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 z 1 = a 1 + b 1 i And z 2 = a 2 + b 2 i.
Let's consider z = a + bi.
The set of complex numbers extends the set of real numbers, which in turn extends the set of rational numbers, etc. This chain of investments can be seen in the figure: N – natural numbers, Z – integers, Q – rational, R – real, C – complex. 
Representation of complex numbers
Algebraic notation.
Consider a complex number z = a + bi, 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 z = a + bi 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.
Let's consider 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, thus we have obtained a new form of writing a complex number: z = reiφ, 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 of this theorem is that there are exactly n different roots of degree n of unity.
Main types of tasks
This section will look at the main types of simple problems involving 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 look at general methods for solving 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, Where a is a certain number, then the discriminant can be represented as D = (ia) 2, hence √D = i|a|, and then you can use the already known formula for the roots of a quadratic equation.
Example. Let's return to the quadratic equation mentioned above 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 for solving them; for a more detailed study of the capabilities of complex numbers, it is recommended to use specialized literature.
Literature
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 school mathematics curriculum, 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, 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 a negative number makes sense.
The choice of the topic “Complex numbers” as the topic of my final qualification 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 of 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” provides solutions to problems with complex numbers in algebraic form, defines 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, and also sets out the rule extracting the square root of 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 the solution of 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 introduction of theoretical foundations, and subsequently their practical application in solving problems.
At the end of the thesis there is a list of references used. Most of them present theoretical material in sufficient detail and in an accessible manner, discuss solutions to some problems, and give practical tasks for independent solution. I would like to pay special attention to such sources as:
1. Gordienko N.A., Belyaeva E.S., Firstov V.E., Serebryakova I.V. Complex numbers and their applications: Textbook. . The material of the textbook is 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 the most important issues in mathematics. For example, a quadratic equation with a negative discriminant has no real roots. 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
are real numbers and, therefore, the set of complex numbers contains the set of real numbers. Complex numbers of the form
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.
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:
Where can I solve a system of equations with complex numbers online?
You can solve the system of equations on our website https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.
Expressions, equations and systems of equations
with complex numbers
Today in class we will practice typical operations with complex numbers, and also master the technique of solving expressions, equations and systems of equations that contain these numbers. This workshop is a continuation of the lesson, and therefore if you are not well versed in the topic, then please follow the link above. Well, for more prepared readers I suggest you warm up right away:
Example 1
Simplify an expression 
, If . Represent the result in trigonometric form and plot it on the complex plane.
Solution: so, you need to substitute the fraction into the “terrible” fraction, carry out simplifications, and convert the result complex number V trigonometric form. Plus a drawing.
What is the best way to formalize the decision? It is more profitable to deal with a “sophisticated” algebraic expression step by step. Firstly, attention is less distracted, and secondly, if the task is not accepted, it will be much easier to find the error.
1) First, let's simplify the numerator. Let's substitute the value into it, open the brackets and fix the hairstyle:
...Yes, such a Quasimodo came from complex numbers...
Let me remind you that during the transformations, completely simple things are used - the rule of multiplying polynomials and the equality that has already become banal. The main thing is to be careful and not get confused by the signs.
2) Now comes the denominator. If , then:
Notice in what unusual interpretation it is used square sum formula. Alternatively, you can perform a rearrangement here 
subformula The results will naturally be the same.
3) And finally, the whole expression. If , then:
To get rid of a fraction, multiply the numerator and denominator by the conjugate expression of the denominator. At the same time, for the purposes of application square difference formulas must first (and already a must!) put the negative real part in 2nd place:
And now the key rule:
WE ARE IN NO RUSH! It's better to play it safe and take an extra step.
In expressions, equations and systems with complex numbers, presumptuous verbal calculations more fraught than ever!
There was a good reduction in the final step and that's just a great sign.
Note : strictly speaking, here the division of a complex number by the complex number 50 occurred (remember that). I have been silent about this nuance until now, and we will talk about it a little later.
Let's denote our achievement with the letter
Let us present the result obtained in trigonometric form. Generally speaking, here you can do without a drawing, but since it is required, it is somewhat more rational to do it right now: 
Let's calculate the modulus of a complex number:
If you draw on a scale of 1 unit. = 1 cm (2 notebook cells), then the obtained value can be easily checked using a regular ruler.
Let's find an argument. Since the number is located in the 2nd coordinate quarter, then:
The angle can be easily checked with a protractor. This is the undoubted advantage of the drawing.
Thus: – the required number in trigonometric form.
Let's check:
, which was what needed to be verified.
It is convenient to find unfamiliar values of sine and cosine using trigonometric table.
Answer: 
A similar example for an independent solution:
Example 2
Simplify an expression 
, Where . Draw the resulting number on the complex plane and write it in exponential form.
Try not to skip the tutorials. They may seem simple, but without training, “getting into a puddle” is not just easy, but very easy. Therefore, we “get our hands on it.”
Often a problem has more than one solution:
Example 3
Calculate if , 
Solution: first of all, let's pay attention to the original condition - one number is presented in algebraic, and the other in trigonometric form, and even with degrees. Let's immediately rewrite it in a more familiar form: 
.
In what form should the calculations be carried out? The expression obviously involves first multiplication and further raising to the 10th power Moivre's formula, which is formulated for the trigonometric form of a complex number. So it seems more logical to convert the first number. Let's find its module and argument: 
We use the rule for multiplying complex numbers in trigonometric form:
if , then
Making the fraction correct, we come to the conclusion that we can “twist” 4 turns ( glad.):
Second solution is to convert the 2nd number into algebraic form 
, perform the multiplication in algebraic form, convert the result to trigonometric form and use Moivre's formula.
As you can see, there is one “extra” action. Those who wish can follow through with the decision and make sure that the results are the same.
The condition says nothing about the form of the final complex number, so:
Answer: 
But “for beauty” or on demand, the result is not difficult to imagine in algebraic form:
On one's own:
Example 4
Simplify an expression
Here we need to remember actions with degrees, although there is not one useful rule in the manual, here it is: .
And one more important note: the example can be solved in two styles. The first option is to work with two numbers and being okay with fractions. The second option is to represent each number as quotient of two numbers: 
And get rid of the four-story structure. From a formal point of view, it doesn’t matter how you decide, but there is a substantive difference! Please think carefully about:
is a complex number;
is the quotient of two complex numbers ( and ), but depending on the context, you can also say this: a number represented as the quotient of two complex numbers.
A short solution and answer at the end of the lesson.
Expressions are good, but equations are better:
Equations with complex coefficients
How do they differ from “ordinary” equations? Odds =)
In light of the above comment, let's start with this example:
Example 5
Solve the equation 
And an immediate preamble “hot on the heels”: initially the right side of the equation is positioned as the quotient of two complex numbers ( and 13), and therefore it would be bad form to rewrite the condition with the number (although this will not cause an error). This difference, by the way, is more clearly visible in the fraction - if, relatively speaking, then this value is primarily understood as "full" complex root of the equation, and not as a divisor of a number, and especially not as a part of a number!
Solution, in principle, can also be done step by step, but in this case the game is not worth the candle. The initial task is to simplify everything that does not contain the unknown "z", resulting in the equation being reduced to the form:
We confidently simplify the middle fraction: 
We transfer the result to the right side and find the difference: 
Note
: and again I draw your attention to the meaningful point - here we did not subtract a number from a number, but brought the fractions to a common denominator! It should be noted that already in the PROGRESS of solving it is not forbidden to work with numbers: 
, however, in the example under consideration this style is more harmful than useful =)
According to the rule of proportion, we express “zet”:
Now you can divide and multiply by the conjugate again, but the suspiciously similar numbers in the numerator and denominator suggest the next move: 
Answer:
To check, let’s substitute the resulting value into the left side of the original equation and carry out simplifications:
– the right side of the original equation is obtained, thus the root is found correctly.
...Now, now... I’ll find something more interesting for you... here you go:
Example 6
Solve the equation 
This equation reduces to the form , which means it is linear. I think the hint is clear - go for it!
Of course... how can you live without him:
Quadratic equation with complex coefficients
At the lesson Complex numbers for dummies we learned that a quadratic equation with real coefficients can have conjugate complex roots, after which a logical question arises: why, in fact, the coefficients themselves cannot be complex? Let me formulate a general case:
Quadratic equation with arbitrary complex coefficients (1 or 2 of which or all three may be, in particular, valid) It has two and only two complex root (possibly one or both of which are valid). At the same time, the roots (both real and with non-zero imaginary part) may coincide (be multiples).
A quadratic equation with complex coefficients is solved using the same scheme as "school" equation, with some differences in the calculation technique:
Example 7
Find the roots of a quadratic equation 
Solution: the imaginary unit comes first, and, in principle, you can get rid of it (multiplying both sides by), however, there is no particular need for this.
For convenience, we write out the coefficients:
Let's not lose the "minus" of a free member! ...It may not be clear to everyone - I’ll rewrite the equation in standard form 
:
Let's calculate the discriminant:
And here is the main obstacle: 
Application of the General Formula for Extracting the Root (see last paragraph of the article Complex numbers for dummies)
complicated by serious difficulties associated with the radical complex number argument (see for yourself). But there is another, “algebraic” way! We will look for the root in the form: 
Let's square both sides: 
Two complex numbers are equal if their real and imaginary parts are equal. Thus, we get the following system: 
The system is easier to solve by selecting (a more thorough way is to express from the 2nd equation - substitute into the 1st, obtain and solve a biquadratic equation). Assuming that the author of the problem is not a monster, we put forward the hypothesis that and are integers. From the 1st equation it follows that “x” modulo more than "Y". In addition, the positive product tells us that the unknowns are of the same sign. Based on the above, and focusing on the 2nd equation, we write down all the pairs that match it: 
It is obvious that the 1st equation of the system is satisfied by the last two pairs, thus: 
An intermediate check wouldn't hurt: 
which was what needed to be checked.
You can choose as a “working” root any meaning. It is clear that it is better to take the version without the “cons”:
We find the roots, not forgetting, by the way, that: 
Answer: 
Let's check whether the found roots satisfy the equation 
:
1) Let's substitute: 
true equality.
2) Let's substitute: 
true equality.
Thus, the solution was found correctly.
Based on the problem we just discussed:
Example 8
Find the roots of the equation
It should be noted that the square root of purely complex numbers can be easily extracted using the general formula 
, Where 
, so both methods are shown in the sample. The second useful remark concerns the fact that preliminary extraction of the root of a constant does not simplify the solution at all.
Now you can relax - in this example you will get away with a slight fright :)
Example 9
Solve the equation and check
Solutions and answers at the end of the lesson.
The final paragraph of the article is devoted to
system of equations with complex numbers
Let's relax and... don't tense up =) Let's consider the simplest case - a system of two linear equations with two unknowns:
Example 10
Solve the system of equations. Present the answer in algebraic and exponential forms, depict the roots in the drawing. 
Solution: the condition itself suggests that the system has a unique solution, that is, we need to find two numbers that satisfy to each equation of the system.
The system can really be solved in a “childish” way (express one variable in terms of another)
, however it is much more convenient to use Cramer's formulas. Let's calculate main determinant systems:
, which means the system has a unique solution.
I repeat that it is better to take your time and write out the steps in as much detail as possible:
We multiply the numerator and denominator by an imaginary unit and get the 1st root:
Likewise: 
The corresponding right-hand sides are obtained, etc.
Let's make the drawing: 
Let's represent the roots in exponential form. To do this, you need to find their modules and arguments:
1) – the arctangent of “two” is calculated “poorly”, so we leave it like this: 