Algebraic and trigonometric form of a complex number. Lecture on the topic: "Trigonometric form of a complex number"

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 as a vector that starts at the pointABOUT and end at a given point (Fig. 2).

Figure 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 the radius - vector with coordinates( a ; b ) (Fig. 4).

Figure 4

Definition . Vector length representing the 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 value 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 multi-valued 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 value belonging to the interval is taken as the main value of the argument .

This formula forr =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) Extracting the square root of a complex number.

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

Ifb > about , That ;

Actions on complex numbers written in algebraic form

The algebraic form of the 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 of addition of polynomials with reduction of similar members.

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 for multiplication of 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 of the divisor.

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

z= .

4. Solve the equation:, x And y Î R.

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

By virtue of 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 is called a plane with Cartesian coordinates ( x, y), if each point with coordinates ( a, b) is assigned a complex number z = a + bi. In this case, the abscissa axis is called real axis, and the y-axis is imaginary. Then every complex number a+bi geometrically represented on a plane as a point A (a, b) or vector .

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

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

From fig. 2 shows that .

The argument of a complex number is defined ambiguously, and up to 2 pk,kÎ Z.

From fig. 2 also shows that if z=a+bi And j=argz, That

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

If zОR And z > 0 then argz = 0 +2pk;

If z ОR And z< 0 then argz = p + 2pk;

If z= 0,argz indefined.

The main value of the argument is determined on the interval 0 £argz£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. Determine on the complex plane the areas specified 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 is the equation of a circle with radius 5 and centered at the origin.

2) Circle with radius 6 centered at the origin.

3) Circle with radius 3 centered at a point z0 = 2 + i.

4) 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) ; A = 1, b = Þ ,

Þ j 1 = .

2) z 2 = –2 – 2i; a =–2, b=-2 Þ ,

.

Note: When defining 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, j4 = , .

To determine the position of a point on a plane, you can use polar coordinates [g, (p), Where G is the distance of the point from the origin, and (R- the angle that the radius makes - 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 relationship between Cartesian and polar coordinates: x \u003d r cos cf, y \u003d r sin (p,

we obtain the trigonometric form of the 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 the 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 / .

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

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

we 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 k.. cos-h; sin

into formula (1) found 27G L

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 designate

argument value enclosed within (p 0 %2 Then

A) ^ r = + 2kk.

Using the well-known Euler formula e, we obtain the exponential form of the complex number.

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

Operations 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)' g
• 2. The operation of subtraction of complex numbers is defined as the operation inverse to addition. Complex number g \u003d g x - g 2, If g 2 + g \u003d 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‘ 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 " * + At1 At2 " ^ =

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

In particular, y-y\u003d (x + y-g) (x-y /) \u003d x 2 + y 2.

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

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

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

If r x -1 2 ? 2 . Then

X + Ті _ (*і + ІU 2 ~ 1 U2 ) x 2 + ІУ2 ( 2 + ^Y 2)( 2 ~ 1 Y 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.sOї ((P - cf 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 forms.

Indeed, if z = ge 1 then

=(ge,) = r p e t = G"(co8 psr + іt gcr).

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

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

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

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 + 2pc, and the argument of the number r, depending on To, denote (p to and boo

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

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

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

(p + 2kp . . cf + 2kp

, To = 0, 1, 77-1,

.(p+2ktg

and in exponential form - according to the formula l[r - y[ge 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 + /)=(- 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 \u003d Uz - / to the fifth power.

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

G = l/3 + 1 =2, CO8 (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
• (s-o "(s-o

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

• (3-i) 'з+/
• 9 + 1 s_±.
• 5 2 1 "

From here O--, A r = 2

Moivre we get: i-2

/ ^ _ 7r, . ?G

• -US-- IBIP -

• --b/-

\u003d - (l / W + g) \u003d -2.

Example 50 Find all values

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

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

• 1 - 2
• ( ( UG L

The root values ​​are found from the expression

V1 - /l/s = l/2

• --+ 2A:/g ---b 2 kk
• 3 . . 3

С08--1- and 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 class

At To= 1 we have one more root value:

• 7G. 7G_
• ---b27g ---b2;g
• 3 . . h

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

• --N-

with? - 7G + / 5Sh - I "

l/3__t_

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