To determine the position of the point on the plane, you can use polar coordinates [g, (p)where g. - distance point from the start of the coordinates, and (R - angle that is radius - vector of this point with a positive axis direction Oh. Positive direction of the corner (R The direction counterclockwise is considered. Taking advantage of the relationship of Cartesian and polar coordinates: x \u003d g cos cp, y \u003d g sin (p,

we obtain the trigonometric form of an integrated number

z - R (SIN (P + I SIN

where g.

XI + U2, (P - an integrated number argument that find from

l X. . u.

formulas cOS (P - -, sin ^ 9 \u003d - or due to the fact that tG (P --, (P-Arctg

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

Example 47. Write in trigonometric form complex number 2 \u003d -1 + l / s /.

Decision. We will find the integrated module and argument:

= yJ 1 + 3 = 2 . Angle cf. Find from relations cOS (R. = -, sin (p \u003d -. Then

receive cOS (P \u003d -, Suup.

y / s g ~

• - -. Obviously, the point z \u003d -1 + v3- / is
• 2 to3

in the second quarter: (R \u003d 120 °

Substituting

2nd . COS - H; sin.

in formula (1) found 27g l

Comment. The integrated number argument is not defined, but with accuracy to the term, multiple 2p. Then through sP ^ G. denote

the value of the argument concluded within (P 0. %2 Then

A) ^ g = + 2kk.

Using the famous Euler formula e, we obtain an indicative form of an integrated number record.

Have g \u003d g (co ^ (p + і?, p (p) \u003d ge,

Actions on complex numbers

• 1. The sum of two complex numbers g, \u003d x] + H. / and g 2 - x 2 + 2 / Determined according to the formula G! +2 2 \u003d (x, + ^ 2) + (^ 1 + ^ 2) 'g
• 2. The operation of subtraction of complex numbers is defined as an operation, reverse addition. Complex number r \u003d g x - g 2, if a g 2 + g \u003d g x,

is the difference of complex numbers 2, and g 2. Then g \u003d (x, - x 2) + (y, - w. 2) /.

• 3. Production of two complex numbers g. \u003d x, + y, -g and 2 2 \u003d x 2 + U2. 'G is determined by the formula
• *1*2 =(* + U."0 (x 2 + T 2. -0 \u003d x 1 x 21 2 -1 + x U2 " * + W.1 W.2 " ^ =

\u003d (XX 2 ~ UU 2) + (x U2 + x 2) - "-

In particular, mr. \u003d (X + U-g) (x-y /) \u003d x 2 + in 2.

You can obtain a formula for multiplication of complex numbers in an indicative and trigonometric forms. We have:

• 1^ 2 - g x E 1 \u003d ) G 2 Е\u003e \u003d g] g 2 cOS ((P + CP 2) + Isin
• 4. The division of complex numbers is defined as an operation, reverse

multiplication, i.e. number g-- is called private from dividing g! on g 2,

if a g x -1 2 ? 2 . Then

H. + TI _ (*і + ІU 2 ~ 1 U2 ) x 2 + ІU2 (2. + ^ U 2) (2 ~ 1 y 2)

x, x 2 + / y, x 2 - ІХ x y 2 - І 2 y x in 2 (x x x 2 + y y) + / (x, 2 + x 2 y])

2 2 x 2 + in 2

1 e.

І (p.

• - 1U E "(1 FG) - I.OSO (((P-SR 1) + І- (R-,)] >2 >2
• 5. The construction of a whole positive degree of integrated number is better to produce if the number is recorded in an indicative or trigonometric forms.

Indeed, if g \u003d 7 then

\u003d (ge,) \u003d g P E T \u003d g " (CO8. pSR + ITT GKR).

Formula g " \u003d G P (Cosn (P + IS N (P) called the Moorem Formula.

6. Removing the root p-th degree from a complex number is defined as an operation, refurbished p, p-1,2,3, ... i.e. Comprehensive number \u003d. in [G. called root p-degree from a complex number

r if g. = g. . From this definition it follows that mr. ", but g. \u003d l / g. (P-PSR X, but cP ^ -SR / Pthat follows from the Moorem formula recorded for the number \u003d g / * + ІІpp (P).

As noted above, the integrated number argument is not defined unequivocally, but with accuracy to the term, multiple 2 g. therefore \u003d (P + 2PK , and the argument of the number G, depending on to, Denote (R K. and B.

dem calculate the formula (R K. \u003d - +. It is clear that there is p com-

plexian numbers p-I degree of which is equal to number 2. These numbers have one

and the same module equal y [g, And the arguments of these numbers are obtained at to = 0, 1, p - 1. Thus, in trigonometric form root and degree Calculate by the formula:

(P + 2KP . . cf + 2kp

, to = 0, 1, 77-1,

. (P + 2ktg

and in an indicative form - by the formula l [g - y [ge p

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

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

• (1 - / l / 2) 3 (z + /) \u003d (1 - CL / 2 / + 6/2 - 2 l / 2 /? 3) (3 + /) \u003d
• (1 - zl / 2 / - 6 + 2l / 2 / dz + /) \u003d (- 5 - l / 2 / dz + /) \u003d

15-CL / 2 / -5 / L / 2/2 \u003d -15 - CL / 2 / -5 / + l / 2 \u003d (-15 + l / 2) - (5 + evil / 2) /;

Example 49. Evaluate the number r \u003d UZ - / in the fifth degree.

Decision. We get the trigonometric form of the record of the number of

R \u003d. l / 3 + 1 \u003d 2, C08 (P --, 5IІ7. (R =

• (1 - 2 / x2 + /)
• (z-,)

O - 2.-x2 + o

• 12+ 4/-9/
• 2 +і - 4/ - 2/ 2 2 - 3/ + 2 4 - 3/ 3 + і
• (h-o "(zh

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

• (3-І) 'z + /
• 9 + 1 z_ ±.
• 5 2 1 "

From here about- -, but r \u003d 2.

Moorov we get: І -2.

/ ^ _ 7g ,. ? G.

• -Sh-- _Bіp -

• - / -

\u003d - (L / s + g) \u003d -2.

Example 50. Find all values

Decision, r \u003d 2, and cf. Find from equation soy (p \u003d -, zt--.

This point is 1 - / d / s is in the fourth quarter, i.e. f \u003d.-. Then

• 1 - 2
• ( ( UG L.

The values \u200b\u200bof the root find from expressing

V1 - / l / s \u003d l / 2

• - + 2a: / g --- b 2 kk
• 3 . . 3

C08--1- І 81p-

For to - 0 We have 2 0 \u003d l / 2

You can find the root values \u200b\u200bof 2, representing the number in the show

- * K / 3 + 2 cL

For to \u003d 1 We have another root value:

• 7g. 7g _
• --- b27g --- b2; g
• 3. . Z.

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

• --N -

cO? - 7g + / 5sh - I "

l / 3__t_

unformed form. As r \u003d.2, A. cf. \u003d, then r \u003d 2e 3, and in [G. = u / 2E. 2

Actions on complex numbers recorded in algebraic form

Algebraic form of a complex number z \u003d(a., B.). Names algebraic appearance

z. = a. + bI.

Arithmetic operations on complex numbers z. 1 \u003d A. 1 + B. 1 i.and z. 2 \u003d A. 2 + B. 2 i.recorded in algebraic form are carried out as follows.

1. Amount (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 the addition of polynomials with bringing such members.

2. Production 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 made according to the usual rule of multiplication of polynomials, taking into account the fact that i. 2 = 1.

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

, (z. 2 ≠ 0),

those. The division is carried out by multiplying the divide and divider by the number associated with the divider.

The construction of complex numbers is determined as follows:

Easy to show that

Examples.

1. Find the amount of integrated 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 a 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 \u003d.7+22i.

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

z \u003d. .

4. Solve Equation: x. and y. Î R..

(2x + Y.) + (x + Y.)i \u003d.2 + 3i.

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

from x \u003d.–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 number the opposite z.=3-I..

Complex numbers in trigonometric form

Complex plane It is called a plane with Cartesian coordinates ( x, Y.), if every point with coordinates ( a, B.) put in line with a complex number z \u003d a + bi. In this case, the abscissa axis is called valid axis, and the ordinate axis - imaginary. Then every integrated number a + BIgeometrically depicted on the plane as a point A (A, B) or vector.

Therefore, the position of the point BUT (and, therefore, a complex number z.) You can set the length of the vector | | \u003d. r. and angle j.Educated vector | | With a positive direction of a valid axis. Vector length is called the complex number moduleand denotes | z | \u003d R, and corner j.called the argument of a complex number And denotes j \u003d arg z.

Clearly | z.| ³ 0 and | z | = 0 Û z \u003d.0.

From fig. 2 It can be seen that.

The argument of a complex number is determined ambiguously, and up to 2 pK, K.Î Z..

From fig. 2 It is also seen that if z \u003d a + bi and j \u003d arg z,that

cos. j \u003d.sin. j \u003d., TG. j \u003d.

If a zîR.and z\u003e0, T. Arg Z \u003d.0 +2pK.;

if a z îR.and z.< 0, T. arg z \u003d p +2pK.;

if a z \u003d.0, ARG Z.not determined.

The main value of the argument is determined on the segment 0 £ Arg Z.£ 2. p,

or -p.£ arg z £ p.

Examples:

1. Find the integrated number module z. 1 = 4 – 3i.and z. 2 = –2–2i.

2. Determine on the complex plane of the region specified by the conditions:

1) | z | \u003d. 5; 2) | z.| £ 6; 3) | z. – (2+i.) | £ 3; 4) 6 £ | z. – i.| £ 7.

Solutions and answers:

1) | z.| \u003d 5 û û - the circle equation with radius 5 and centered at the beginning of the coordinates.

2) Circle with a radius 6 centered at the beginning of the coordinates.

3) Circle Radius 3 centered at point z 0 = 2 + i..

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

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

1) ; but = 1, b. = Þ ,

Þ J 1 \u003d .

2) z. 2 = –2 – 2i.; a \u003d.–2, b \u003d.-2 þ ,

.

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

In this way: z. 1 = .

2) , r. 2 = 1, J 2 \u003d, .

3) , r. 3 \u003d 1, J 3 \u003d .

4) , r. 4 \u003d 1, J 4 \u003d, .

Lecture

Trigonometric form of a complex number

Plan

1.Gometric image of complex numbers.

2.Trigonometric recording of complex numbers.

3. Aims on complex numbers in trigonometric form.

Geometric image of complex numbers.

a) integrated numbers depict the points of the plane at the following rule: a. + bI = M. ( a. ; b. ) (Fig.1).

Picture 1

b) a complex number can be depicted with a vector that has the beginning at the pointABOUT and the end at this point (Fig. 2).

Figure 2.

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

Figure 3.

Trigonometric recording of complex numbers.

Complex numberz. = a. + bI You can specify with a radius - vector With coordinates( a. ; b. ) (Fig.4).

Figure 4.

Definition . Length vector depicting a complex numberz. , is called the module of this number and is indicated orr. .

For any integrated numberz. His moduler. = | z. | determined uniquely by the formula .

Definition . The magnitude of the angle between the positive direction of the actual axis and the vector depicting a complex number is called the argument of this integrated number and is indicatedBUT rg. z. orφ .

An argument of a complex numberz. = 0 not determined. An argument of a complex numberz. ≠ 0 - the value is multi-valued and determined with accuracy to the foundation2πK (K \u003d 0; - 1; 1; - 2; 2; ...): Arg z. = arg. z. + 2πk wherearg. z. - The main value of the argument concluded in the interval(-π; π] , i.e-π < arg. z. ≤ π (Sometimes the initial value of the argument take the value belonging to the gap .

This formula forr. =1 often refer to the formula Moava:

(cos φ + i sin φ) n. \u003d cos (nφ) + i sin (nφ), n  n .

Example 11. Calculate(1 + i. ) 100 .

We write a complex number1 + i. in trigonometric form.

a \u003d 1, b \u003d 1 .

cos φ \u003d. , sin φ \u003d , φ = .

(1 + i) 100 = [ (COS. + I SIN )] 100 = ( ) 100 (COS. · 100 + i sin · 100) \u003d \u003d 2 50 (COS 25π + I SIN 25π) \u003d 2 50 (COS π + i sin π) \u003d - 2 50 .

4) Extraction square root From the integrated number.

When removing a square root from a complex numbera. + bI We have two cases:

if ab. \u003e O. T. ;

3.1. Polar coordinates

The plane often applies polar coordinate system . It is defined if a point O, called pole, and outgoing from the pole ray (for us this is the axis OX) - Polar axis. Position point M is fixed by two numbers: radius (or radius-vector) and an angle φ between the polar axis and the vector.The angle φ is called polar corner; It is measured in radians and is counted from the polar axis counterclockwise.

The position of the point in the polar coordinate system is set by an ordered pair of numbers (R; φ). At Pole R \u003d 0,a φ is not defined. For all other points r\u003e 0, A φ is determined with an accuracy of the term of multiple 2π. At the same time, pairs of numbers (R; φ) and (R 1; φ 1) compares the same point if.

For a rectangular coordinate system xoy. Cartesian coordinates of the point are easily expressed through its polar coordinates as follows:

3.2. Geometric interpretation of a complex number

Consider on the decartian plane rectangular coordinate system xoy..

Any integrated number z \u003d (a, b) is put in accordance with the plane with coordinates ( x, Y.), where the coordinate X \u003d A, i.e. The actual part of the complex number, and the coordinate Y \u003d Bi - the imaginary part.

The plane whose points are complex numbers - a complex plane.

In the figure in a complex number z \u003d (a, b)corresponds to the point M (X, Y).

The task.Picture complex numbers on the coordinate plane:

3.3. Trigonometric form of a complex number

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

Record integrated number - trigonometric form of a complex number.

The number R is called module integrated number z. and is denoted. The module is a non-negative real number. For .

The module is zero if and only when z \u003d 0, i.e. a \u003d b \u003d 0.

The number φ is called argument Z. and denotes. The z argument is defined ambiguously, as well as the polar angle in the polar coordinate system, namely, with an accuracy of the term of multiple 2π.

Then we accept:, where φ - the 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.

Decision. 1) We consider the module :;

2) We are looking for φ: ;

3) Trigonometric shape:

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

It is enough to substitute the values trigonometric functions and convert the expression:

Example 3.Find a module and an integrated number argument;

1) ;

2); φ - at 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 - with the help of uncomplicated trigonometric transformations You can show that when multiplying the modules of numbers are multiplied, and the arguments are folded: ;