Actions on complex numbers, written in algebraic form

Algebraic form of a 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 for adding polynomials with reduction of similar terms.

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 carried out according to the usual rule for multiplying 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 number of the divisor.

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 the quotient z from division z 1 = 3 – 2na z 2 = 3 – i.

z = .

4. Solve the equation: , x And y Î R.

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

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

Therefore, the position of the point A(and, therefore, a complex number z) can be specified 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 the vector is called modulus of a complex number and is denoted by | z |=r, and the angle j called complex number argument and is designated j = arg z.

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

From Fig. 2 it is clear that .

The argument of a complex number is determined ambiguously, but with an accuracy of 2 pk,kÎ Z.

From Fig. 2 it is also clear that if z=a+bi And j=arg z, That

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

If zÎR And z> 0,then arg z = 0 +2pk;

If z ОR And z< 0,then arg z = p + 2pk;

If z = 0,arg z indefined.

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

or -p£ arg z £ p.

Examples:

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

2. Define areas on the complex plane defined by the conditions:

1) | z | = 5; 2) | z| £6; 3) | z – (2+i) | £3; 4) £6 | z – i| £7.

Solutions and answers:

1) | z| = 5 Û Û - equation of a circle with radius 5 and center at the origin.

2) A circle with radius 6 with center at the origin.

3) Circle with radius 3 with center at point z 0 = 2 + i.

4) A ring bounded by circles with radii 6 and 7 with a center at a point z 0 = i.

3. Find the modulus and argument of the numbers: 1) ; 2) .

1) ; A = 1, b = Þ ,

Þ j 1 = .

2) z 2 = –2 – 2i; a =–2, b =-2 Þ ,

.

Hint: When determining the main argument, use the complex plane.

Thus: z 1 = .

2) , r 2 = 1, j 2 = , .

3) , r 3 = 1, j 3 = , .

4) , r 4 = 1, j 4 = , .

COMPLEX NUMBERS XI

§ 256. Trigonometric form of complex numbers

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

Let us 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, from where r = √a 2 + b 2

So, any complex number a + bi can be represented in the form :

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 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 accurate to an angle divisible by 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 null argument is undefined.

Modulus of a complex number r sometimes denoted | z |, and the argument arg z . Let's look at a few examples of representing 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 the infinite set of values ​​of 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 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 divisible by 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 corresponds vector O.A.> , 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 x-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 O.A. > 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 the complex number -5 in trigonometric form.

The complex number -5 corresponds to a vector O.A.> ending at an axis point X with abscissa -5 (Fig. 335). The length of such a vector is 5, and the angle it forms with the x-axis is equal to 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 a set of points representing complex numbers whose moduli 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 simultaneously be the modulus of a complex number? r And - r ?

2050. Can the argument of a complex number simultaneously be angles? φ And - φ ?

Present 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°).

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 on a 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 begins at the pointABOUT and the end at a given point (Fig. 2).

Figure 2

Example 7. Construct 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 specified using the 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 uniquely 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 – a multi-valued quantity and is determined to within a term2πk (k = 0; - 1; 1; - 2; 2; …): Arg z = arg z + 2πk , Wherearg z – the main value of the argument contained in the interval(-π; π] , that is-π < arg z ≤ π (sometimes a value belonging to the interval is taken as the main value of the argument .

This formula whenr =1 often called 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) Extracting the square root of a complex number.

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

Ifb >o , That ;

2.3. Trigonometric form of complex numbers

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

Let us denote by φ the angle between the positive semi-axis Ox and the vector (the angle φ is considered positive if it is measured counterclockwise, and negative otherwise).

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

Writing a non-zero 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 form of writing a complex number - (Euler's formula) - 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 using 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, satisfying the inequalities, is called the main value and is denoted by arg z.

The arguments Arg z and arg z are related by

, (4)

Formula (5) is a consequence of system (3), therefore all arguments of a 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 according to the formulas:

Formulas for multiplying and dividing 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 the root of a complex number, the formula is used:

, (9)

where k=0, 1, 2, …, n-1.

Problem 54. Calculate where .

Let us present the solution to this expression in 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) ; d) ; 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 , , .

Since and , , then , and

Therefore, , therefore

Answer: , Where .

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

Let's imagine the numbers and in trigonometric form.

1) , where Then

Find the value of the main argument:

Let's substitute the values ​​and into the expression, we get

2) , where then

Then

3) Let's find the quotient

Assuming 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 holds:

a) Let us represent these complex numbers in trigonometric form:

Because .

Let's pretend that . Then

.

The last expression is a positive number, since the sine signs contain 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 proven.

Problem 59. Write the number in algebraic form .

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

This implies the equality: .

Applying Moivre's formula: ,

we get

The trigonometric form of the given number is found.

Let us now write this number in algebraic form:

.

Answer: .

Problem 60. Find the sum , ,

Let's consider the amount

Applying Moivre's formula, we find

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

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

Isolating the imaginary part in the last expression, we find

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

Problem 61. Find the sum:

A) ; b) .

According to Newton's formula for exponentiation, we have

Using Moivre's formula we find:

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

And .

These formulas can be written in compact form as follows:

,

, Where - whole part numbers a.

Problem 62. Find all , for which .

Because the , then, using 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 with the center at the point (0;0) (Fig. 30).

Answer: , ,

, .

Problem 63. Solve the equation , .

By condition ; therefore, this equation does not have a root, and therefore it is equivalent to the equation.

In order for the number z to be the root of a given equation, the number must be the root nth degree from number 1.

From here 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 having identical modules correspond to points in 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 module several times smaller than the module w0, and an argument greater than the argument w0. WITH geometric point From the point of view, the point corresponding to w1 can be obtained using homothety with the center at the origin and coefficient , as well as a rotation relative to the origin by an angle counterclockwise. As a result of applying these two transformations to the points of the ring (Fig. 31), the latter will transform into a ring bounded by circles with the same center and radii 1 and 2 (Fig. 32).

Conversion implemented using parallel transfer to a vector. By transferring the ring with the center at the point to the indicated vector, we obtain a ring of the same size with the center at the point (Fig. 22).

The proposed method, which uses the idea of ​​geometric transformations of a plane, is probably less convenient to describe, but is very elegant and effective.

Problem 66. Find if .

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

Let's write the number z in trigonometric form:

, Where , . According to 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:

. From here, . For the number we get , can be equal to or .

In the first case , in the second

.

Answer: , .

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

Note that 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 for , taken with the opposite sign (generalized Vieta’s theorem), i.e.

Students, school documentation, draw conclusions about the degree of mastery of this concept. Summarize the study of features 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 had passed since 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 recommendations legal psychology(the lawyer takes into account the psychological aspects of the professional actions performed - professional and psychological preparedness). Let's now consider psychological analysis legal facts. ...

Mathematics of trigonometric substitution and testing the effectiveness of the developed teaching methodology. Stages of work: 1. Development of an optional course on the topic: “Application of trigonometric substitution for solving algebraic problems” with students in classes with in-depth study mathematics. 2. Conducting the developed elective course. 3. Carrying out a diagnostic test...

Cognitive tasks are intended only to complement existing teaching aids and must be in appropriate combination with all traditional means and elements educational process. Difference educational tasks in teaching humanities from exact, from mathematical problems is only that in historical problems there are no formulas, strict algorithms, etc., which complicates their solution. ...

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

we obtain the trigonometric form of writing a complex number

z - r(sin (p + i sin

Where G

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

l X . y y

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. Let's find the modulus and argument of a complex number:

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

we get cos(p = -,suup

u/z g~

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

in the second quarter: (R= 120°

Substituting

2 k.. cos--h; sin

into formula (1) found 27Г L

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

argument value enclosed within (p 0 %2 Then

A)^r = + 2kk.

Using the famous Euler formula e, we obtain the exponential form of writing a complex number.

We have r = g(co^(p + i?,p(p)=ge,

Operations on complex numbers

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

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

• 3. Product of two complex numbers g x= x, +y, -z and 2 2 = x 2+ U2‘ r is determined by the formula
• *1*2 =(* +U"0(X 2+ T 2 -0= X 1 X 2 Y 1 2 -1 +x Y2 " * + U1 U2 " ^ =

= (хх 2 ~УУ 2)+(Х У2 + Х 2У)-"-

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

You can obtain formulas for multiplying complex numbers in exponential and trigonometric forms. We have:

• 1^ 2 - G x e 1 = )G 2 e > = G]G 2 cOs((P + avg 2) + isin
• 4. Division of complex numbers is defined as the inverse operation

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

If g x -1 2 ? 2 . Then

X + Ti _ (*і + IU 2 ~ 1 U2 ) x 2 + ІУ2 ( 2 + ^У 2)( 2 ~ 1 У 2)

x, x 2 + /y, x 2 - ix x y 2 - i 2 y x y 2 (x x x 2 + y x y 2)+ /(- x,y 2 + X 2 Y])

2 2 x 2 + Y 2

1 e

i(r g

• - 1U e "(1 Fg) - I.сОї((Р -ср 1)+ І- (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 forms.

Indeed, if g = ge 1 then

=(ge,) = g p e t = G"(co8 psr+іт gkr).

Formula g" =r n (cosn(p+is n(p) called Moivre's formula.

6. Root extraction P- th power of a complex number is defined as the inverse operation of raising to a power p, p- 1,2,3,... i.e. complex number = y[g called a root P- th power of a complex number

g, if G = g x. From this definition it follows that g - g", A g x= l/g. (r-psr x, A sr^-sr/p, which follows from Moivre’s formula written for the number = r/*+ іьіпп(р).

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

dem calculate using the formula (r k= - + . It is clear that there is P com-

complex numbers, P-th power of which is equal to the number 2. These numbers have one

and the same module equal y[g, and the arguments of these numbers are obtained by To = 0, 1, P - 1. Thus, in trigonometric form root i-th degrees are calculated using the formula:

(p + 2kp . . Wed + 2kp

, To = 0, 1, 77-1,

.(p+2ktg

and in exponential form - according to the formula l[g - y[ge p

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

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

• (1 - /l/2) 3 (z + /) = (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. Raise the number r = Uz - / to the fifth power.

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

G = l/3 + 1 =2, C08 (p --, 5ІІ7 (R =

• (1 - 2/X2 + /)
• (z-,)

O - 2.-X2 + o

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

Z/ 2 12-51 + 3 15 - 5/

• (3-i) ’з+/
• 9 + 1 z_±.
• 5 2 1 "

From here O--, A r = 2

We get Moivre: i -2

/ ^ _ 7G, . ?G

• -SS-- ІБІП -

• --b / -

= -(l/w + g)= -2 .

Example 50: Find all values

Solution, r = 2, a Wed we find from the equation sob(p = -,zt--.

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

• 1 - 2
• ( ( UG L

We find the root values ​​from the expression

V1 - /l/z = l/2

• --+ 2А:/г ---ь 2 kk
• 3 . . 3

S08--1- and 81P-

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

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

-* TO/ 3 + 2 cl

At To= 1 we have another root value:

• 7G. 7G_
• ---ь27г ---ь2;г
• 3. . h

7G . . 7G L-С05- + 181П - 6 6

• --N-

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

l/3__t_

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