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Functions of a complex variable examples. Derivative of the FKP

Let the function W = f(Z) is given on some set and Z 0 , belonging to E, the limit point of this set. Let's add Z 0 = x 0 + i· y 0 increment Δ Z = Δ x+ i· Δ y to point Z = Z 0 + Δ Z belonged to many E. Then the function W = u+ i· v = f(Z) = u(x, y)+ i· v(x, y). We get the increment Δ W = Δ u+ i· Δ v = f(Z 0 + Δ Z) - f(Z 0 ) = Δ f(Z 0 ) ,
.

If there is a finite limit
, then it is called derivative of a functionf(Z) at pointZ 0 by manyE, and is denoted
,
,
, W" .

Formally, the derivative function of a complex variable is defined in exactly the same way as the derivative function of a real variable, but their content is different.

In the definition of the derivative of a function f(x) real variable at a point X 0 , x→ x 0 along a straight line. In the case of a function of a complex variable f(Z), Z may strive for Z 0 along any plane path leading to a point Z 0 .

Therefore, the requirement for the existence of a derivative of a function of a complex variable is very strict. This explains that even simple functions of a complex variable do not have a derivative.

Example.

Consider the function W = = x- i· y. Let us show that this function has no derivative at any point. Let's take any point Z 0 = x 0 + i· y 0 , let's give it an increment Δ Z = Δ x+ i· Δ y, then the function will be incremented. Means

,
,

We will first consider Δ Z = Δ x + i· Δ y such that Δ x → 0 , and Δ y = 0 , i.e. point Z 0 + Δ Z → Z 0 along a horizontal straight line. In this case we get that

We will now consider the increment ∆ Z such that ∆ x = 0 , and ∆ y → 0 , i.e. When Z 0 + ∆ Z→ Z 0 along a vertical straight line, and it will be obvious
.

The resulting limits are different, so the ratio has no limit at ∆ Z → 0 , that is, the function
has no derivative at any point Z 0 .

Let us find out the meaning of the derivative with respect to a set. Let E is the real axis, and W = f(Z) = x, then this is an ordinary real function of a real variable f(x) = x and its derivative will be equal 1 (
).

Let it now E- this is the entire plane (Z). Let us show that the function f(Z) = x in this case it has no derivative at any point. Indeed, in this case
.It is clear from this that if
A
, That
. If
, A
, That
.Hence, the attitude has no limit at
, so the function f(Z) = x has no derivative at any point
.

Note that if a complex-valued function of a real variable is considered, then it immediately follows from the definition of the derivative that
, therefore, (this is the derivative along the real axis).

Formula for incrementing functions.

Let the function W = f(Z) has at the point Z 0 derivative
. Let us show that representation (1) holds, where the quantity
, When
.

Indeed, by definition of derivative we have
, therefore, the value
, When
. Therefore, representation (1) takes place (multiply both sides by
and move it
to the left side).

Lecture No. 8 Differentiability and differential of a function of a complex variable

Function W = f(Z) called differentiable at the pointZ 0 , if at this point representation (2) takes place, where A is a fixed complex number, and the quantity
tends to zero when
.

If the function W = f(Z) differentiable at the point Z 0 , then the principal linear relative to
part of it A·
increment
at the point Z 0 called differential function f(Z) at the point and is designated
.

The theorem holds.

Theorem.

In order for the functionW = f(Z) was differentiable at the pointZ 0 , it is necessary and sufficient that it has a finite derivative at this point
, and it always turns out that in representation (2)
.

Proof.

Necessity. Let the function be differentiable at the point Z 0 . Let us show that it has a finite derivative at this point, and that this derivative is equal to the number A. Due to differentiation f(Z) at the point Z 0 representation (2) takes place, which means
(3). Passing to the limit here at
we get that
, Means
.

Adequacy. Let the function f(Z) has at the point Z 0 final derivative
. Let us show that representation (2) holds. Due to the existence of the derivative
representation (1) takes place, but this is also representation (2), in which A =
. Sufficiency has been established.

As we know, the differential, taking as the differential of the independent variable Z its increment
, that is, assuming
, we can write
and therefore
(this is a relation of differentials, not a single symbol).

Theorem

In order for the function w = f(z) , defined in some area D complex plane, was differentiable at the point z 0 = x 0 + iy 0 as a function of a complex variable z, it is necessary and sufficient that its real and imaginary parts u And v were differentiable at the point ( x 0 ,y 0) as functions of real variables x And y and that, in addition, at this point the Cauchy-Riemann conditions are satisfied:

; ;

If the Cauchy-Riemann conditions are satisfied, then the derivative f"(z) can be represented in any of the following forms:

Proof

Consequences

Story

These conditions first appeared in the work of d'Alembert (1752). In the work of Euler, reported to the St. Petersburg Academy of Sciences in 1777, the conditions first received the character of a general sign of the analyticity of functions. Cauchy used these relations to construct the theory of functions, starting with the memoir, presented to the Paris Academy of Sciences in 1814. Riemann's famous dissertation on the foundations of the theory of functions dates back to 1851.

Literature

Shabbat B.V. Introduction to comprehensive analysis. - M.: Science, . - 577 p.
Titchmarsh E. Theory of functions: Transl. from English - 2nd ed., revised. - M.: Science, . - 464 s.
Privalov I. I. Introduction to the theory of functions of a complex variable: A manual for high school. - M.-L.: State Publishing House, . - 316 p.
Evgrafov M. A. Analytical functions. - 2nd ed., revised. and additional - M.: Science, . - 472 s.

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Let's consider some complex quantity $w$, which is given by the expression $w(z)=u(x,y)+v(x,y)\cdot i$, where $u(x,y),\, \, \, v(x,y)$ are real functions of a real variable, $z=x+yi$.

This quantity is a complex function of a real variable.

Definition 1

A function $w(z)$ is called analytic at some point z if this function differentiable in some neighborhood of a given point z.

Definition 2

A function is called analytic in some domain D if it is analytic at every point in this domain.

Let the functions $u(x),\, \, \, v(x)$ be differentiable.

Definition 3

The expression $w_(x) "=u"_(x) (x,y)+i\cdot v"_(x) (x,y)$ is called the derivative complex function real variable by real argument $x$.

The derivative with respect to the real argument $y$ is defined similarly.

To calculate the derivative, we use the following formula:

\ \

1) For the function $w=(3x+2)+(x^(3) +2y)\cdot i$ we obtain:

\ \

2) For the function $w=(x+e^(y))+(3y^(2) +\ln x)\cdot i$ we obtain:

\ \

In order for some function $w(z)$ to be differentiable at some point $z_(0) =x_(0) +y_(0) \cdot i$, it is necessary and sufficient that $u(x,y)$ and $v(x,y)$ were differentiable at the point $(x_(0) ;y_(0))$ and satisfied following conditions:

\[\begin(array)(l) (\frac(\partial u(x,y))(\partial x) =\frac(\partial v(x,y))(\partial y) ) \\ ( \frac(\partial u(x,y))(\partial y) =-\frac(\partial v(x,y))(\partial x) ) \end(array).\]

These conditions are called Cauchy-Riemann conditions.

Note 1

The Cauchy-Riemann conditions are relations that connect the real and imaginary parts of the differentiable function $w(z)=u(x,y)+v(x,y)\cdot i$, where $u(x,y),\, \, \, v(x,y)$ are real functions of a real variable, $z=x+yi$.

Let us select the real and imaginary parts of the function. Let's put $z=x+yi$ and get:

Therefore, $u(x,y)=e^(1+2y) \cdot \cos (-2x);\, \, \, \, v(x,y)=e^(1+2y) \cdot \sin (-2x)$ - the required real and imaginary parts of the function.

Let's use the Cauchy-Riemann conditions: $\frac(\partial u)(\partial x) =\frac(\partial v)(\partial y) ;\frac(\partial u)(\partial y) =-\frac( \partial v)(\partial x) $.

\[\begin(array)(l) (\frac(\partial u)(\partial x) =2e^(1+2y) \cdot \sin (-2x);\frac(\partial v)(\partial y) =2e^(1+2y) \cdot \sin (-2x)) \\ (2e^(1+2y) \cdot \sin (-2x)=2e^(1+2y) \cdot \sin ( -2x)) \end(array)\] \[\begin(array)(l) (\frac(\partial u)(\partial y) =2e^(1+2y) \cdot \cos (-2x) ;\frac(\partial v)(\partial x) =-2e^(1+2y) \cdot \cos (-2x)) \\ (2e^(1+2y) \cdot \cos (-2x)= -(-2e^(1+2y) \cdot \cos (-2x))) \end(array)\]

The Cauchy-Riemann conditions are satisfied for any real $x,y$. Therefore, the function is analytic for any real $x,y$.

Let's find the derivative of the function and calculate the value of the derivative of the function at a given point $z_(0) =\frac(\pi )(6) $.

The derivative of the function has the form:

Let's calculate the value of the derivative of the function at a given point

In practice, you can encounter the following problems.

Problem 1

Given the real part $u(x,y)$ of some function of a complex variable $w(z)$, it is necessary to find the imaginary part $v(x,y)$ of this function. Reconstruct the function $w(z)$ from the known real and imaginary parts.

Problem 2

Given the imaginary part $v(x,y)$ of some function of a complex variable $w(z)$, it is necessary to find the imaginary part $u(x,y)$ of this function. Reconstruct the function $w(z)$ from the known real and imaginary parts.

The algorithm for solving problem 2 will be as follows:

find the real part using the Cauchy-Riemann conditions;
compose the function $w(z)=u(x,y)+v(x,y)\cdot i$;
perform transformations and select the variable $z=x+yi$ or $\overline(z)=x-yi$.

Note 1

When solving practical problems, the following relationships may be useful:

\ \ \

Note 2

The operation of division by the imaginary unit $i$ is equivalent to the operation of multiplication by $-i$.

Example 3

From the real part $u(x,y)=-x^(2) +y^(2) -5y$ of some function of a complex variable, restore its imaginary part $v(x,y)$ and restore this function, while the function satisfies the initial condition $w(0)=0$.

Let us find the imaginary part $v(x,y)$ of the desired function $w(z)$. Let's use the first Cauchy-Riemann condition:

\[\frac(\partial u(x,y))(\partial x) =\frac(\partial v(x,y))(\partial y) .\]

Let's substitute the original values and get:

\[\frac(\partial v(x,y))(\partial y) =\frac(\partial (-x^(2) +y^(2) -5y))(\partial x) =-2x \] \ \

Let's find the unknown function $\phi (x)$.

Let's use the second Cauchy-Riemann condition:

\[\frac(\partial u(x,y))(\partial y) =-\frac(\partial v(x,y))(\partial x).\] \ \[\phi "(x) =5\Rightarrow \phi (x)=\int 5dx =5x+C\]

Hence,

The imaginary part of the desired function $w(z)$ is restored, then we can write the function itself:

Let's transform the resulting expression:

\ \[=-x^(2) +y^(2) -5y+-2xyi+5xi+Ci=(-x^(2) +y^(2) -2xyi)+(-5y+5xi)+Ci =\] \[=-(x^(2) +2xyi-y^(2))+5i\cdot (x-\frac(y)(i))+Ci\] \

Using initial condition$w(0)=0$, let's find the value of the constant $C$.

Therefore, the required function has the form:

The imaginary part of the function will take the form.

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. From this fact The following point logically follows:

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 common 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 independent decision. Before you rush into battle on the complex plane with your checkers drawn, 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.
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 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.

Bad news is that for many functions of a complex variable the derivative does not exist at all, and one has to find 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 part 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 the Cauchy integral formula or Cauchy integral. If as a contour in ( 52 ) choose a circle , then, replacing and taking into account that is the differential of the arc length , the Cauchy integral can be represented as a formula for the average value:

In addition to the independent meaning of the Cauchy integral formula, ( 52 ), (54 ) actually provide a very convenient way to calculate contour integrals, which, as you can see, will be expressed in terms of the “remainder” value integrand function at the point where this function has a singularity.

Example 3-9. Calculate the integral of a function along the contour (Fig.20).

Solution. The point at which the function has a singularity, unlike Example 4-1, is located inside the circle. Let us represent the integral in the form ( 52 ):

Cauchy's formula.

Let be a region on the complex plane with a piecewise smooth boundary, the function is holomorphic in and be a point inside the region. Then the following Cauchy formula is valid:

The formula is also valid if we assume that it is holomorphic inside and continuous on the closure, and also if the boundary is not piecewise smooth, but only rectifiable. (Holomorphic function is a function of a complex number, piecewise smooth is a function of a real number)

Elementary FKP: Taylor function, trigonometric functions, hyperbolic functions, inverse trigonometric functions, logarithmic functions, Cauchy formula.

1. Derivative and differential. The definitions of the derivative and differential of a function of a complex variable coincide literally with the corresponding definitions for functions of a single real variable.

Let the function w = f(z) = and + iv defined in some neighborhood U points zo. Let's give the independent variable z = x + gu increment A z= A.g + gau, does not lead outside the surrounding area U. Then the function w = f(z) will receive the corresponding increment Aw = = f(z 0 + Dg) - f(z 0).

Derivative of the function w = f(z) at the point zq is called the limit of the function increment ratio Aw to the increment of argument A z while striving Az to zero (in an arbitrary manner).

The derivative is denoted f"(z Q), w or y-. The definition of derivative can be written as

The limit in (6.1) may not exist; then they say that the function w = f(z) has no derivative at the point zq.

Function w = f(z) called differentiable about the point Zq, if it is defined in some neighborhood U points zq and its increment Aw can be represented in the form

where is a complex number L does not depend on A g, and the function a(Ag) is infinitesimal at Az-» 0, i.e. Pm a(Ag) = 0.

Just as for functions of a real variable, it is proved that the function f(z) differentiable at the point zq if and only if it has a derivative in zo. and A = f"(zo). Expression f"(zo)Az called differential of the function f(z) at the point Zqand is designated dw or df(zo). In this case, the increment Az of the independent variable -r is also called the differential of the variable r and

denoted by dz. Thus,

The differential is the main linear part of the increment of the function.

Example 6.1. Investigate whether the function has w= /(r) = R ez derivative in arbitrary point Zq.

Solution. By condition, w = Rea = X. Due to the definition of the derivative, the limit (C.1) should not depend on which path

dot z = Zq + Az approaching th at A z-? 0. Let us first take A z - Ah(Fig. 15, a). Because Aw = Ah. then = 1. If

take A z = iAy(Fig. 15, b), That Oh= 0 and therefore Aw = 0.

This means u = 0. Therefore, the relationship will betray when Az-> 0 not A z A z

exists and therefore the function w= Re g = X has no derivative at any point.

At the same time the function w = z = X + iy, obviously has a derivative at any point r, and /"(th) = 1. From here it is clear that the real and imaginary parts of the differentiable function f(r) cannot be arbitrary; they must be connected by some additional relations. These relations arise because the condition for the existence of the derivative /"(0) is significantly more restrictive than the condition for the existence of the derivative of functions of one real variable or partial derivatives of functions of several real variables: it is required that the limit in (6.1) exists and does not depend on the path, according to to which the point r = r + Ar approaches r as Ar 0. To derive these relations, recall the definition of differentiability of a function of two variables.

Real function u = u(x,y) real variables X And at called differentiable at a point Ro(ho,oo), if it is defined in some neighborhood of point D> and its total increment is A And = their o + Oh, oh+ A y) - and (ho, Uo) representable in the form

Where IN And WITH- real numbers independent of J , Ay, A {3 Oh And Ay, tending to zero at Oh -» 0, Ay-> 0.

If the function And is differentiable at the point Po, then it has a

G, " di(P 0)^ di(Ro) gt ,

ny derivatives in Po, and IN= ---, C = ---. But (different

oh oh

from functions of one variable) from the existence of partial derivatives of the function u(x,y) its differentiability does not yet follow.

2. Cauchy-Riemann conditions.

Theorem 6.1. Let the function w = f(z) of complex variable z= (f, y) is defined in the neighborhood of the point, zq= (jo, y o) and f(z) = u(x,y) +iv(x, y). In order for f(z) to be differentiable at the point Zq, it is necessary and sufficient that the functions u(x, y) XI v(x, y) be differentiable at the point(jo, oo) and that at this point the conditions are met

Equalities (6.4) are called Cauchy-Riemann conditions .

Proof. Necessity. Let the function w = f(z) is differentiable at the point zq, i.e.

Let's denote f"(zo) = a + ib a(Dg) = fi(Ax, Ау)+ g7(J, Ay); Az = Ah + (Ay, Where /3 and 7 - real functions of variables Ah, oh, tending to zero as J -> 0, Au -> 0. Substituting these equalities into (6.5) and separating the real and imaginary parts, we obtain:

Since equality complex numbers is equivalent to the equality of their real and imaginary parts, then (6.6) is equivalent to the system of equalities

Equalities (6.7) mean that the functions u(x,y), v(x,y) satisfy condition (6.3) and, therefore, are differentiable. Since the coefficients for J and Ay are equal to the partial derivatives with respect to w and at accordingly, then from (6.7) we obtain

whence conditions (6.4) follow.

Adequacy. Let us now assume that the functions u(x, y) And v(x,y) differentiable at a point (ho.oo) And u(x,y) and conditions (6.4) are satisfied.

Denoting a = ^, 6 = -^ and applying (6.4), we arrive at equalities (6.8). From (6.8) and the condition of differentiability of functions u(x,y), v(x,y) we have

where ft, 7i, ft, d-2 - functions tending to zero as Ah -> 0, Au ->-> 0. From here

An + iAv= (o + ib)(Ah + i.Ay)+ (ft + ift)Ax + (71 + *72) Ay.(6.9) Let us define the function a(Dr) by the equality

and put A = A 4- ib. Then (6.9) will be rewritten as the equality

which coincides with (6.2). Day of proof of differentiability

functions f(z) It remains to show that lim a(Az) = 0. From the equality

it follows that Oh^ |Dg|, Ay^ |Dg|. That's why

If Az-? 0, then Oh-? 0, Ay-> 0, which means that the functions ft, ft, 71, 72 tend to zero. Therefore a(Dr) -> 0 at Az-> 0, and the proof of Theorem 6.1 is complete.

Example 6.2. Find out if a function is w = z 2 differentiable; if so, at what points?

Solution, w = u + iv = (x + iy) 2 = x 2 - y 2 + 2ixy, where and = = x 2 - y 2, V = 2xy. Hence,

Thus, the Cauchy-Riemann conditions (6.4) are satisfied at each point; that means the function w = g 2 will be differentiable in C.

Example 6.3. Investigate the differentiability of a function w = - z - x - iy.

Solution. w = u + iv = x - iy, where u = x, v = -y And

Thus, the Cauchy-Riemann conditions are not satisfied at any point, and therefore the function w = z is not differentiable anywhere.

You can check the differentiability of a function and find derivatives directly using formula (6.1).

Example 6.4. Using formula (6.1), investigate the differentiability of the function IV = z 2.

Solution. A w- (zq + A z) 2- Zq = 2 zqAz -I- (A z) 2 , where

Therefore, the function w = zr is differentiable at any point 2o, and its derivative f"(zo) =2 zo-

Since the main theorems on limits are preserved for functions of a complex variable, and the definition of the derivative of a function of a complex variable also does not differ from the corresponding definition for functions of a real variable, then the well-known rules for differentiating the sum, difference, product, quotient and complex function remain valid for functions of a complex variable . Similarly, it can also be proven that if the function f(z) differentiable at the point zo. then it is continuous at this point; the converse is not true.

3. Analytical functions. Function w= /(^differentiable only at the point itself zq, but also in some neighborhood of this point, is called analytical at point zq. If f(z) is analytic at every point of the region D, then it's called analytic (regular, holomorphic) in domain D.

From the properties of derivatives it immediately follows that if f(z) And g(z)- analytical functions in the field D, then the functions f(z) + g(z), f(z) - g(z), f(z) g(z) also analytical in the field D, and the quotient f(z)/g(z) analytical function at all points of the region D. in which g(z) f 0. For example, function

is analytic in the C plane with dropped points z= = 1 and z - i.

The following statement follows from the theorem on the derivative of a complex function: if the function And = u(z) is analytical in the domain D and displays D to the region D" variable and, and function w = f(u) analytical in the field D", then a complex function w = f(u(z)) variable z analytical in D.

Let us introduce the concept of a function that is analytic in a closed domain D. Difference from open area here is that boundary points are added that do not have a neighborhood belonging to D; therefore the derivative at these points is not defined. Function f(z) called analytical (regular, holomorphic) in a closed region D, if this function can be extended into some wider area D i containing D, to analytical D functions.

Conditions (6.4) were studied back in the 18th century. d'Alembert and Euler. Therefore, they are sometimes also called the d'Alembert-Euler conditions, which is more correct from a historical point of view.