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°).
3.1. Polar coordinates
Often used on a plane polar coordinate system . It is defined if a point O is given, called pole, and the ray emanating from the pole (for us this is the axis Ox) – polar axis. The position of point M is fixed by two numbers: radius (or radius vector) and angle φ between the polar axis and 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 given by an ordered pair of numbers (r; φ). At the Pole r = 0, and φ is not defined. For all other points r > 0, and φ is defined up to a term that is 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 complex number
Let us consider a Cartesian rectangular coordinate system on the plane xOy.
Any complex number z=(a, b) is associated with a point on the plane with coordinates ( x, y), Where coordinate x = a, i.e. the real part of the complex number, and the coordinate y = bi is the imaginary part.
A plane whose points are complex numbers is a complex plane.
In the figure, a complex number z = (a, b) corresponds to point M(x, y).
Exercise.Draw complex numbers on the coordinate plane:
3.3. Trigonometric form of a complex number
A complex number on the plane has the coordinates of a point M(x;y). Wherein:
Writing a complex number 
- trigonometric form of a complex number.
The number r is called module
complex number z and is designated . Modulus is a non-negative real number. For 
.
The modulus is zero if and only if z = 0, i.e. a = b = 0.
The number φ is called argument z and is designated. The argument z is defined ambiguously, like the polar angle in the polar coordinate system, namely up to a term that is a multiple of 2π.
Then we accept: , where φ is the smallest value of the argument. It's obvious that
.
When studying the topic in more depth, an auxiliary argument φ* is introduced, such that
Example 1. Find the trigonometric form of a complex number.
Solution. 1) consider the module: ;
2) looking for φ: 
;
3) trigonometric form: 
Example 2. Find the algebraic form of a complex number 
.
Here it is enough to substitute the values of 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. Operations with complex numbers in trigonometric form
· Addition and subtraction It’s more convenient to do with complex numbers in algebraic form:
· Multiplication– using simple trigonometric transformations it can be shown that When multiplying, the modules of numbers are multiplied, and the arguments are added: ;
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 of a geometric progression with the 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 is the integer part of the number 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 this equation, the number must be the nth root of the 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. From a geometric point of view, the point corresponding to w1 can be obtained using a homothety with a center at the origin and a 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.
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LectureTrigonometric 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 
;