Operations on complex numbers written in algebraic form
Algebraic form of a 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 for adding polynomials with 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 carried out according to the usual rule for multiplying 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 number of the divisor.
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 the quotient z from division z 1 = 3 – 2na z 2 = 3 – i.
z = .
4. Solve the equation: , x And y Î R.
(2x+y) + (x+y)i = 2 + 3i.
Due to 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 called a plane with Cartesian coordinates ( x, y), if each point with coordinates ( a, b) is associated with a complex number z = a + bi. In this case, the abscissa axis is called real axis, and the ordinate axis is imaginary. Then every complex number a+bi geometrically depicted on a plane as a point A (a, b) or vector.
Therefore, the position of the point A(and, therefore, a complex number z) can be specified 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 the vector is called modulus of a complex number and is denoted by | z |=r, and the angle j called complex number argument and is designated j = arg z.
It is clear that | z| ³ 0 and | z | = 0 Û z = 0.
From Fig. 2 it is clear that .
The argument of a complex number is determined ambiguously, but with an accuracy of 2 pk,kÎ Z.
From Fig. 2 it is also clear that if z=a+bi And j=arg z, That
cos j =,sin j =, tg j = .
If zÎR And z> 0,then arg z = 0 +2pk;
If z ОR And z< 0,then arg z = p + 2pk;
If z = 0,arg z indefined.
The main value of the argument is determined on the interval 0 £ arg z£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. Define areas on the complex plane defined 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 - equation of a circle with radius 5 and center at the origin.
2) A circle with radius 6 with center at the origin.
3) Circle with radius 3 with center at point z 0 = 2 + i.
4) A ring bounded by circles with radii 6 and 7 with a center at a point z 0 = i.
3. Find the modulus and argument of the numbers: 1) ; 2) .
1) ; A = 1, b = Þ 
,
Þ j 1 = 
.
2) z 2 = –2 – 2i; a =–2, b =-2 Þ 
,
.
Hint: When determining 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, j 4 = , 
.
In this section we will talk more about the trigonometric form of a complex number. The demonstrative form is much less common in practical tasks. I recommend downloading and printing if possible. trigonometric tables, methodological material can be found on the page Mathematical formulas and tables. You can't go far without tables.
Any complex number (except zero) can be written in trigonometric form:
Where is it modulus of a complex number, A - complex number argument.
Let us represent the number on the complex plane. For definiteness and simplicity of explanation, we will place it in the first coordinate quadrant, i.e. we believe that:
Modulus of a complex number is the distance from the origin to the corresponding point in the complex plane. Simply put, module is the length radius vector, which is indicated in red in the drawing.
The modulus of a complex number is usually denoted by: or
Using the Pythagorean theorem, it is easy to derive a formula for finding the modulus of a complex number: . This formula is correct for any meanings "a" and "be".
Note : The modulus of a complex number is a generalization of the concept modulus of a real number, as the distance from a point to the origin.
Argument of a complex number called corner between positive semi-axis the real axis and the radius vector drawn from the origin to the corresponding point. The argument is not defined for singular:.
The principle under consideration is actually similar to polar coordinates, where the polar radius and polar angle uniquely define a point.
The argument of a complex number is standardly denoted: or
From geometric considerations, we obtain the following formula for finding the argument:
. Attention! This formula only works in the right half-plane! If the complex number is not located in the 1st or 4th coordinate quadrant, then the formula will be slightly different. We will also analyze these cases.
But first, let's look at the simplest examples when complex numbers are located on coordinate axes.
Example 7
Represent complex numbers in trigonometric form: ,,,. Let's make the drawing:
In fact, the task is oral. For clarity, I will rewrite the trigonometric form of a complex number:
Let us remember firmly, the module – length(which is always non-negative), argument – corner
1) Let's represent the number in trigonometric form. Let's find its modulus and argument. It's obvious that. Formal calculation using the formula:. It is obvious that (the number lies directly on the real positive semi-axis). Thus, the number in trigonometric form:.
The reverse check action is as clear as day:
2) Let us represent the number in trigonometric form. Let's find its modulus and argument. It's obvious that. Formal calculation using the formula:. Obviously (or 90 degrees). In the drawing, the corner is indicated in red. So the number in trigonometric form is: 
.
Using , it’s easy to get back the algebraic form of the number (at the same time performing a check):
3) Let us represent the number in trigonometric form. Let's find its module and
argument. It's obvious that . Formal calculation using the formula:
Obviously (or 180 degrees). In the drawing the corner is indicated in blue. Thus, the number in trigonometric form:.
Examination:
4) And the fourth interesting case. It's obvious that. Formal calculation using the formula:.
The argument can be written in two ways: First way: (270 degrees), and, accordingly: 
. Examination:
However, the following rule is more standard: If the angle is greater than 180 degrees, then it is written with a minus sign and the opposite orientation (“scrolling”) of the angle: (minus 90 degrees), in the drawing the angle is marked in green. It's easy to notice
which is the same angle.
Thus, the entry takes the form: 
Attention! In no case should you use the parity of the cosine, the oddness of the sine, and further “simplify” the notation:
By the way, it’s useful to remember appearance and properties of trigonometric and inverse trigonometric functions, reference materials are located in the last paragraphs of the page Graphs and properties of basic elementary functions. And complex numbers will be learned much easier!
In the design of the simplest examples, this is how you should write it: : "it is obvious that the modulus is... it is obvious that the argument is...". This is really obvious and easy to solve verbally.
Let's move on to consider more common cases. There are no problems with the module; you should always use the formula. But the formulas for finding the argument will be different, it depends on which coordinate quarter the number lies in. In this case, three options are possible (it is useful to rewrite them):
1) If (1st and 4th coordinate quarters, or right half-plane), then the argument must be found using the formula.
2) If (2nd coordinate quarter), then the argument must be found using the formula 
.
3) If (3rd coordinate quarter), then the argument must be found using the formula 
.
Example 8
Represent complex numbers in trigonometric form: ,,,.
Since there are ready-made formulas, it is not necessary to complete the drawing. But there is one point: when you are asked to represent a number in trigonometric form, then It’s better to do the drawing anyway. The fact is that a solution without a drawing is often rejected by teachers; the absence of a drawing is a serious reason for a minus and failure.
We present the numbers in complex form, and the first and third numbers will be for independent solution.
Let's represent the number in trigonometric form. Let's find its modulus and argument.
Since (case 2), then
– this is where you need to take advantage of the oddity of the arctangent. Unfortunately, the table does not contain the value , so in such cases the argument has to be left in a cumbersome form: – numbers in trigonometric form.
Let's represent the number in trigonometric form. Let's find its modulus and argument.
Since (case 1), then (minus 60 degrees).
Thus:
– a number in trigonometric form.
But here, as already noted, are the disadvantages don't touch.
In addition to the fun graphical verification method, there is also an analytical verification, which was already carried out in Example 7. We use table of values of trigonometric functions, while taking into account that the angle is exactly the table angle (or 300 degrees): – numbers in the original algebraic form.
Present the numbers in trigonometric form yourself. A short solution and answer at the end of the lesson.
At the end of the section, briefly about the exponential form of a complex number.
Any complex number (except zero) can be written in exponential form:
Where is the modulus of a complex number, and is the argument of the complex number.
What do you need to do to represent a complex number in exponential form? Almost the same: execute a drawing, find a module and an argument. And write the number in the form .
For example, for the number in the previous example we have found the module and argument:,. Then this number will be written in exponential form as follows:.
The number in exponential form will look like this:
Number 
- So:
The only advice is don't touch the indicator exponents, there is no need to rearrange factors, open parentheses, etc. A complex number is written in exponential form strictly according to form.
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’s 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 a given equation, the number must be nth root degrees from 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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Mathematics of trigonometric substitution and testing the effectiveness of the developed teaching methodology. Stages of work: 1. Development elective course on the topic: “Use of trigonometric substitution for solving algebraic problems” with students in classes with advanced mathematics. 2. Conducting the developed elective course. 3. Carrying out a diagnostic test...
Cognitive tasks are intended only to complement existing teaching aids and must be in an appropriate combination with all traditional means and elements of the educational process. The difference between educational tasks in teaching the humanities and the exact sciences, from mathematical problems The only problem is that historical problems lack formulas, strict algorithms, etc., which complicates their solution. ...
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°).
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 
;