and circles into itself. To verify this, it is convenient to go to the real (coordinate) description of the mapping

The proof requires a description of the inverse mapping

Consider the equation if
, then it will work out general equation straight. If
, That

Therefore, when
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 equation under consideration will be rewritten in the form

, (
)

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

Power functions

The main difference between these functions and those discussed earlier is that they are not one-to-one (
). We can say that the function
transforms a complex plane into two copies of the same plane. An accurate treatment of this topic requires the use of the cumbersome apparatus of Riemann surfaces and goes beyond the scope of the issues 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
. Geometric 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 transforms into when displaying

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

So the function
It has various inverse functions,

specified in various sectors of the plane

In such cases the mapping is said to be multi-sheeted.

Zhukovsky function

The function has its own name because it formed the basis of the wing theory aircraft, 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
.

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

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

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

,
.

If
, then these ellipses fill the entire plane. It can be verified in a similar way that the images of segments are hyperbolas

.

Exponential function

The function can be expanded into a power series that is absolutely convergent 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 mapped one-to-one by a function onto the entire complex plane. This partition is essential for understanding how the inverse function works, more precisely inverse functions. On each of the stripes there is a naturally defined inverse mapping

The inverse function in this case is also multivalent, and the number of inverse functions is infinite.

The geometric description of the mapping is quite simple: straight lines
turn into rays
, segments

turn into circles
.

Where
are real numbers, and - a special character called imaginary unit . For an 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. Amount
complex numbers And is called a complex number

2. By difference
complex numbers And is called a complex number

3. The work
complex numbers And is called a complex number

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

.

Solution. 1) .

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

Trigonometric form complex number:

Where
- modulus of a complex number,
is the argument of a complex number. Corner not uniquely defined, up to a term
:

,
.

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

, (or
).

Demonstrative form complex number:

.

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

,

Where
.

Points corresponding to values
, are the vertices of the correct
a square inscribed in a circle of radius
with center at the origin.

Example 4.2. Find all root values
.

Solution. Let's imagine a complex number
in trigonometric form:

,

, where
.

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

,
.

Believing
, we find

,
,

, .

Here we 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 with center at a point radius . Inequality
defines a half-plane located to the right of the straight line
, and the inequality
- half-plane located above the straight line
. In addition, the system of inequalities
sets the angle between the rays
And
emanating from the origin.

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

Solution. The first inequality corresponds to a ring with center at the point
and two radii 1 and 2, the 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 the single-valued function
defined and continuous in the region
, 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 calculating the usual curvilinear integrals, namely

.

If the function
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 make a change of 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 path of integration is part of a line emanating from a point , or part of a circle centered at a point , then it is useful to make a variable replacement of the form
. In the first case
, A - real integration variable; in the second case
, A - real integration variable.

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

Solution. Let us rewrite the integrand in the form Then , . Let's apply formula (4.3): Because , That , . That's why

Example 4.5. Calculate integral
, Where - arc of a circle
,
(Fig. 4.5) .

Solution. Let's say , 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's series, and the series
called the right part Laurent series.

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

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

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

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

2) contains a finite number of terms with negative powers 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 essentially a special point functions
.

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

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 essentially singular point of the function
, if at
a function has no limit, neither finite nor infinite.

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

…,

.

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

,

where is the function
analytic at a point And

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

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

At
dot is the pole of order
functions
.

At
dot is a removable singular point of the function
.

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

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

Consider the point
. At this point we get:

,
,

,
.

The order of zero is
.

,
,

,
,

,
,

,
.

.

So, period
is a pole of second order (
).

. Then

,
.

The order of the zero numerator is
.

,
,
.

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

Theorem 4.1. (Cauchy's theorem on residues ). If the function
is analytic on the boundary region
and everywhere inside the region, except for a finite number of singular points
, That

.

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

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

1) residue of a function at a point equal to the coefficient for minus the first degree in the Laurent expansion
in the vicinity of a 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 a removable singular point is equal to zero.

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

.

In particular, if
, That
.

4) if - simple pole, then

5) if - pole
th order function
, That

Example 4.7. Calculate integral
.

Solution. Finding singular points integrand function . Function has two singular points And Only a point falls inside the contour (Fig. 4.6). Dot - pole of the second order, since is a zero of multiple 2 for the function .

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

By Theorem 4.1 we find

Lecture No. 4.

Geometrically, a function of a complex variable w=f(z) specifies the display of a certain set z– planes to a certain set w-plane. Dot wÎ G called way points z when displayed w=f(z), dot zÎ D – prototype points w.

If everyone z only one value matches w=f(z), then the function is called unambiguous (w=|z|,w=,w= Re z etc.) If some z matches more than one value w, the function is called polysemantic (w= Arg z).

If (i.e. in various points region D function takes different meanings), then the function w=f(z) is called unifoliate in area D.

In other words, the univalent function w=f(z) one-to-one maps the area D on G. With single-sheet display w=f(z) inverse image of any point wÎ G consists of a single element: : . That's why z can be considered as a function of a variable w, defined on G. It is designated and called inverse function .

If in the area D there is at least one pair of points, then the function f(z) are called multileaf in area D.

If display w=f(z) is multileaf on D(For example, w=z n), then in this case some values wÎ G matches more than one point zÎ D:f(z)=w. Therefore, the inverse mapping is not single-valued, it is a multi-valued function.

Single digit on area D function w=f(z) is called branch of a multivalued function F, if value f at any point zÎ D matches one of the values F at this point.

In order to isolate single-valued branches of a multi-valued function, proceed as follows: area D divide functions into domains of univalence w=f(z) so that no two of the regions have common interior points and so that each point zÎ D belonged to one of these areas or the border of some of them. In each of these domains of univalence one defines a function inverse to w=f(z). It is the single-valued branch of the multi-valued function.

The concept of conformal mapping

Example. Find the stretch coefficient and rotation angle at a point z=2i when displaying .

■ Finding the derivative and its value at a given point .

Stretch ratio k equal to the modulus of the derivative: .

Angle of rotation j is equal to the argument of the derivative. The point lies in the fourth quarter, therefore, . ■

Example 3.5. Determine which part of the plane when displayed w=z 2 is stretched, and which one is compressed.

■ Finding the derivative w¢=2 z. Tension factor at any point z equals k=|w¢( z)|=2|z|. The set of points in the complex plane for which k>1, that is 2| z|>1 or , forms part of the plane, which is stretched when displayed. Therefore, when displaying w=z 2 the outside of the circle is stretched, and inner part- shrinks. ■

Display w=f(z) is called conformal (i.e., preserves its shape) at a point if it preserves the angles between curves and has the property of constant extension of the neighborhood of the point.

Any mapping established by means of an analytic function f(z) is conformal at all points where .

The mapping is called conformal in the region , if it is conformal at every point of this region.

A conformal mapping in which the direction of reference of the angles is preserved is called conformal mapping of the first kind . A conformal mapping in which the direction of angles is reversed is called conformal mapping of the ΙΙ genus (For example, ).

In the theory and practice of conformal mappings, two problems are posed and solved.

The first task is to find the image of a given line or area under a given mapping - direct task .

The second is to find a function that maps a given line or area to another given line or area - inverse problem .

When solving a direct problem, it is taken into account that the image of a point z 0 when displayed w=f(z) is a point w 0 , such that w 0 =f(z 0), that is, the result of the substitution z 0 in f(z). Therefore, to find the image of a set, you need to solve a system consisting of two relations. One of them specifies the mapping function w=f(z), the other is the equation of the line, if the problem of finding the image of the line is being solved, or the inequality that determines the set of points of the preimage, if the problem of mapping areas is being solved. In both cases, the solution procedure is reduced to eliminating the variable z from two given ratios.

Rule 3.3. To find the image of the line given by the equation F(x,y)=0 (or explicitly y=j(x)), when displaying w=f(z) necessary:

1. Select the real and imaginary parts of the function f(z): u=Re f(z), v=Im f(z).

2. Exclude from the system X And u. The resulting relationship is the equation of the image of this line.

Rule 3.4. To find the image of a given line when displaying w=f(z) necessary:

1. Write the equation of the line in parametric form z=z(t) or in complex form .

2. Depending on the type of line equation, consider the corresponding case:

If the line is given in parametric form, substitute the expression z(t) V w=f(z);

If the line is given in complex form, then express z from w=f(z), that is, and . Then you should substitute z and in the equation of the line. The resulting relationship is the equation of the image of this line.

Rule 3.5. To find an image of a given area, you should use one of two methods.

First way.

1. Write down the equation of the boundary of this area. Find the image of the boundary of a given area using rules 3.3 or 3.4.

2. Select an arbitrary internal point of a given area and find its image under the given mapping. The region to which the resulting point belongs is the desired image of the given region.

Second way.

1. Express z from the ratio w=f(z).

2. Substitute what you received in step 1. an expression in an inequality that defines a given region. The resulting ratio is the desired image.

Example. Find the image of a circle | z|=1 when displayed using a function w=z 2 .

■ 1 way(according to rule 3.3).

1. Let z=x+iy, w=u+iv. Then u+iv =x 2 -y 2 +i 2xy. We get:

2. Let's exclude X And at from these equations. To do this, let’s square the first and second equations and add:

u 2 +v 2 =x 4 -2x 2 y 2 +y 4 +2x 2 y 2 =x 4 +2x 2 y 2 +y 4 =(x 2 +y 2) 2 .

Taking into account the third equation of the system, we obtain: u 2 +v 2 =1 or | w| 2 =1, that is | w|=1. So, the image of the circle | z|=1 is a circle | w|=1, traversable twice. This follows from the fact that since w=z 2 then Arg w=2Arg z+2pk. So when the point z describes full circle |z|=1, then its image describes the circle | w|=1 twice.

Method 2(according to rule 3.4).

1. Let us write the equation of the unit circle in parametric form: z=e it (0£ t£2 p).

2. Let's substitute z=e it in ratio w=z 2: w=e i 2 t=cos2 t+i sin2 t. Therefore, | w| 2 = cos 2 2 t+sin 2 2 t=1, that is | w|=1 – image equation. ■

Example. Find the equation of the image of a line y=x when displayed w=z 3 .

■ Since the curve is given explicitly, we apply rule 3.3.

1. w=z 3 =(x+iy) 3 =x 3 +3x 2 iy+3x(iy) 2 +(iy) 3 =x 3 - 3xy 2 +i(3x 2 y-y 3).

Means,

2. In the resulting system we substitute y=x: Excluding X from these equations, we get v=-u.

So, the image of the bisector of the I and III coordinate angles of the system xOy is the bisector of the II and IV coordinate angles of the system uOv. ■

1. Linear function

Linear function called a function of the form

w=az+b, (4.1)

Where A, b- complex constants.

This function is defined . Therefore, if , then the linear function produces a conformal mapping of the entire plane of the complex variable. In this case, the tangents to all curves are rotated by the same angle Arg a, and the tension at all points is equal. If a= 1, then , which means there is no stretching or rotation. In this case we get w=z+b. This mapping shifts the entire plane by a vector.

In the general case, moving to the exponential form of writing a complex number, we obtain. Therefore, a linear mapping is a composition of three geometric transformations:

w 1 =rz- similarity with coefficient r=|a|;

w 2 =e i j w 1 =rze i j- turn at an angle j=arg a around the point ABOUT;

w=w 2 +b=re i j z+b- parallel transfer to a vector.

Therefore, the mapping w=az+b changes the linear dimensions of any plane figure in | a| once, rotates this figure by an angle j=arg a around the origin and shifts it in the direction of the vector by its value.

A linear mapping has a circular property, that is, it maps circles z-planes in a circle w-plane (and vice versa); converts straight lines into straight lines.

Example. Find the image of the axis OU when displayed w=2iz-3i.

■ 1 way(according to rule 3.4). We choose the axis equation in parametric form.

1. Since in real form the equation of the axis Oy: x=0, -¥<y<+¥, то в комплексной форме запишется как z=iy, -¥<y<+¥. Это параметрическое уравнение, в качестве параметра выбран at.

2. Let's substitute z=iy into expression w=2iz-3i: w=-2y-3i, -¥<y<+¥. Это уравнение образа в параметрической форме (at- parameter). Having isolated the real and imaginary parts, we obtain the image equation in real form: u=-2y, v=-3 or v=-3, -¥<u<+¥. Это есть уравнение прямой в плоскости uOv, parallel to the real axis.

Method 2. We use the circular property of a linear transformation - the image of a straight line is a straight line. Since a straight line is defined by specifying two points, it is sufficient on the axis OU select any two points and find their images. The straight line passing through the found points will be the required one. Let's select points z 1 =0, z 2 =i, their images w 1 =-3i, w 2 =-2-3i when mapped, lie on the line Im w= -3. Therefore, the image of the axis OU is a straight line v=-3.

3 way(geometric). From the relation w=2iz-3i follows that a=2i, b=-3i, |a|=2, . This means that the given straight line (axis OU) must be rotated by an angle relative to the origin, and then shifted down 3 units. Stretching by 2 times does not change the geometric appearance of the original line, since it passes through the origin. ■

Example. Find some linear function representing a circle | z-i|=1 per circumference | w- 3|=2.

■ The posed problem is the inverse problem of the theory of mappings - given a given image and preimage, find the corresponding mapping. Without additional conditions, the problem does not have a unique solution. Let us present a geometric solution.

1. Move the center of the circle to the origin. To do this, we apply the mapping w 1 =z-i.

2. In plane w 1 let us apply a mapping that gives a 2-fold stretch, that is w 2 =2w 1 .

3. Shift the circle 3 units to the right: w=w 2 +3. Finally we get: w=2(z-i)+3, w= 2z+3-2i– the required function.

You can choose a different order for performing geometric operations - do not shift first, but rotate or stretch. ■

2. Fractional linear function

Fractional linear called a function of the form

, (4.2)

Where a, b,c,d- complex numbers such that , .

Properties of fractional linear transformation

1°Conformity

Display w=L(z) is conformal at all endpoints of the complex plane except .

2° Circular property

The image of a straight line or a circle in a fractional linear mapping w=L(z) is a straight line or a circle (and the image of a straight line can be either a circle or a straight line, and the image of a circle can be both a straight line and a circle). It is easy to establish that when displaying w=L(z) all straight lines and circles passing through the point go into straight planes ( w), and all straight lines or circles not passing through the point d, - in the circumference of the plane ( w).

3°Double relation invariance

Attitude is preserved under a fractional linear mapping, i.e. it is its invariant. This relationship is called double ratio of four points. Thus, the fractional linear transformation is uniquely determined by specifying three points and their images: . Using these pairs, you can find a fractional linear function using the formula:

. (4.3)

This formula can also be applied in the case when some of the numbers z k And wk turn into ¥, if you use the rule: the difference in which the symbol ¥ occurs should be replaced by 1.

4°Maintaining symmetry

If points z 1 and z 2 are symmetrical about some line or circle g, then for any fractional linear mapping w=L(z) their images w 1 and w 2 will be symmetrical relative to the image g: .

Symmetry about a straight line is understood in the usual sense.

Points z And z* are called symmetrical about the circle |z-z 0 |=R, if they lie on the same ray emerging from the center of the circle, and the product of their distances from the center of the circle is equal to the square of its radius, that is

|z-z 0 |×| z*-z 0 |=R 2 . (4.4)

A point symmetrical to a point z 0 – the center of the circle is obviously the point at infinity.

5°Boundary traversal matching principle (displaying areas bounded by lines or circles)

If, in a fractional linear mapping, a straight line or a circle g turns into a straight line or circle g¢, then the area D, which is limited g, is transformed into one of two areas that are bounded by g¢. In this case, the principle of correspondence of border bypass takes place: if during some line bypass g region D turns out to be on the left (right), then with the corresponding traversal of the line g¢ region D¢ should also be on the left (right).

Example. Find the fractional linear function w=L(z), such that w(i)=2i, w(¥)=1, w(-1)=¥.

■ Let us denote z 1 =i, z 2 =¥, z 3 =-1 and w 1 =2i, w 2 =1, w 3 =¥. Let us apply formula (4.3), replacing the differences containing z 2 and w 3 to ¥:

or .

Let's convert: - w-wi+ 2i- 2=wz-wi-z+i Û w(z+1)=z-2+iÛ is the required function. ■