FEDERAL AGENCY FOR EDUCATION
STATE EDUCATIONAL INSTITUTION
HIGHER PROFESSIONAL EDUCATION
"VORONEZH STATE PEDAGOGICAL UNIVERSITY"
DEPARTMENT OF AGLEBRA AND GEOMETRY
Complex numbers
(selected tasks)
GRADUATE QUALIFYING WORK
specialty 050201.65 mathematics
(with additional specialty 050202.65 computer science)
Completed by: 5th year student
physical and mathematical
faculty
Scientific adviser:
VORONEZH – 2008
1. Introduction……………………………………………………...…………..…
2. Complex numbers (selected problems)
2.1. Complex numbers in algebraic form….……...……….….
2.2. Geometric interpretation of complex numbers…………..…
2.3. Trigonometric form of complex numbers
2.4. Application of the theory of complex numbers to the solution of equations of the 3rd and 4th degree……………..……………………………………………………………
2.5. Complex numbers and parameters…………………………………...….
3. Conclusion……………………………………………………………………………….
4. List of references………………………….………………………......
1. Introduction
In the mathematics program school course number theory is introduced using examples of sets of natural numbers, integers, rationals, irrationals, i.e. on the set of real numbers, the images of which fill the entire number line. But already in the 8th grade there is not enough supply of real numbers when solving quadratic equations with a negative discriminant. Therefore, it was necessary to replenish the stock of real numbers with the help of complex numbers, for which the square root of negative number has the meaning.
Choosing the topic “Complex Numbers” as my graduation topic qualifying work, is that the concept of a complex number expands students’ knowledge about number systems, about solving a wide class of problems of both algebraic and geometric content, about solving algebraic equations any degree and about solving problems with parameters.
This thesis examines the solution to 82 problems.
The first part of the main section “Complex numbers” contains solutions to problems with complex numbers in algebraic form, the operations of addition, subtraction, multiplication, division, the conjugation operation for complex numbers in algebraic form, the power of an imaginary unit, the modulus of a complex number are defined, and the extraction rule is also stated square root from a complex number.
In the second part, problems on the geometric interpretation of complex numbers in the form of points or vectors of the complex plane are solved.
The third part examines operations on complex numbers in trigonometric form. The formulas used are: Moivre and extracting the root of a complex number.
The fourth part is devoted to solving equations of the 3rd and 4th degrees.
When solving problems in the last part, “Complex numbers and parameters,” the information given in the previous parts is used and consolidated. A series of problems in the chapter are devoted to determining families of lines in the complex plane defined by equations (inequalities) with a parameter. In part of the exercises you need to solve equations with a parameter (over field C). There are tasks where a complex variable simultaneously satisfies a number of conditions. A special feature of solving problems in this section is the reduction of many of them to solving equations (inequalities, systems) of the second degree, irrational, trigonometric with a parameter.
A feature of the presentation of the material in each part is the initial input theoretical foundations, and subsequently their practical application in solving problems.
At the end thesis a list of used literature is presented. Most of them present theoretical material in sufficient detail and in an accessible manner, consider solutions to some problems, and give practical tasks For independent decision. Special attention I would like to refer to such sources as:
1. Gordienko N.A., Belyaeva E.S., Firstov V.E., Serebryakova I.V. Complex numbers and their applications: Textbook. . Material teaching aid presented in the form of lectures and practical exercises.
2. Shklyarsky D.O., Chentsov N.N., Yaglom I.M. Selected problems and theorems of elementary mathematics. Arithmetic and algebra. The book contains 320 problems related to algebra, arithmetic and number theory. These tasks differ significantly in nature from standard school tasks.
2. Complex numbers (selected problems)
2.1. Complex numbers in algebraic form
The solution of many problems in mathematics and physics comes down to solving algebraic equations, i.e. equations of the form
,where a0, a1, …, an are real numbers. Therefore, the study of algebraic equations is one of critical issues in mathematics. For example, the quadratic equation with negative discriminant. The simplest such equation is the equation
.In order for this equation to have a solution, it is necessary to expand the set of real numbers by adding to it the root of the equation
.Let us denote this root by
. Thus, by definition, or,hence,
.called the imaginary unit. With its help and with the help of a pair of real numbers, an expression of the form is compiled.
The resulting expression was called complex numbers because they contained both real and imaginary parts.
So, complex numbers are expressions of the form, and are real numbers, and is a certain symbol that satisfies the condition . The number is called the real part of a complex number, and the number is its imaginary part. The symbols , are used to denote them.
Complex numbers of the form , and are real numbers, and is a certain symbol that satisfies the condition . The number is called the real part of a complex number, and the number is its imaginary part. The symbols , are used to denote them.
are real numbers and, therefore, the set of complex numbers contains the set of real numbers.are called purely imaginary. Two complex numbers of the form and are said to be equal if their real and imaginary parts are equal, i.e. if equalities , .
Algebraic notation of complex numbers allows operations on them according to the usual rules of algebra. To solve problems with complex numbers, you need to understand the basic definitions. The main goal of this review article is to explain what complex numbers are and present methods for solving basic problems with complex numbers. So, a complex number will be called a number of the form z = a + bi , Where a, b - real numbers, which are called the real and imaginary parts of a complex number, respectively, and denote.
a = Re(z), b=Im(z) i called the imaginary unit. i 2 = -1 . In particular, any real number can be considered complex: a = a + 0i , where a is real. If a = 0 And b ≠ 0
, then the number is usually called purely imaginary.
Now let's introduce operations on complex numbers. Consider two complex numbers a = 0 z 1 = a 1 + b 1 i.
z 2 = a 2 + b 2 i To solve problems with complex numbers, you need to understand the basic definitions. The main goal of this review article is to explain what complex numbers are and present methods for solving basic problems with complex numbers. So, a complex number will be called a number of the form.
Let's consider The set of complex numbers extends the set of real numbers, which in turn extends the set rational numbers integers, Z - integers, Q - rational, R - real, C - complex. 
Representation of complex numbers
Algebraic notation.
Consider a complex number To solve problems with complex numbers, you need to understand the basic definitions. The main goal of this review article is to explain what complex numbers are and present methods for solving basic problems with complex numbers. So, a complex number will be called a number of the form, this form of writing a complex number is called algebraic. We have already discussed this form of recording in detail in the previous section. The following visual drawing is used quite often 
Trigonometric form.
From the figure it can be seen that the number To solve problems with complex numbers, you need to understand the basic definitions. The main goal of this review article is to explain what complex numbers are and present methods for solving basic problems with complex numbers. So, a complex number will be called a number of the form can be written differently. It's obvious that a = rcos(φ), b = rsin(φ), r=|z|, hence z = rcos(φ) + rsin(φ)i, φ ∈ (-π; π)
is called the argument of a complex number. This representation of a complex number is called trigonometric form. The trigonometric form of notation is sometimes very convenient. For example, it is convenient to use it to raise a complex number to an integer power, namely, if z = rcos(φ) + rsin(φ)i, That z n = r n cos(nφ) + r n sin(nφ)i, this formula is called Moivre's formula.
Demonstrative form.
z 2 = a 2 + b 2 i z = rcos(φ) + rsin(φ)i- a complex number in trigonometric form, write it in another form z = r(cos(φ) + sin(φ)i) = re iφ, the last equality follows from Euler’s formula, so we get new uniform complex number notation: z = re iφ, which is called indicative. This form of notation is also very convenient for raising a complex number to a power: z n = r n e inφ, Here n not necessarily an integer, but can be an arbitrary real number. This form of notation is quite often used to solve problems.
Fundamental theorem of higher algebra
Let's imagine that we have a quadratic equation x 2 + x + 1 = 0. Obviously, the discriminant of this equation is negative and it has no real roots, but it turns out that this equation has two different complex roots. So, the fundamental theorem of higher algebra states that any polynomial of degree n has at least one complex root. It follows from this that any polynomial of degree n has exactly n complex roots, taking into account their multiplicity. This theorem is a very important result in mathematics and is widely used. A simple corollary to this theorem is that there are exactly n different roots degree n of unity.
Main types of tasks
This section will cover the main types simple tasks to complex numbers. Conventionally, problems involving complex numbers can be divided into the following categories.
- Performing simple arithmetic operations on complex numbers.
- Finding the roots of polynomials in complex numbers.
- Raising complex numbers to powers.
- Extracting roots from complex numbers.
- Using complex numbers to solve other problems.
Now let's consider general techniques solutions to these problems.
The simplest arithmetic operations with complex numbers are performed according to the rules described in the first section, but if complex numbers are presented in trigonometric or exponential forms, then in this case you can convert them into algebraic form and perform operations according to known rules.
Finding the roots of polynomials usually comes down to finding the roots of a quadratic equation. Suppose that we have a quadratic equation, if its discriminant is non-negative, then its roots will be real and can be found according to a well-known formula. If the discriminant is negative, that is, D = -1∙a 2 z = a + bi a is a certain number, then the discriminant can be represented as D = (ia) 2, hence √D = i|a|, and then you can use well-known formula for the roots of a quadratic equation.
Example. Let's go back to what was mentioned above. quadratic equation x 2 + x + 1 = 0 .
Discriminant - D = 1 - 4 ∙ 1 = -3 = -1(√3) 2 = (i√3) 2.
Now we can easily find the roots: 
Raising complex numbers to powers can be done in several ways. If you need to raise a complex number in algebraic form to a small power (2 or 3), then you can do this by direct multiplication, but if the power is larger (in problems it is often much larger), then you need to write this number in trigonometric or exponential forms and use already known methods.
Example. Consider z = 1 + i and raise it to the tenth power.
Let's write z in exponential form: z = √2 e iπ/4.
Then z 10 = (√2 e iπ/4) 10 = 32 e 10iπ/4.
Let's return to algebraic form: z 10 = -32i.
Extracting roots from complex numbers is the inverse operation of exponentiation and is therefore performed in a similar way. To extract roots, the exponential form of writing a number is often used.
Example. Let's find all roots of degree 3 of unity. To do this, we will find all the roots of the equation z 3 = 1, we will look for the roots in exponential form.
Let's substitute into the equation: r 3 e 3iφ = 1 or r 3 e 3iφ = e 0 .
Hence: r = 1, 3φ = 0 + 2πk, therefore φ = 2πk/3.
Different roots are obtained at φ = 0, 2π/3, 4π/3.
Therefore 1, e i2π/3, e i4π/3 are roots.
Or in algebraic form: 
The last type of problems includes a huge variety of problems and there are no general methods for solving them. Let's give a simple example of such a task:
Find the amount sin(x) + sin(2x) + sin(2x) + … + sin(nx).
Although the formulation of this problem does not involve complex numbers, it can be easily solved with their help. To solve it, the following representations are used: 
If we now substitute this representation into the sum, then the problem is reduced to summing the usual geometric progression.
Conclusion
Complex numbers are widely used in mathematics, this review article examined the basic operations on complex numbers, described several types of standard problems, and briefly described general methods their solutions, for a more detailed study of the capabilities of complex numbers, it is recommended to use specialized literature.
Literature
The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. For clarity, let’s solve the following problem:
Calculate \[ (z_1\cdot z_2)^(10),\] if \
First of all, let's pay attention to the fact that one number is presented in algebraic form, the other in trigonometric form. It needs to be simplified and brought to the following form
\[ z_2 = \frac(1)(4) (\cos\frac(\pi)(6)+i\sin\frac(\pi)(6)).\]
The expression \ says that first of all we do multiplication and raising to the 10th power using the Moivre formula. This formula is formulated for the trigonometric form of a complex number.
We get:
\[\begin(vmatrix) z_1 \end(vmatrix)=\sqrt ((-1)^2+(\sqrt 3)^2)=\sqrt 4=2\]
\[\varphi_1=\pi+\arctan\frac(\sqrt 3)(-1)=\pi\arctan\sqrt 3=\pi-\frac(\pi)(3)=\frac(2\pi)( 3)\]
Following the rules for multiplying complex numbers in trigonometric form, we do the following:
In our case:
\[(z_1+z_2)^(10)=(\frac(1)(2))^(10)\cdot(\cos (10\cdot\frac(5\pi)(6))+i\sin \cdot\frac(5\pi)(6)))=\frac(1)(2^(10))\cdot\cos \frac(25\pi)(3)+i\sin\frac(25\ pi)(3).\]
Making the fraction \[\frac(25)(3)=8\frac(1)(3)\] correct, we come to the conclusion that we can “twist” 4 turns \[(8\pi rad.):\]
\[ (z_1+z_2)^(10)=\frac(1)(2^(10))\cdot(\cos \frac(\pi)(3)+i\sin\frac(\pi)(3 ))\]
Answer: \[(z_1+z_2)^(10)=\frac(1)(2^(10))\cdot(\cos \frac(\pi)(3)+i\sin\frac(\pi) (3))\]
This equation can be solved in another way, which boils down to bringing the 2nd number into algebraic form, then performing the multiplication in algebraic form, converting the result to trigonometric form and applying Moivre’s formula:
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