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 \u003d 0,and φ is undefined. For all other points r\u003e 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 \u003d (a, b) is assigned a point on the plane with coordinates ( x, y), where coordinate x \u003d a, i.e. the real part of the complex number, and the coordinate y \u003d bi is the imaginary part.

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

In the figure, the complex number z \u003d (a, b)match point M (x, y).

The task.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 \u003d 0, i.e. a \u003d b \u003d 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.

Decision. 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 \u200b\u200bof 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: ;

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 pointABOUT 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 uniquely determined 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 denotedAND rg z orφ .

Complex number argumentz = 0 not determined. Complex number argumentz ≠ 0 is a multivalued quantity and is determined up to the term2πk (k \u003d 0; - 1; 1; - 2; 2; ...): Arg z = arg z + 2πk wherearg z - the main value of the argument enclosed in the interval(-π; π] , i.e-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 \u003d cos (nφ) + i sin (nφ), n  N .

Example 11. Calculate(1 + i ) 100 .

Let's 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) Extracting the square root of a complex number.

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

if ab \u003e about then ;

Actions on complex numbers written in algebraic form

The algebraic form of the complex number z \u003d(a, b). is called an algebraic expression of the form

z = a + bi.

Arithmetic operations on complex numbers z 1 \u003d a 1 + b 1 iand z 2 \u003d a 2 + b 2 iwritten in algebraic form are carried out as follows.

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

3. 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.

Exponentiation of complex numbers is defined as follows:

It is easy to show that

Examples of.

1. Find the sum of complex numbers z 1 = 2 – iand 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 \u003d7+22i.

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

z \u003d .

4. Solve the equation:, x and y Î R.

(2x + y) + (x + y)i \u003d2 + 3i.

Due to the equality of the complex numbers, we have:

from where 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 reciprocal of the 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 assigned a complex number z \u003d a + bi... In this case, the abscissa axis is called real axis, and the ordinate axis is imaginary... Then each complex number a + biis geometrically depicted on a plane as a point A (a, b) or vector.

Therefore, the position of the point AND (and, therefore, the complex number z) can be specified by the length of the vector | | \u003d r and angle jformed by vector | | with a positive direction of the real axis. The length of the vector is called modulus of a complex numberand denoted by | z | \u003d rand the angle jcalled complex number argument and denoted j \u003d arg z.

It is clear that | z| ³ 0 and | z | = 0 Û z \u003d0.

Fig. 2 shows that.

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

Fig. 2 it is also seen that if z \u003d a + bi and j \u003d arg z,then

cos j \u003d, sin j \u003d, tg j \u003d.

If a zÎRand z\u003e0, then arg z \u003d0 +2pk;

if a z ÎRand z< 0, then arg z \u003d p +2pk;

if a z \u003d0, arg znot 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 modulus of complex numbers z 1 = 4 – 3iand z 2 = –2–2i.

2. Determine on the complex plane the areas 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 is the equation of a circle with radius 5 and center at the origin.

2) A circle of radius 6 centered at the origin.

3) A circle of radius 3 centered on a point z 0 = 2 + i.

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

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

1) ; and = 1, b = Þ ,

Þ j 1 \u003d .

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

.

Note: Use the complex plane when defining the main argument.

Thus: 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, .

To determine the position of a point on a plane, you can use polar coordinates [r, (p)where r 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 \u003d r cos cf, y \u003d r sin (p,

we get the trigonometric form of writing a complex number

z - r (sin (p + i sin

where r

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

l X . at

formulas cos (p - -, sin ^ 9 \u003d - 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 \u003d -1 + l / Z /.

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

= yj 1 + 3 = 2 . Angle wed find from the relations cos (p = -, sin (p \u003d -. Then

get cos (p \u003d -, suup

u / z g ~

• - -. Obviously, the point z \u003d -1 + V3- / is
• 2 to3

in the second quarter: (R \u003d 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 for a complex number.

We have r \u003d r (ω ^ (p + i?, n (p) \u003d r,

Actions on complex numbers

• 1. The sum of two complex numbers r, \u003d X] + y x / u r 2 - x 2 + y 2 / is determined according to the formula r! +2 2 \u003d (x, + ^ 2) + (^ 1 + ^ 2) 'g
• 2. The operation of subtracting complex numbers is defined as the inverse of addition. Complex number r \u003d rx - r 2, if a z 2 + z \u003d z x,

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

• 3. Product of two complex numbers r x \u003d x, + y, -z 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 2 Y1 2 -1 + x Y2 " * + Have1 Have2 " ^ =

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

In particular, yr \u003d (x + y-z) (x-y /) \u003d 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 \u003d ) Г 2 е\u003e \u003d Г] Г 2 cOs ((P + cf 2) + isin
• 4. Division of complex numbers is defined as the inverse operation

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

if a r x -1 2 ? 2 . Then

X + 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 \u003d rt 1 then

\u003d (re,) \u003d r n e t \u003d r " (co8 psr + іьт гкр).

Formula g " \u003d rn (cosn (p + is n (p) called the Moivre formula.

6. Extracting the root p-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 \u003d y [g called root p-th degree of a complex number

d if r = r x ... It follows from this definition that g - g ", and r x \u003d l / g. (p-psr x, and cf-cp / n, which follows from the Moivre formula written for the number \u003d r / * + ilipn (p).

As noted above, the argument of a complex number is not uniquely determined, but up to a term multiple of 2 g. therefore \u003d (p + 2pc , and the argument of the number r, depending on to, denote (p to and boo

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

plex numbers, p-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, p - 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 + /) \u003d (1 - Zl / 2 / + 6/2 - 2 l / 2 /? 3) (3 + /) \u003d
• (1 - Zl / 2 / - 6 + 2l / 2 / DZ + /) \u003d (- 5 - l / 2 / DZ + /) \u003d

15-Zl / 2 / -5 / -l / 2/2 \u003d -15 - Zl / 2 / -5 / + l / 2 \u003d (-15 + l / 2) - (5 + Zl / 2) /;

Example 49. Construct the number r \u003d Uz - / to the fifth power.

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

Г \u003d l / 3 + 1 \u003d 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 about- -, and r \u003d 2

Moivre we get: i -2

/ ^ _ 7Г,. ? D

• -US-- ІБІП -

• --B / -

\u003d - (l / Z + z) \u003d -2.

Example 50. Find all values

Solution, r \u003d 2, and wed find from the equation soy (p \u003d -, zt -.

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

• 1 - 2
• ( ( UG L

We find the values \u200b\u200bof the root from the expression

V1 - / l / s \u003d l / 2

• - + 2A: / g --- b 2 kk
• 3 . . 3

С08-1- і 81П-

When to - 0 we have 2 0 \u003d l / 2

You can find the values \u200b\u200bof the root of the number 2 by presenting the number in the display

- * K / 3 + 2 cl

When to \u003d 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 \u003d2, a wed \u003d, then r \u003d 2e 3, and y [g = y / 2e 2