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

This article begins a series of lessons in which I will look at typical tasks, 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 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” are 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. In class 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. From this fact The following point logically follows:

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 rushing into battle with swords drawn, complex plane, let me give you the most important advice on 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. Don't rush!

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 are valid elementary functions. 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 to this function was differentiable 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.

Let me touch on 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's 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:

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

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 by normal rules:

Imaginary unit when differentiated, it is considered a constant.

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

To be decided inverse problem– in the resulting expression you 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 parentheses 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 approximate sample finishing 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 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 the numbers are fair following formulas:

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 complex expression, function, it is only important that they accept only valid meanings. Actually, we will see this right now:

Example 7

Find the derivative.

Solution: General line 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.

Next steps are standard, let us 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 initial 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, in this case .

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

As a result:
– 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 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 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:

Concept of a function of a complex variable

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

A 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 specified similarly:

A single-valued function of a complex variable is a rule according to which each complex value of the independent variable (from the domain of definition) corresponds to one and only one complex value of the function. 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. In class Complex numbers for dummies we considered a complex number in the form . Since now the letter “z” has become a variable, we will denote it as follows: , while “x” and “y” can take on different real values. 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:

Real and imaginary part of a function of a complex variable

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

The function is called the real part of the function.
The function is called the imaginary part of the function.

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

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) Regrouping of terms: first we rewrite the terms in which there is no imaginary unit (first group), then the terms where there is (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).

Find the real and imaginary part of the function

This is an example for you to solve on your own. Before you rush into battle on the complex plane with your checkers drawn, 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 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. Don't rush!

Full solution and 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.
Cauchy-Riemann conditions

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 functions of a complex variable there is no derivative at all, and you have to figure out whether a particular function is differentiable. 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) So that the so-called Cauchy-Riemann conditions are satisfied:

Only in this case will the derivative exist!

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.

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

Let me dwell on one more technical point: in what order should we 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: .

3) 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, .

FKP integral. Cauchy's theorem.

Formula ( 52 ) is called integral formula Cauchy or Cauchy integral. If as a contour in ( 52 ) select a circle, then, replacing and given that is the arc length differential , the Cauchy integral can be represented as a mean value formula.

Let function = u(x,y)+iv(x,y) is defined in the vicinity of the point z = x+iy. If the variable z increment  z= x+i y, then the function
will receive an increment


= (z+ z)–
=u(x+ x, y+ y)+

+ iv(x+ x, y+ y) - u(x,y) - iv(x,y) = [u(x+ x, y+ y) –

– u(x,y)] + i[v(x+ x, y+ y) - v(x,y)] =

= u(x,y) + i v(x,y).

Definition. If there is a limit

=

,

then this limit is called the derivative of the function
at the point z and is denoted by f(z) or
. Thus, by definition,

=

=

. (1.37)

If the function
has a derivative at the point z, then they say that the function
differentiable at the point z. Obviously, for the differentiability of the function
it is necessary that the functions u(x,y) And v(x,y) were differentiable. However, this is not enough for the derivative to exist f(z). For example, for the function w== x–iy functions u(x,y)=x

And v(x,y)=–y differentiable at all points M( x,y), but the limit of the ratio
at  x0,  y0 does not exist, because if  y= 0,  x 0, then  w/ z= 1,

if  x = 0,  y 0, then  w/z = -1.

There is no single limit. This means that the function

w= has no derivative at any point z. For the existence of a derivative of a function of a complex variable, additional conditions are required. Which ones exactly? The answer to this question is given by the following theorem.

Theorem. Let the functions u(x,y) And v(x,y) are differentiable at the point M( x,y). Then in order for the function

= u(x,y) + iv(x,y)

had a derivative at the point z = x+iy, it is necessary and sufficient for the equalities to hold

Equalities (1.38) are called Cauchy-Riemann conditions.

Proof. 1) Necessity. Let the function
has a derivative at point z, that is, there is a limit

=

=
.(1.39)

The limit on the right side of equality (1.39) does not depend on which path the point takes  z =  x+i y strives

to 0. In particular, if y = 0, x  0 (Fig. 1.10), then

If x = 0, y  0 (Fig. 1.11), then

(1.41)

Fig.1.10 Fig. 1.11

The left sides in equalities (1.40) and (1.41) are equal. This means that the right sides are also equal

It follows that

Thus, from the assumption of the existence of the derivative f(z) equality (1.38) follows, that is, the Cauchy-Riemann conditions are necessary for the existence of the derivative f(z).

1) Sufficiency. Let us now assume that equalities (1.38) are satisfied:

and prove that in this case the function
has a derivative at the point z= x+iy, that is, the limit (1.39)

=

exists.

Since the functions u(x,y) And v(x,y) are differentiable at the point M( x,y), then the total increment of these functions at the point M( x,y) can be represented in the form

,

where  1 0,  2 0,  1 0,  2 0 at  x0, y0.

Since, by virtue of (1.38),

Hence,

=
,

 1 =  1 +i 1 0,  2 =  2 +i 2 0 at z =  x+iy0.

Thus,

Since  z 2 =  x 2 + y 2 , then  x/ z1,  y/z1. That's why

at  z  0.

It follows that right side equality (1.42) has a limit at  z 0, therefore, and left side has a limit at  z 0, and this limit does not depend on which path  z tends to 0. Thus, it has been proven that if at the point M(x,y) conditions (1.38) are satisfied, then the function
has a derivative at the point z = x+iy, and

.

The theorem is completely proven.

In the process of proving the theorem, two formulas (1.40) and (1.42) were obtained for the derivative of a function of a complex variable

,

.

Using formulas (1.38) we can obtain two more formulas

, (1.43)

. (1.44)

If the function f(z) has a derivative at all points of the region D, then we say that the function
is differentiable in domain D. For this, it is necessary and sufficient that the Cauchy-Riemann conditions be satisfied at all points of domain D.

Example. Check the Cauchy-Riemann conditions for

functions e z .

Because e z = e x+iy = e x(cos y + i sin y),

That u(x, y) = Re e z = e x cos y, v(x, y) = Im e z = e x sin y,

,
,

,
,

hence,

Cauchy-Riemann conditions for a function e z fulfilled at all points z. So the function e z is differentiable on the entire plane of complex variable, and

Differentiability is proved in exactly the same way

functions z n , cos z, sin z, ch z,sh z, Ln z, and the validity of the formulas

(z n) = n z n-1, (cos z) = -sin z, (sin z) = cos z,

(ch z) = sh z, (sh z) = ch z, (Ln z) = 1/z.

For functions of a complex variable, all rules for differentiating functions of a real variable remain in force. The proof of these rules follows from the definition of the derivative in the same way as for functions of a real variable.

Let w=f (z) is a single-valued function defined in the domain.

Definition 1.Derivative from function f(z) at point
is called the limit of the ratio of the increment of a function to the increment of the argument when the latter tends to zero:

A function that has a derivative at a point z, called differentiable at this point.

It is obvious that all arithmetic properties of derivatives are satisfied.

Example.

Using Newton's binomial formula, it is similarly deduced that

The series for the exponential, sine and cosine satisfy all the conditions for term-by-term differentiation. By direct verification it is easy to obtain that:

Comment. Although the definition of the derivative of the FKP formally completely coincides with the definition for the FKP, it is essentially more complex (see the remark in §5).

Definition 2. Function f (z) , continuously differentiable at all points of the region G, called analytical or regular on this area.

Theorem 1. If the function f(z) is differentiable at all points of the region G, then it is analytical in this area. (b/d)

Comment. In fact, this theorem establishes the equivalence of the regularity and differentiability of the FKP on a domain.

Theorem 2. A function that is differentiable in some domain has infinitely many derivatives in that domain. (n/d. Below (in §13) this statement will be proven under certain additional assumptions)

Let us represent the function as the sum of the real and imaginary parts: Theorem 3. (Cauchy – Riemann conditions). Let the function f(z) is differentiable at some point
. Then the functions u (x,y) And v (x,y) have partial derivatives at this point, and

And
, called Cauchy–Riemann conditions.

(Since the value of the derivative does not depend on the way the quantity tends
to zero, choose the following path: We get:

Likewise, when
we have:
, which proves the theorem.)

The converse is also true:

Theorem 4. If the functions u(x,y) And v (x,y) have continuous partial derivatives at some point that satisfy the Cauchy–Riemann conditions, then the function itself f (z) – differentiable at this point. (b/d)

Theorems 1 – 4 show the fundamental difference between PKP and FDP.

Theorem 3 allows you to calculate the derivative of a function using any of the following formulas:

In this case it can be considered X And at arbitrary complex numbers and calculate the derivative using the formulas:

Examples. Check the function for regularity. If the function is regular, calculate its derivative.

the function is regular;

2. the function is not differentiable.

Comment. It is not difficult to see that any real function complex argument – ​​not differentiable.

§9.Harmonic functions.

Let us recall the definition of harmonic functions given in the course “Field Theory”:

Definition. Function u(x,y) is called harmonic, if it satisfies the Laplace equation:

Let on the area G an analytical function is given. This function satisfies the Cauchy–Riemann conditions:
,
(§8). Since the analytic function is infinitely differentiable, then the functions u And v are also infinitely differentiable. Let us differentiate the first condition with respect to x, second in y and add the resulting equalities:

those. the real part of the analytical function is harmonic. If the conditions are differentiated by at, By X and subtract, it is easy to verify that the imaginary part is harmonic. Thus it is proven

Theorem. The real and imaginary parts of the analytical function are harmonic:

It is clear that two arbitrary harmonic functions, generally speaking, will not be the real and imaginary parts of some analytic function. To do this, they must also satisfy the Cauchy–Riemann conditions. However, from any harmonic function it is possible to determine the second part of the analytical function (i.e., the analytical function itself) up to a constant.

Example. Prove what can be a real part of an analytic function and define this function.

From the 2nd condition K – R:

CAUCHY - RIEMANN EQUATIONS

- differential ur-nia, Crimea satisfy substances. and imaginary parts analytical function. Function f(z) = u(x, y)+i(x, y), z=x+iy, continuously differentiable in the domain D complex plane , is analytic in D if and only if the K.-R. equations are valid:

K. - R. u. first introduced by J. L. D'Alembert in 1752 and L. Euler in 1777 and used by O. Cauchy and B. Riemann. Formally, K.-R. u. can also be written in the form

As a consequence of K.-R. u. is the fact that u(x, y)And ( x, y) - harmonic functions V D. Two harmonious functions called mutually conjugate if they satisfy the K.-R. For any function that is harmonic in a simply connected domain, there is a conjugate harmonic. function determined accurate to post. term. In the case of non-simply connected domains, the last statement is, generally speaking, not true.

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• - famous French mathematician. His first teacher and educator was his father, a passionate Latinist and zealous Catholic. At the age of 13, Augustin K. was assigned to the central school...

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• - Augustin Louis, French mathematician, member of the Paris Academy of Sciences. Graduated from the Ecole Polytechnique and the School of Bridges and Roads in Paris. In 1810-13 he worked as an engineer in Cherbourg...
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37. Koshas and chakras