Represent complex numbers in trigonometric form online. Trigonometric and exponential forms of a complex number. Complex numbers xi

2.3. Trigonometric form of complex numbers

Let the vector be given on complex plane number .

Denote by φ the angle between the positive semi-axis Ox and the vector (the angle φ is considered positive if it is counted counterclockwise, and negative otherwise).

Denote the length of the vector by r. Then . We also denote

Writing a non-zero complex number z as

is called the trigonometric form of the complex number z. The number r is called the modulus of the complex number z, and the number φ is called the argument of this complex number and is denoted by Arg z.

The trigonometric form of writing a complex number - (Euler's formula) - an exponential form of writing a complex number:

The complex number z has infinitely many arguments: if φ0 is any argument of the number z, then all the others can be found by the formula

For a complex number, the argument and trigonometric form are not defined.

Thus, the argument of a non-zero complex number is any solution to the system of equations:

(3)

The value φ of the argument of a complex number z that satisfies the inequalities is called the main value and is denoted by arg z.

Arguments Arg z and arg z are related by equality

, (4)

Formula (5) is a consequence of system (3), so all arguments of the complex number satisfy equality (5), but not all solutions φ of equation (5) are arguments of the number z.

The main value of the argument of a non-zero complex number is found by the formulas:

Multiplication and division formulas for complex numbers in trigonometric form have the following form:

. (7)

When raising a complex number to a natural power, de Moivre's formula is used:

When extracting a root from a complex number, the formula is used:

, (9)

where k=0, 1, 2, …, n-1.

Problem 54. Calculate , where .

Let's represent the solution of this expression in the exponential form of writing a complex number: .

If , then .

Then , . Therefore, then And , Where .

Answer: , at .

Problem 55. Write complex numbers in trigonometric form:

A) ; b) ; V) ; G) ; e) ; e) ; and) .

Since the trigonometric form of a complex number is , then:

a) In a complex number: .

That's why

b) , Where ,

G) , Where ,

e) .

and) , A , That .

That's why

Answer: ; 4; ; ; ; ; .

Problem 56. Find the trigonometric form of a complex number

Let , .

Then , , .

Because and , , then , and

Therefore, therefore

Answer: , Where .

Problem 57. Using the trigonometric form of a complex number, perform the following actions: .

Imagine numbers and in trigonometric form.

1) , where Then

Finding the value of the main argument:

Substitute the values and into the expression , we get

2) where then

Then

3) Find the quotient

Assuming k=0, 1, 2, we get three different meanings desired root:

If , then

if , then

if , then .

Answer: :

: .

Problem 58. Let , , , be different complex numbers and . Prove that

a) number is a real positive number;

b) the equality takes place:

a) Let's represent these complex numbers in trigonometric form:

Because .

Let's pretend that . Then

The last expression is a positive number, since there are numbers from the interval under the sine signs.

because the number real and positive. Indeed, if a and b are complex numbers and are real and greater than zero, then .

Besides,

hence the required equality is proved.

Problem 59. Write down the number in algebraic form .

We represent the number in trigonometric form, and then find its algebraic form. We have . For we get the system:

From this follows the equality: .

Applying De Moivre's formula:

we get

Trigonometric form found given number.

We now write this number in algebraic form:

Answer: .

Problem 60. Find the sum , ,

Consider the sum

Applying the De Moivre formula, we find

This sum is the sum of n terms geometric progression with denominator and first member .

Applying the formula for the sum of the terms of such a progression, we have

Highlighting imaginary part in the last expression, we find

Separating the real part, we also obtain the following formula: , , .

Problem 61. Find the sum:

A) ; b) .

According to Newton's formula for raising to a power, we have

According to De Moivre's formula, we find:

Equating the real and imaginary parts of the obtained expressions for , we have:

And .

These formulas can be written in a compact form as follows:

, Where - whole part numbers a.

Problem 62. Find all for which .

Because the , then, applying the formula

, To extract the roots, we get ,

Hence, , ,

, .

The points corresponding to the numbers are located at the vertices of a square inscribed in a circle of radius 2 centered at the point (0;0) (Fig. 30).

Answer: , ,

, .

Problem 63. Solve the equation , .

By condition ; therefore, this equation has no root, and, therefore, it is equivalent to the equation.

In order for the number z to be the root of this equation, it is necessary that the number be the root nth degree from number 1.

Hence we conclude that the original equation has roots determined from the equalities

Thus,

i.e. ,

Answer: .

Problem 64. Solve the equation in the set of complex numbers.

Since the number is not the root of this equation, then for this equation is equivalent to the equation

That is, the equation.

All roots of this equation are obtained from the formula (see problem 62):

; ; ; ; .

Problem 65. Draw on the complex plane a set of points that satisfy the inequalities: . (2nd way to solve problem 45)

Let .

Complex numbers with the same modules correspond to points of the plane lying on a circle centered at the origin, so the inequality satisfy all points open ring, bounded by circles with a common center at the origin and radii and (Fig. 31). Let some point of the complex plane correspond to the number w0. Number , has a modulus times smaller than the modulus w0, an argument that is greater than the argument w0. From a geometric point of view, the point corresponding to w1 can be obtained using a homothety centered at the origin and coefficient , as well as a counterclockwise rotation relative to the origin. As a result of applying these two transformations to the points of the ring (Fig. 31), the latter will turn into a ring bounded by circles with the same center and radii 1 and 2 (Fig. 32).

transformation is implemented using parallel translation on the vector . Transferring the ring centered at a point to the indicated vector, we obtain a ring of the same size centered at a point (Fig. 22).

The proposed method, which uses the idea of geometric transformations of the plane, is probably less convenient in description, but it is very elegant and efficient.

Problem 66. Find if .

Let , then and . The original equality will take the form . From the condition of equality of two complex numbers, we obtain , , whence , . Thus, .

Let's write the number z in trigonometric form:

, Where , . According to De Moivre's formula, we find .

Answer: - 64.

Problem 67. For a complex number, find all complex numbers such that , and .

Let's represent the number in trigonometric form:

. Hence , . For a number we get , can be equal to either .

In the first case , in the second

Answer: , .

Problem 68. Find the sum of numbers such that . Specify one of these numbers.

Note that already from the very formulation of the problem it can be understood that the sum of the roots of the equation can be found without calculating the roots themselves. Indeed, the sum of the roots of the equation is the coefficient of , taken with the opposite sign (the generalized Vieta theorem), i.e.

Students, school documentation, draw conclusions about the degree of assimilation this concept. Summarize the study of the features of mathematical thinking and the process of forming the concept of a complex number. Description of methods. Diagnostic: I stage. The interview was conducted with a mathematics teacher who teaches algebra and geometry in the 10th grade. The conversation took place after some time had elapsed...

Resonance" (!)), which also includes an assessment of one's own behavior. 4. Critical assessment of one's understanding of the situation (doubts). 5. Finally, the use of recommendations legal psychology(accounting by a lawyer psychological aspects performed professional actions - professional and psychological readiness). Consider now the psychological analysis of legal facts. ...

Mathematics of trigonometric substitution and verification of the effectiveness of the developed teaching methodology. Stages of work: 1. Development of an optional course on the topic: "The use of trigonometric substitution for solving algebraic problems" with students from classes with in-depth study mathematics. 2. Conducting a developed optional course. 3. Carrying out a diagnostic control...

Cognitive tasks are intended only to supplement the existing teaching aids and should be in an appropriate combination with all traditional means and elements. educational process. The difference between learning objectives in teaching humanities from exact, from math problems consists only in the fact that there are no formulas, rigid algorithms, etc. in historical problems, which complicates their solution. ...

Lecture

Trigonometric form of a complex number

Plan

1.Geometric representation of complex numbers.

2.Trigonometric notation of complex numbers.

3. Actions on complex numbers in trigonometric form.

Geometric representation of complex numbers.

a) Complex numbers are represented by points of the plane according to the following rule: a + bi = M ( a ; b ) (Fig. 1).

Picture 1

b) A complex number can be represented as a vector that starts at the pointABOUT and end at a given point (Fig. 2).

Figure 2

Example 7. Plot points representing complex numbers:1; - i ; - 1 + i ; 2 – 3 i (Fig. 3).

Figure 3

Trigonometric notation of complex numbers.

Complex numberz = a + bi can be set using the radius - vector with coordinates( a ; b ) (Fig. 4).

Figure 4

Definition . Vector length representing the complex numberz , is called the modulus of this number and is denoted orr .

For any complex numberz its moduler = | z | is determined uniquely by the formula .

Definition . The value of the angle between the positive direction of the real axis and the vector representing a complex number is called the argument of this complex number and is denotedA rg z orφ .

Complex number argumentz = 0 indefined. Complex number argumentz≠ 0 is a multi-valued quantity and is determined up to the term2πk (k = 0; - 1; 1; - 2; 2; ...): Arg z = arg z + 2πk , Wherearg z - the main value of the argument, enclosed in the interval(-π; π] , that is-π < arg z ≤ π (sometimes the value belonging to the interval is taken as the main value of the argument .

This formula forr =1 often referred to as De Moivre's formula:

(cos φ + i sin φ) n = cos (nφ) + i sin (nφ), n  N .

Example 11 Calculate(1 + i ) 100 .

Let's write a complex number1 + i in trigonometric form.

a = 1, b = 1 .

cos φ = , sin φ = , φ = .

(1+i) 100 = [ (cos + i sin )] 100 = ( ) 100 (cos 100 + i sin 100) = = 2 50 (cos 25π + i sin 25π) = 2 50 (cos π + i sin π) = - 2 50 .

4) Extraction square root from a complex number.

When extracting the square root of a complex numbera + bi we have two cases:

Ifb > about , That ;

3.1. Polar coordinates

Often used on the plane polar coordinate system . It is defined if a point O is given, called pole, and a beam emanating from the pole (for us, this is the axis Ox) is the polar axis. The position of point M is fixed by two numbers: radius (or radius vector) and the angle φ between the polar axis and the vector . The angle φ is called polar angle; measured in radians and counted counterclockwise from the polar axis.

The position of a point in the polar coordinate system is given by an ordered pair of numbers (r; φ). At the pole r = 0 and φ is not defined. For all other points r > 0 and φ is defined up to a multiple of 2π. In this case, the pairs of numbers (r; φ) and (r 1 ; φ 1) are assigned the same point if .

For a rectangular coordinate system xOy Cartesian coordinates points are easily expressed in terms of its polar coordinates as follows:

3.2. Geometric interpretation of a complex number

Consider on the Cartesian plane rectangular system coordinates xOy.

Any complex number z=(a, b) is assigned a point of the plane with coordinates ( x, y), Where coordinate x = a, i.e. the real part of the complex number, and the coordinate y = bi is the imaginary part.

A plane whose points are complex numbers is a complex plane.

In the figure, the complex number z = (a, b) match point M(x, y).

Exercise.Picture on coordinate plane complex numbers:

3.3. Trigonometric form of a complex number

A complex number in the plane has the coordinates of a point M(x; y). Wherein:

Writing a complex number - trigonometric form of a complex number.

The number r is called module complex number z and is denoted. Module is a non-negative real number. For .

The modulus is zero if and only if z = 0, i.e. a=b=0.

The number φ is called argument z and denoted. The argument z is defined ambiguously, like the polar angle in the polar coordinate system, namely, up to a multiple of 2π.

Then we accept: , where φ is smallest value argument. It's obvious that

With a deeper study of the topic, an auxiliary argument φ* is introduced, such that

Example 1. Find the trigonometric form of a complex number.

Solution. 1) we consider the module: ;

2) looking for φ: ;

3) trigonometric form:

Example 2 Find the algebraic form of a complex number .

Here it suffices to substitute the values trigonometric functions and transform the expression:

Example 3 Find the modulus and argument of a complex number ;

1) ;

2) ; φ - in 4 quarters:

3.4. Operations with complex numbers in trigonometric form

· Addition and subtraction it is more convenient to perform with complex numbers in algebraic form:

· Multiplication– with the help of simple trigonometric transformations, it can be shown that when multiplying, the modules of numbers are multiplied, and the arguments are added: ;

COMPLEX NUMBERS XI

§ 256. Trigonometric form of complex numbers

Let the complex number a + bi corresponds vector OA> with coordinates ( a, b ) (see Fig. 332).

Denote the length of this vector by r , and the angle it makes with the axis X , through φ . By definition of sine and cosine:

a / r = cos φ , b / r = sin φ .

That's why A = r cos φ , b = r sin φ . But in this case the complex number a + bi can be written as:

a + bi = r cos φ + ir sin φ = r (cos φ + i sin φ ).

As you know, the square of the length of any vector is equal to the sum of the squares of its coordinates. That's why r 2 = a 2 + b 2 , whence r = √a 2 + b 2

So, any complex number a + bi can be represented as :

a + bi = r (cos φ + i sin φ ), (1)

where r = √a 2 + b 2 , and the angle φ determined from the condition:

This form of writing complex numbers is called trigonometric.

Number r in formula (1) is called module, and the angle φ - argument, complex number a + bi .

If a complex number a + bi is not equal to zero, then its modulus is positive; if a + bi = 0, then a = b = 0 and then r = 0.

The modulus of any complex number is uniquely determined.

If a complex number a + bi is not equal to zero, then its argument is determined by formulas (2) definitely up to an angle multiple of 2 π . If a + bi = 0, then a = b = 0. In this case r = 0. From formula (1) it is easy to understand that as an argument φ in this case, you can choose any angle: after all, for any φ

0 (cos φ + i sin φ ) = 0.

Therefore, the zero argument is not defined.

Complex number modulus r sometimes denote | z |, and the argument arg z . Let's look at a few examples of the representation of complex numbers in trigonometric form.

Example. 1. 1 + i .

Let's find the module r and argument φ this number.

r = √ 1 2 + 1 2 = √ 2 .

Therefore sin φ = 1 / √ 2 , cos φ = 1 / √ 2 , whence φ = π / 4 + 2nπ .

Thus,

1 + i = √ 2 ,

Where P - any integer. Usually, from an infinite set of values \u200b\u200bof the argument of a complex number, one is chosen that is between 0 and 2 π . In this case, this value is π / 4 . That's why

1 + i = √ 2 (cos π / 4 + i sin π / 4)

Example 2 Write in trigonometric form a complex number √ 3 - i . We have:

r = √ 3+1 = 2 cos φ = √ 3 / 2 , sin φ = - 1 / 2

Therefore, up to an angle divisible by 2 π , φ = 11 / 6 π ; hence,

√ 3 - i = 2(cos 11 / 6 π + i sin 11 / 6 π ).

Example 3 Write in trigonometric form a complex number i .

complex number i corresponds vector OA> ending at point A of the axis at with ordinate 1 (Fig. 333). The length of such a vector is equal to 1, and the angle that it forms with the abscissa axis is equal to π / 2. That's why

i = cos π / 2 + i sin π / 2 .

Example 4 Write the complex number 3 in trigonometric form.

The complex number 3 corresponds to the vector OA > X abscissa 3 (Fig. 334).

The length of such a vector is 3, and the angle it makes with the x-axis is 0. Therefore

3 = 3 (cos 0 + i sin 0),

Example 5 Write in trigonometric form the complex number -5.

The complex number -5 corresponds to the vector OA> ending at the axis point X with abscissa -5 (Fig. 335). The length of such a vector is 5, and the angle it makes with the x-axis is π . That's why

5 = 5(cos π + i sin π ).

Exercises

2047. Write these complex numbers in trigonometric form, defining their modules and arguments:

1) 2 + 2√3 i , 4) 12i - 5; 7).3i ;

2) √3 + i ; 5) 25; 8) -2i ;

3) 6 - 6i ; 6) - 4; 9) 3i - 4.

2048. Indicate on the plane the sets of points representing complex numbers whose modules r and arguments φ satisfy the conditions:

1) r = 1, φ = π / 4 ; 4) r < 3; 7) 0 < φ < π / 6 ;

2) r =2; 5) 2 < r <3; 8) 0 < φ < я;

3) r < 3; 6) φ = π / 3 ; 9) 1 < r < 2,

10) 0 < φ < π / 2 .

2049. Can numbers be the module of a complex number at the same time? r And - r ?

2050. Can the argument of a complex number be angles at the same time φ And - φ ?

Present these complex numbers in trigonometric form by defining their modules and arguments:

2051*. 1 + cos α + i sin α . 2054*. 2(cos 20° - i sin 20°).

2052*. sin φ + i cos φ . 2055*. 3(- cos 15° - i sin 15°).

Actions on complex numbers written in algebraic form

The algebraic form of the complex number z =(a,b). is called an algebraic expression of the form

z = a + bi.

Arithmetic operations on complex numbers z 1 = a 1 +b 1 i And z 2 = a 2 +b 2 i, written in algebraic form, are carried out as follows.

1. Sum (difference) of complex numbers

z 1 ±z 2 = (a 1 ± a 2) + (b 1 ±b 2)∙i,

those. addition (subtraction) is carried out according to the rule of addition of polynomials with reduction of similar members.

2. Product of complex numbers

z 1 ∙z 2 = (a 1 ∙a 2 -b 1 ∙b 2) + (a 1 ∙b 2 +a 2 ∙b 1)∙i,

those. multiplication is performed according to the usual rule for multiplication of polynomials, taking into account the fact that i 2 = 1.

3. The division of two complex numbers is carried out according to the following rule:

, (z 2 ≠ 0),

those. division is carried out by multiplying the dividend and the divisor by the conjugate of the divisor.

The exponentiation of complex numbers is defined as follows:

It is easy to show that

Examples.

1. Find the sum of complex numbers z 1 = 2 – i And z 2 = – 4 + 3i.

z 1 +z 2 = (2 + (–1)∙i)+ (–4 + 3i) = (2 + (–4)) + ((–1) + 3) i = –2+2i.

2. Find the product of complex numbers z 1 = 2 – 3i And z 2 = –4 + 5i.

= (2 – 3i) ∙ (–4 + 5i) = 2 ∙(–4) + (-4) ∙(–3i)+ 2∙5i– 3i∙ 5i = 7+22i.

3. Find private z from division z 1 \u003d 3 - 2 z 2 = 3 – i.

z= .

4. Solve the equation:, x And y Î R.

(2x+y) + (x+y)i = 2 + 3i.

By virtue of the equality of complex numbers, we have:

where x=–1 , y= 4.

5. Calculate: i 2 ,i 3 ,i 4 ,i 5 ,i 6 ,i -1 , i -2 .

6. Calculate if .

7. Calculate the reciprocal of a number z=3-i.

Complex numbers in trigonometric form

complex plane is called a plane with Cartesian coordinates ( x, y), if each point with coordinates ( a, b) is assigned a complex number z = a + bi. In this case, the abscissa axis is called real axis, and the y-axis is imaginary. Then every complex number a+bi geometrically represented on a plane as a point A (a, b) or vector .

Therefore, the position of the point A(and hence the complex number z) can be set by the length of the vector | | = r and angle j formed by the vector | | with the positive direction of the real axis. The length of a vector is called complex number modulus and is denoted by | z|=r, and the angle j called complex number argument and denoted j = argz.

It is clear that | z| ³ 0 and | z | = 0 Û z= 0.

From fig. 2 shows that .

The argument of a complex number is defined ambiguously, and up to 2 pk,kÎ Z.

From fig. 2 also shows that if z=a+bi And j=argz, That

cos j =, sin j =, tg j = .

If zОR And z > 0 then argz = 0 +2pk;

If z ОR And z< 0 then argz = p + 2pk;

If z= 0,argz indefined.

The main value of the argument is determined on the interval 0 £argz£2 p,

or -p£ arg z £ p.

Examples:

1. Find the modulus of complex numbers z 1 = 4 – 3i And z 2 = –2–2i.