2.3. Trigonometric form of complex numbers

Let the vector be specified on the complex plane by a number.

Let us denote by φ the angle between the positive semiaxis Ox and the vector (the angle φ is considered positive if it is counted counterclockwise, and negative otherwise).

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

Writing a nonzero complex number z in the form

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.

Trigonometric notation of a complex number - (Euler's formula) - exponential notation of 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 nonzero 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 principal and is denoted by arg z.

Arg z and arg z are related by the equality

, (4)

Formula (5) is a consequence of system (3), therefore 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 can be found by the formulas:

The formulas for multiplication and division of complex numbers in trigonometric form are as follows:

. (7)

When raising a complex number to a natural power, the Moivre 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 notation of a complex number:.

If, then.

Then , ... Therefore, then and , where .

Answer: , at .

Problem 55. Write down complex numbers in trigonometric form:

a) ; b); v) ; G) ; e); e) ; g).

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

a) In a complex number:.

,

That's why

b) , where ,

G) , where ,

e) .

g) , a , then .

That's why

Answer: ; 4; ; ; ; ; .

Problem 56. Find the trigonometric form of a complex number

.

Let be , .

Then , , .

Since and ,, then, and

Therefore, therefore

Answer: , where .

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

Let's represent numbers and in trigonometric form.

1), where then

Find the value of the main argument:

Substitute the values ​​and into the expression, we get

2) where then

Then

3) Find the quotient

Setting k = 0, 1, 2, we get three different values ​​of the 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) We represent these complex numbers in trigonometric form:

Because .

Let's pretend that . Then

.

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

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

Besides,

therefore, the required equality is proved.

Problem 59. Write down the number in algebraic form .

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

This implies the equality: .

Applying the Moivre formula:,

we get

Found the trigonometric form of the given number.

We now write this number in algebraic form:

.

Answer: .

Problem 60. Find the sum,,

Consider the amount

Applying the Moivre formula, we find

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

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

Separating the 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

Using the Moivre 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 is the integer part of the number a.

Problem 62. Find everyone for whom.

Insofar as , 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 an equation.

In order for the number z to be the root of this equation, the number must be the nth root of the 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 a 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 the set of points satisfying the inequalities: ... (2nd way to solve problem 45)

Let be .

Complex numbers with the same moduli correspond to points of the plane lying on a circle centered at the origin, therefore, the inequality satisfy all points of an 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 that is times smaller than the modulus w0, and an argument that is larger than the argument w0. Geometrically, the point corresponding to w1 can be obtained using a homothety with a center at the origin and a coefficient, as well as rotation about the origin by an angle counterclockwise. As a result of applying these two transformations to the points of the ring (Fig. 31), the latter transforms into a ring bounded by circles with the same center and radii 1 and 2 (Fig. 32).

Transformation implemented using parallel translation to a vector. Moving 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, using the idea of ​​geometric transformations of the plane, is probably less convenient in description, but very elegant and effective.

Problem 66. Find if .

Let, then and. The original equality takes 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 the Moivre 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 the 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. Enter one of these numbers.

Note that already from the very formulation of the problem, one can understand 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 at taken with the opposite sign (generalized Vieta's theorem), i.e.

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

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 the recommendations of legal psychology (taking into account the psychological aspects of professional actions performed by a lawyer - professional psychological preparedness). Let us now consider the psychological analysis of legal facts. ...

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

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

COMPLEX NUMBERS XI

§ 256. Trigonometric form of complex numbers

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

We denote the length of this vector by r , and the angle that it forms with the axis NS , across 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 is 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 the 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 the complex number a + bi is not equal to zero, then its argument is determined by the formulas (2) unequivocally accurate 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 undefined.

Complex number module r sometimes denote | z | and the arg argument z ... Let's look at some examples of how complex numbers can be represented in trigonometric form.

Example. 1. 1 + i .

Find the module r and the 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 NS - any integer. Usually, from an infinite set of values ​​of the argument of a complex number, one is chosen that lies 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 a complex number in trigonometric form √ 3 - i ... We have:

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

Therefore, up to an angle multiple of 2 π , φ = 11 / 6 π ; hence,

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

Example 3 Write a complex number in trigonometric form i.

Complex number i matches vector OA> ending at point A of the axis at with ordinate 1 (fig. 333). The length of such a vector is 1, and the angle it makes with the abscissa is 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 > NS abscissa 3 (Fig. 334).

The length of such a vector is 3, and the angle it forms with the abscissa is 0. Therefore,

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

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

The complex number -5 corresponds to the vector OA> ending at an axis point NS with an abscissa -5 (Fig. 335). The length of such a vector is 5, and the angle it makes with the abscissa 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 set of points representing complex numbers, the moduli r and arguments φ of which 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 the module of a complex number be numbers at the same time? r and - r ?

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

To represent these complex numbers in trigonometric form, 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 °).

3.1. Polar coordinates

On a plane is often used polar coordinate system ... It is defined if a point O is given, called pole, and a ray emanating from the pole (for us, this is the axis Ox) is the polar axis. The position of point M is fixed with two numbers: the 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 specified by an ordered pair of numbers (r; φ). At the pole r = 0, and φ is undefined. For all other points r> 0, and φ is defined up to a multiple of 2π. In this case, pairs of numbers (r; φ) and (r 1; φ 1) are associated with the same point if.

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

3.2. Geometric interpretation of a complex number

Consider on the plane a Cartesian rectangular coordinate system xOy.

Any complex number z = (a, b) is assigned a point on 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.

The plane whose points are complex numbers is the complex plane.

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

Exercise.Draw complex numbers on the coordinate plane:

3.3. Trigonometric form of a complex number

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

Complex number notation - trigonometric form of a complex number.

The number r is called module complex number z and is indicated by. Modulus 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, as well as the polar angle in the polar coordinate system, namely, up to a multiple of 2π.

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

.

For 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) consider the module:;

2) we are looking for φ: ;

3) trigonometric form:

Example 2. Find the algebraic form of a complex number .

Here it is enough to substitute the values ​​of 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. Actions with complex numbers in trigonometric form

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

· Multiplication- using simple trigonometric transformations, one can show that when multiplying, the modules of numbers are multiplied, and the arguments are added: ;

To determine the position of a point on a plane, you can use polar coordinates [r, (p), where G is the distance of a point from the origin, and (R- the angle that makes up the radius - the vector of this point with the positive direction of the axis Oh. Positive direction of angle change (R counterclockwise direction is considered. Using the connection between Cartesian and polar coordinates: x = r cos cf, y = r sin (p,

we get the trigonometric form of writing a complex number

z - r (sin (p + i sin

where G

Xі + y2, (p is the argument of a complex number, which is found from

l X . at

formulas cos (p --, sin ^ 9 = - or due to the fact that tg (p --, (p-arctg

Note that when choosing values Wed from the last equation it is necessary to take into account the signs x and y.

Example 47. Write a complex number in trigonometric form 2 = -1 + l / Z /.

Solution. Find the modulus and argument of a complex number:

= yj 1 + 3 = 2 . Injection Wed find from the relations cos (p = -, sin (p = -. Then

get cos (p = -, suup

u / z g ~

• - -. Obviously, the point z = -1 + V3- / is
• 2 To 3

in the second quarter: (R= 120 °

Substituting

2 r.... cos - h; sin

into formula (1) found 27Г Л

Comment. The argument of a complex number is not uniquely defined, but up to a term that is a multiple of 2p. Then through cn ^ r denote

the argument value enclosed within (p 0 %2 Then

A) ^ r = + 2kk.

Using the well-known Euler formula That is, we get an exponential notation of a complex number.

We have r = r (ω ^ (p + i?, n (p) = r,

Actions on complex numbers

• 1. The sum of two complex numbers r, = X] + y x/ and r 2 - x 2 + y 2 / is determined according to the formula r! +2 2 = (x, + ^ 2) + (^ 1 + ^ 2) ’
• 2. The operation of subtracting complex numbers is defined as the inverse of addition. Complex number r = rx - r 2, if z 2 + z = z x,

is the difference of complex numbers 2, and d 2. Then r = (x, - x 2) + (y, - at 2) /.

• 3. Product of two complex numbers r x= x, + y, -z and 2 2 = x 2+ U2‘G is determined by the formula
• *1*2 =(* + U"0 (X 2+ T 2 -0 = X 1 X 2 Y 1 2 -1 + x Y2 " * + Have1 Have2 " ^ =

= (xx 2 ~ YY 2) + (X Y2 + X 2Y) - "-

In particular, yr= (x + y-z) (x-y /) = x 2 + y 2.

You can get multiplication formulas for complex numbers in exponential and trigonometric forms. We have:

• 1^ 2 - Г х е 1 = ) Г 2 е> = Г] Г 2 cOs ((P + cf 2) + isin
• 4. Division of complex numbers is defined as the inverse operation

multiplication, i.e. number G-- is called the quotient of dividing! on r 2,

if r x -1 2 ? 2 . Then

NS + Ті _ (*і + IU 2 ~ 1 U2 ) x 2 + ІУ2 (2 + ^ Y 2) (2 ~ 1 Y 2)

x, x 2 + / y, x 2 - іх х у 2 - і 2 y x y 2 (x x x 2 + y x y 2) + / (- x, y 2 + X 2 Y])

2 2 x 2 + U 2

1 e

i (p g

• - 1U e "(1 Fg) - I.сОї ((R -cr 1) + I- (R-,)] >2 >2
• 5. Raising a complex number to a positive integer power is best done if the number is written in exponential or trigonometric form.

Indeed, if r = rt 1 then

= (re,) = r n e t = G"(co8 psr + іт гкр).

Formula g " = rn (cosn (p + is n (p) called the Moivre formula.

6. Extracting the root NS- th power of a complex number is defined as the inverse operation of raising to a power n, n- 1,2,3, ... i.e. complex number = y [g called root NS- th degree of a complex number

d if G = r x... It follows from this definition that g - g ", a r x= l / g. (p-psr x, a cf-cp / n, which follows from the Moivre formula written for the number = r / * + ilipn (p).

As noted above, the argument of a complex number is not uniquely determined, but up to a term multiple of 2 f. That's why = (p + 2pk, and the argument of the number r, depending on To, denote (p to and boo

dem calculate by the formula (p to= - +. It is clear that there is NS com-

plex numbers, NS-th power of which is equal to 2. These numbers have one

and the same module equal to y [r, and the arguments of these numbers are obtained when To = 0, 1, NS - 1. Thus, in trigonometric form, the root of the i-th power is calculated by the formula:

(p + 2kp . . Wed + 2kp

, To = 0, 1, 77-1,

. (p + 2ktg

and in exemplary form - according to the formula l [z - y [z n

Example 48. Perform operations on complex numbers in algebraic form:

a) (1- / H / 2) 3 (3 + /)

• (1 - / l / 2) 3 (s + /) = (1 - Zl / 2 / + 6/2 - 2 l / 2 /? 3) (3 + /) =
• (1 - Zl / 2 / - 6 + 2l / 2 / DZ + /) = (- 5 - l / 2 / DZ + /) =

15-Zl / 2 / -5 / -l / 2/2 = -15 - Zl / 2 / -5 / + l / 2 = (-15 + l / 2) - (5 + Zl / 2) /;

Example 49. Construct the number r = Uz - / to the fifth power.

Solution. We obtain the trigonometric form of writing the number r.

Г = l / 3 + 1 = 2, C08 (p --, 5ІІ7 (R =

• (1 - 2 / X2 + /)
• (s-,)

O - 2.-X2 + o

• 12+ 4/-9/
• 2 +і - 4/ - 2/ 2 2 - 3/ + 2 4 - 3/ 3 + і
• (z-O "(z-O

З / 2 12-51 + 3 15 - 5 /

• (3-i) 's + /
• 9 + 1 s_ ±.
• 5 2 1 "

From here O--, a r = 2

Moivre we get: i -2

/ ^ _ 7Г,. ?G

• -USH-- IBIP -

• --B / -

= - (l / Z + z) = -2.

Example 50. Find all values

Solution, r = 2, and Wed find from the equation soy (p = -, zt -.

This point 1 - / d / z is in the fourth quarter, i.e. f =-. Then

• 1 - 2
• ( ( UG L

We find the values ​​of the root from the expression

V1 - / l / s = l / 2

• - + 2A: / g --- b 2 kk
• 3 . . 3

С08-1- і 81П-

At To - 0 we have 2 0 = l / 2

You can find the values ​​of the root of the number 2 by presenting the number in the display

-* TO/ 3 + 2 cl

At To= 1 we have one more root value:

• 7G. 7G _
• --- b27g --- b2; g
• 3. ... s

7G ... ... 7G L-C05- + 181P - 6 6

• --Н -

co? - 7G + / 5SH - Z "

l / 3__t_

tel form. Because r = 2, a Wed=, then r = 2e 3, and y [g = y / 2e 2

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 by a vector that starts at the pointO and the end at this point (Fig. 2).

Picture 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 radius vector with coordinates( a ; b ) (fig. 4).

Figure 4

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

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

Definition . The magnitude 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 multivalued 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 main value of the argument is taken as a value belonging to the interval .

This formula forr =1 often referred to as the Moivre 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) Extracting the square root of a complex number.

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

ifb > about , then ;