Translation from complex form to trigonometric. The trigonometric form of complex numbers. Complex numbers xi

2.3. Trigonometric form of complex numbers

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

Significantly through φ cut between the positive pіvvіssyu Ox і vector (kut φ is considered positive, as if it were reversed by the anti-year arrow, and negative otherwise).

The length of the vector is significant in terms of r. Todi. Significantly so

Writing a vіdmіnnogo vіd zero of the complex number z vvglyadі

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 is called the argument of that complex number and is denoted by Arg z.

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

The complex number z has no number of arguments: if φ0 is any argument of the number z, then everything else can be known from the formula

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

In this order, the argument of the nominal zero of the complex number is the solution of the system of equals:

(3)

The value of the argument of the complex number z, which satisfies the irregularities, is called the head and is denoted by arg z.

Arguments Arg z and arg z prove jealousy

, (4)

Formula (5) is the result of system (3), so all the arguments of the complex number are satisfied with the equality (5), but not all solutions of the equality φ (5) are the arguments of the number z.

The headline value of the argument of the nominal zero of a complex number is behind the formulas:

The multiplication formulas for complex numbers in the trigonometric form may look like this:

. (7)

When taken up to a natural degree of a complex number, de Moivre's formula is victorious:

When the root of the complex number is changed, the formula is victorious:

, (9)

de k = 0, 1, 2, …, n-1.

Task 54. Calculate, de.

Let's give a solution to this virase in the display form of the complex number: .

Yakscho something.

Todi , . That's why і de .

Suggestion: , per .

Task 55. Write complex numbers in trigonometric form:

a); b); in); G); e); e) ; and).

Oskіlki trigonometric form of a complex number can be seen, then:

a) For a complex number: .

Tom

b) de,

G) de,

e) .

and) , a , then.

Tom

Suggestion: ; 4; ; ; ; ; .

Task 56. Find the trigonometric form of a complex number

Come on .

Todi , , .

Oskilki i , , then , and

Father, to that

Suggestion: de .

Task 57. Vicorist trigonometric form of a complex number

Imagine the numbers in trigonometric form.

1) , de also

We know the value of the head argument:

Let's assume the value and the difference, take it away

2) , de todi

Todi

3) Know privately

In addition, k=0, 1, 2, we take three different values shukany root:

Yakscho something

like that

like that .

Suggestion:

: .

Problem 58 . Bring what

a) number є diysnim positive number;

b) may be equal:

a) Give the given complex numbers in trigonometric form:

So yak.

Let's say what. Todi

The rest is a positive number, so that under the signs of sinuses there are numbers in the interval.

because the number verbally and positively. Actually, like a and b are complex numbers and speech and more than zero, then .

Cream of that

later, the necessary equivalence has been brought.

Task 59. Write a number in the form of algebra .

We represent the number in trigonometric form, and then we know the form of algebra. Maymo . For take the system:

Sounds of jealousy: .

Zastosovuyuchi De Moivre's formula:

acceptable

The trigonometric form of the given number is found.

Now let's write the number in the form of algebra:

Suggestion: .

Task 60. Find out the amount , ,

Let's look at the sum

Zastosovuyuchi formula de Moivre, we know

Tsya sum є sum n members of the geometric progression z znamennik the first member .

Zastosovuyuchi formula for the sum of the members of such a progression, perhaps

Seeing a part in the rest of the view, we know

Seeing the actual part, we take the following formula: , , .

Task 61. Find out the amount:

a) ; b).

Behind Newton's formula for the link at the feet is possible

Behind De Moivre's formula we know:

Pririvnyuyuchi speech and obvious parts of otrimanih viraziv for, maybe:

і .

Qi formulas for a compact view can be written as follows:

de - tsila part of number a.

Task 62

Oskilki , then, fixing the formula

, For the cultivation of roots, it is necessary ,

Otzhe, , ,

, .

The points, which correspond to numbers, are shaded at the vertices of a square inscribed in a circle of radius 2 with the center at the point (0; 0) (Fig. 30).

Suggestion: , ,

, .

Office 63 , .

For the mind; to that Dane equal I don’t have a root, and, later, it’s more equal than equal.

In order for the number z to be the root of this equation, it is necessary that the number was roots n-th step from number 1.

Zvіdsi stowed, scho vhіdne rіvnyannja maє korіnnya, vznachene z іvіrnosti

in such a manner,

tobto. ,

Suggestion: .

Task 64. Razvnyanya at impersonal complex numbers.

Since the number is not the root of a given equal, then with a given equal equal equal

That is, equal.

The mustache of the root of this equation comes from the formula (div. problem 62):

; ; ; ; .

Task 65. Paint on a complex plane a faceless dot that pleases the unevenness: . (2nd way to resolve tasks 45)

Come on .

Complex numbers, which can have the same modules, are given points of the plane, which lie on the stake with the center on the cob of coordinates, to that unevenness satisfies all points of the open circle, surrounded by stakes from the head center on the cob of coordinates and radii (Fig. 31). Let the point of the complex plane correspond to the number w0. Number , maє module, times less than the module w0, argument, by bigger argument w0. From the geometrical point to the point, which corresponds to w1, you can take a vicorist homothety with the center on the cob of coordinates and the coefficient , as well as the rotation of the cob of coordinates on the coot of the anti-Godinnikov arrow. As a result, zastosuvannya tsikh dvoh reversal to the point of the ring (Fig. 31), the remaining transition in the ring, surrounded by kіl with the same center and radii 1 and 2 (Fig. 32).

Reincarnation be realized for help parallel transfer to the vector. Transferring the ring with the center at the point to the designation vector, take the ring of the same size with the center at the point (Fig. 22).

The proponation method, which victorious idea of the geometric transformations of the area, is sing-song, less easy in the description, but even more subtle and effective.

Zavdannya 66. Know, like .

Let me go. Vihіdna jealousy in the future I see . Mind the equivalence of two complex numbers is taken away, stars. In such a manner,

We write the number z in the trigonometric form:

de , . Zgidno with the formula of De Moivre, we know.

Vidpovid: - 64.

Task 67. For a complex number, find u .

Imagine a number in trigonometric form:

. Zvіdsi,. For the number we take away, we can add either.

In the first moment , for another

Suggestion: , .

Task 68. Find the sum of such numbers, scho. Enter one of these numbers.

Respectfully, you can already understand from the formula itself that you can know the sum of the root equals without counting the root itself. True, the sum of roots is equal є coefficient at , taking іz with the opposite sign (the theorem of Вієт is pointed out), then.

Learned, school documentation, worked out visnovki about the rhіvnja svoєnnya tsgogo opredelya. Pіdbiti pіdbumok vyvchennya osobennosti mathematicheskoj smylennya and the process of forming the understanding of a complex number. Description of methods. Diagnostic: I stage. Rozmov was held with a teacher of mathematics, as in the 10th grade he taught algebra and geometry. Rozmova wilted on the cob in a day's time...

Resonance "(!)), which also includes an assessment of the state of mind. 4. Critically assessing one's own understanding of the situation (doubtful). 5. Nareshti, voicing the recommendation of legal psychology (the appearance of a lawyer of psychological aspects in the preparation of professional research - professionalism). psychological analysis of legal facts.

Mathematics of the trigonometric substitution and re-verification of the efficiency of the divided methodology of the experiment. Stages of work: 1. Development of an optional course on the topic: “Studying a trigonometric substitution on top algebraic problems” with classes and perverted mathematics. 2. Conducting a divided optional course. 3. Carrying out a diagnostic control...

The appointments of the zavdannya poklikany only to supplement the essentials of the beginning and the responsibilities are known to the docile due to the traditional methods and elements of the initial process. Vіdmіnіst navchalnyh zavdannya vkladannі humanіtarnіh nauki vіd exact, mathematichnіh zavdanі scho scho vіdnіnіh zavdannya vіdsutnі formulas, zhorstkі algorithms just. ...

To determine the position of the point on the plane, you can speed up the polar coordinates [g, (p), de G- Move points to the cob of coordinates, and (R- cut, which is to become a radius - vector of a point on a positive direct axis Oh. Positive direct change (R vvazhayetsya directly against the year's arrows. Having speeded up with a link of Cartesian and polar coordinates: x = r cos cf, y = r sin (p,

take the trigonometric form of the complex number

z - r(sin (p + i sin

de G

Xi + y2, (p is the argument of a complex number, which one knows

l X . y y

formulas cos(p --, sin^9 = - or through those tg(p --, (p-arctg

We respect that when choosing a value Wed from the rest of the year it is necessary to insure the signs x and y

Butt 47 2 \u003d -1 + l / Z /.

Solution. We know the modulus and argument of a complex number:

= yj 1 + 3 = 2 . Kut Wed we know from spіvvіdnoshen cos(p = -, sin(p = - . Todi

taken cos(p = -,suup

u/z g~

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

at the other quarter: (R= 120°

Substituting

2 to.. cos-h; sin

formula (1) is known 27G L

Respect. The argument of a complex number is assigned not unambiguously, but exactly up to a warehouse, multiple 2p. Go through cn^r signify

the value of the argument, placed in the boundaries (P 0 %2 Todi

A) ^ r = + 2kk.

Vikoristovuyuchi euler's formula That is, we need to show the form of writing a complex number.

Maymo r = r (s ^ (p + i?, n (p) = r,

Dії over complex numbers

1. The sum of two complex numbers r = X] + y x/ i r 2 - x 2+y 2 / signifies zgidno with the formula r! +2 2 = (x, + ^ 2) + (^ 1 + ^ 2) ' g
2. The operation of looking at complex numbers is considered to be an operation that reverses before folding. Complex number r \u003d r x - r 2 yakscho g 2 + g \u003d g x,

є difference of complex numbers 2, that g 2 . Todi g \u003d (x, - x 2) + (y, - at 2) /.

3. Dobutok two complex numbers g x\u003d x, + y, -r and 2 2 \u003d x 2+ U2'g is assigned to the formula
*1*2 =(* +U"0(X 2+ T 2 -0 = X 1 X 2 Y 1 2 -1 + x Y2 " * + At1 At2 " ^ =

\u003d (xx 2 ~ YY 2) + ( X Y2 + X 2Y) - "-

Zokrema, y-y\u003d (x + y-z) (x-y /) \u003d x 2 + y 2.

You can take the formulas for the multiplication of complex numbers in display and trigonometric forms. Maemo:

1^ 2 - r x e 1 = )Г 2 e > = Г]Г 2 cOs((P + cp 2) + isin
4. The subdivision of complex numbers is shown as an operation, reverse

plural, tobto. number G-- called private in the form of a rozpodіlu g! on p 2,

yakscho r x -1 2 ? 2 . Todi

X + Ті _ (*і + ІU 2 ~ 1 U2 ) x 2 + ІУ2 ( 2 + ^Y 2)( 2 ~ 1 Y 2)

x, x 2 + / y, x 2 - x 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) - І.сОї ((Р -СР 1) + І- (R-,)] >2 >2
5. Climbing into the positive step of a complex number is better than robust, as the number is written in illustrative trigonometric forms.

Right, that's right z = ge 1 then

=(ge,) = r p e t = G"(Co8 psr + іt gcr).

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

6. Root fork P- th step from a complex number is shown as an operation, returning to the step p, p- 1,2,3,... then. complex number = y[g called the root P- step of a complex number

g, yakscho G = g x. Why g - g ", a g x= l/r. (p-psr x, a Wed-Wed/R, which is evident from the Moivre formula, written for the number = r / * + iipp(p).

As it was assigned more, the argument of the complex number was not assigned unambiguously, but exactly up to an additional multiple of 2 and. Tom = (p + 2pc, and the argument of the number r, what to deposit vіd before, meaningfully (R to i boo

dem calculate by formula (R to= - +. It's clear what you know P com-

plex numbers, P-th step of any of them is closer to the number 2. Qi of the number may be one

and the same module, equal y[r, and the arguments of the numbers go beyond before = 0, 1, P - 1. In this order, in trigonometric form root i-th the steps are calculated according to the formula:

(p + 2kp) . . cf + 2kp

, before = 0, 1, 77-1,

.(r+2ktg

and in the show form - behind the formula l[r - y[ge n

Applied 48

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

(1 - /l/2) 3 (s + /) \u003d (1 - Zl / 2 / + 6 / 2 - 2 l / 2 / ? 3) (3 + /) \u003d
(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)/;

Butt 49. Write the number r \u003d Uz - / five steps.

Solution. We take the trigonometric form of writing the number p.

G = l/3+1 =2, CO8 (p --, 5ІІ7 (R =

(1 - 2/X2+/)
(Z-,)

Pro - 2.-x2 + pro

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 s_±.
5 2 1 "

Zvіdsi pro--, a r = 2

Moivre is taken: i-2

/ ^ _ 7r, . ?G

-US-- IBIP -

--b/-

\u003d - (L / Z + r) \u003d -2.

Example 50. Know all the meanings

Solution, r = 2, a Wed we know from equal coy(p = -, zt--.

Tsya point 1 - / d / s rebuy at the fourth quarter, tobto. f =--. Todi

1 - 2
( ( UG L

The value of the root is known by viraz

V1 - /l/s = l/2

--+ 2A:/g ---b 2 kk
3 . . 3

С08--1-i 81П-

At before - 0 may 2 0 = l/2

You can know the value of the root z of the number 2 by presenting the number in the display

-* TO/ 3 + 2 class

At before= 1

7G. 7G_
---b27g ---b2;g
3 . . h

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

--N-

zі? - 7G + / 5Sh - I "

l/3__t_

no form. so yak r= 2, a Wed= , then r = 2e 3 a 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.Dії over 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).

Malyunok 1

b) A complex number can be represented by a vector, which can be an ear at a pointPro і end at tsіy point (Fig. 2).

Malyunok 2

Example 7. Look for points that represent complex numbers:1; - i ; - 1 + i ; 2 – 3 i (Fig.3).

Baby 3

Trigonometric notation of complex numbers.

Complex numberz = a + bi you can ask for an additional radius. with coordinates( a ; b ) (Fig.4).

Malyunok 4

Appointment . Dovzhina vector , which represents a complex numberz , is called the modulus of which number it is designated orr .

For any complex numberz yoga moduler = | z | stands out unambiguously for the formula .

Appointment . The value of the kuta between the positive direction of the functional axis and the vector , which represents a complex number, is called the argument of that complex number and is assignedBUT rg z orφ .

Complex number argumentz = 0 no appointments. Complex number argumentz≠ 0 - the value is richly significant2πk (k = 0; - 1; 1; - 2; 2; ...): Arg z = arg z + 2πk , dearg z - Golovne znachennya argument, stowed in the gap(-π; π] , then-π < arg z ≤ π (Sometimes the main value of the argument is to take the value that needs to be .

Qiu formula atr =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) Examination of the square root of a complex number.

When the square root of a complex number is changeda + bi there are two fluctuations:

yakschob > pro , then ;

3.1. Polar coordinates

Often zastosovuetsya on the flat polar coordinate system . Vaughn is assigned, as the point O is given, as it is called pole, and promise, what to go out of the pole (for us, all Ox) - polarity. The position of the point M is fixed by two numbers: radius (or radius-vector) and φ between the polar vector . Kut φ is called polar kutom; vimiryuєtsya in radians and vіdrakhovuєtsya in the polar axis of the anti-annual arrow.

The position of the point in the polar coordinate system is determined by an ordered pair of numbers (r; φ). Bіlya pole r = 0 and φ is not assigned. For all other points r > 0 and φ is assigned exactly to the folded multiple of 2π. For each pair of numbers (r; φ) i (r 1 ; φ 1) one and the same point is set, which is .

For a rectangular coordinate system xOy The Cartesian coordinates of a point are easily inverted through її polar coordinates in the following way:

3.2. Geometric interpretation of a complex number

Let's take a look at the Cartesian rectangular coordinate system xOy.

Whether or not a complex number z=(a, b) should be assigned a point of the plane with coordinates ( x, y), de coordinate x = a, then. the real part of the complex number, and the coordinate y = bi is the explicit part.

The area, with dots as a complex number, is a complex area.

On a small complex number z = (a, b) point M(x, y).

Manager.Display complex numbers on the coordinate plane:

3.3. Trigonometric form of a complex number

Complex number on the plane can coordinate points M(x; y). With whom:

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

The number r is called module complex number z i signify. The module is not a speech number. For .

The module is more equal to zero then and less then, if z = 0, then. a=b=0.

The number φ is called argument z and signify. The argument z is ambiguous, since it is the polar coot of the polar coordinate system, and most accurately up to a multiple, which is supposed to be 2π.

Todi accept: de? least value argument. Obviously what

For more deep learning, they introduce an additional argument φ *, such that

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

Solution. 1) Important module: ;

2) jokingly φ: ;

3) trigonometric form:

butt 2. Know the form of complex number algebra .

Here it’s enough to give the values of trigonometric functions and rewrite the viraz:

example 3. Know the modulus and argument of a complex number;

1) ;

2); φ - in 4 quarters:

3.4. Dії with complex numbers in trigonometric form

· Additional information easier to converse with complex numbers in the form of algebra:

· plural- with the help of clumsy trigonometric transformations, you can show that when multiplying, the modules of numbers are multiplied, and the arguments are added: ;

Дії over complex numbers written in the form of algebra

Algebraic form of the complex number z =(a,b). is called algebraically in terms of

z = a + bi.

Arithmetic operations on complex numbers z 1 = a 1 +b 1 iі z 2 = a 2 +b 2 i, Recorded in the form of algebra, zdіysnyuyutsya coming rank.

1. Sum (cost) of complex numbers

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

tobto. additions (withdrawals) are subject to the rule of folding rich members with the given similar members.

2. Doboot of complex numbers

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

tobto. multiplication is carried out according to the universal rule of multiplication of rich terms, in order to ensure that i 2 = 1.

3. Rozpodіl two complex numbers zdіysnyuєtsya following this rule:

, (z 2 ≠ 0),

tobto. rozpodіl zdіysnyuєtsya multiplied by the dilemma of that dilnik by the number connected to the dilnik.

The grading to the stage of complex numbers is defined as follows:

It's easy to show that

Apply.

1. Know the sum of complex numbers z 1 = 2 – iі z 2 = – 4 + 3i.

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

2. Know the doboot of complex numbers z 1 = 2 – 3iі z 2 = –4 + 5i.

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

3. Know privately z according to the rozpodіlu z 1 = 3 - 2on z 2 = 3 – i.

z= .

4. Razvyazati equalization:, xі y Î R.

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

Due to the equality of complex numbers, it is possible:

stars x=–1 , y= 4.

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

6. Calculate, yakscho.

7. Calculate the number opposite to the number z=3-i.

Complex numbers in trigonometric form

complex plane called the plane with Cartesian coordinates ( x, y), which is a skin point with coordinates ( a, b) set y to a complex number z = a + bi. When all abscissa is called right hand, and all ordinates - manifest. Same skin complex number a+bi geometrically depicted on the plane like a point A (a, b) or a vector.

Otze, the position of the point BUT(i, aka, complex number z) can be inserted with a double vector | | = r ta kutom j, let's make a vector | | from a positive direct action axis. The valley of the vector is called complex number modulus that signifies | z|=r, and kut j called complex number argument and signify j = argz.

It dawned on me that | z| ³ 0 ta | z | = 0 Û z= 0.

3 fig. 2 shows that .

The argument of a complex number is ambiguous, but exactly up to 2 pk, kÎ Z.

3 fig. 2 it is also clear that it is z=a+biі j = argz, then

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

Yakscho zОRі z > 0 then argz = 0 +2pk;

yakscho z ОRі z< 0 then argz = p + 2pk;

yakscho z= 0,argz no appointments.

The head value of the argument is assigned to the value 0 £argz£2 p,

or -p£ arg z £ p.

Apply:

1. Know the modulus of complex numbers z 1 = 4 – 3iі z 2 = –2–2i.