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– complex plane.
In the figure, the complex number z = (a, b) corresponds to a 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 
.
Module equal to zero then and only when 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 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: ;
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 
;
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 = , 
.
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°).
