Trigonometric form of complex numbers. Complex numbers in trigonometric form Trigonometric form of a complex number property
COMPLEX NUMBERS XI
§ 256. Trigonometric form of complex numbers
Let the complex number a + bi corresponds vector OA> with coordinates ( a, b ) (see Fig. 332).
Denote the length of this vector by r , and the angle it makes with the axis X , across φ . By definition of sine and cosine:
a / r = cos φ , b / r = sin φ .
That's why but = 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 , whence r = √a 2 + b 2
So, any complex number a + bi can be represented as :
a + bi = r (cos φ + i sin φ ), (1)
where r = √a 2 + b 2 , and the angle φ 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 up to an angle multiple of 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 zero argument is not defined.
Complex number modulus r sometimes denote | z |, and the argument arg z . Let's look at a few examples of the representation of complex numbers in trigonometric form.
Example. one. 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 .
In this way,
1 + i = √ 2 ,
where P - any integer. Usually, from an infinite set of values \u200b\u200bof 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 in trigonometric form a complex number √ 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 ; Consequently,
√ 3 - i = 2(cos 11 / 6 π + i sin 11 / 6 π ).
Example 3 Write in trigonometric form a complex number i .
complex number i corresponds vector OA> ending at point A of the axis at with ordinate 1 (Fig. 333). The length of such a vector is equal to 1, and the angle that it forms with the abscissa 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 OA > 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 in trigonometric form the complex number -5.
The complex number -5 corresponds to the vector OA> ending at the axis point X with abscissa -5 (Fig. 335). The length of such a vector is 5, and the angle it makes with the x-axis is 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 the sets of points representing complex numbers whose modules 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 be the module of a complex number at the same time? r And - r ?
2050. Can the argument of a complex number be angles at the same time φ And - φ ?
Present these complex numbers in trigonometric form by 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 the plane polar coordinate system . It is defined if a point O is given, called pole, and a beam emanating from the pole (for us, this is the axis Ox) is the polar axis. The position of point M is fixed by two numbers: radius (or radius vector) and the angle φ between the polar axis and the vector . The angle φ is called polar angle; It is 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 multiple of 2π. In this case, the pairs of numbers (r; φ) and (r 1 ; φ 1) are assigned 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 a complex number
Consider on the plane the Cartesian rectangular coordinate system xOy.
Any complex number z=(a, b) is assigned a point of 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, the complex number z = (a, b) match point M(x, y).
The task.Draw complex numbers on the coordinate plane:
3.3. Trigonometric form of a complex number
A complex number in 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 denoted. Module 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 denoted. The argument z is defined ambiguously, like the polar angle in the polar coordinate system, namely, up to a multiple of 2π.
Then we accept: , where φ is the smallest value of the argument. It's obvious that
.
With a deeper study of the topic, an auxiliary argument φ* is introduced, such that
Example 1. Find the trigonometric form of a complex number.
Solution. 1) we 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 is more convenient to perform with complex numbers in algebraic form:
· Multiplication– with the help of simple trigonometric transformations, it can be shown that when multiplying, the modules of numbers are multiplied, and the arguments are added: ;
In this section, we will focus more on the trigonometric form of a complex number. The exponential form in practical tasks is much less common. Please download and print if possible. trigonometric tables, methodological material can be found on the Mathematical formulas and tables page. You can't go far without tables.
Any complex number (except zero) can be written in trigonometric form:
Where is it complex number modulus, but - complex number argument.
Draw a number on the complex plane. For definiteness and simplicity of explanations, we will place it in the first coordinate quarter, i.e. we believe that:
The modulus of a complex number is the distance from the origin of coordinates to the corresponding point of the complex plane. Simply put, modulus is the length radius vector, which is marked 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 valid for any meanings "a" and "be".
Note : the modulus of a complex number is a generalization of the concept real number modulus, as the distance from the point to the origin.
The argument of a complex number called injection between positive 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 usually denoted by: or
From geometric considerations, the following formula for finding the argument is obtained:
. Attention! This formula works only 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 consider these cases.
But first, consider the simplest examples, when complex numbers are located on the coordinate axes.
Example 7
Express complex numbers in trigonometric form: ,,,. Let's execute the drawing:
In fact, the task is oral. For clarity, I will rewrite the trigonometric form of a complex number:
Let's remember tightly, the module - length(which is always non-negative), the argument is injection
1) Let's represent the number in trigonometric form. Find its modulus and argument. It's obvious that. Formal calculation according to the formula:. It is obvious that (the number lies directly on the real positive semiaxis). So the number in trigonometric form is:
Clear as day, reverse check action:
2) Let's represent the number in trigonometric form. Find its modulus and argument. It's obvious that. Formal calculation according to the formula:. Obviously (or 90 degrees). In the drawing, the corner is marked in red. So the number in trigonometric form is: 
.
Using , it is easy to get back the algebraic form of the number (at the same time by checking):
3) Let's represent the number in trigonometric form. Find its module and
argument. It's obvious that . Formal calculation according to the formula:
Obviously (or 180 degrees). In the drawing, the angle is indicated in blue. So the number in trigonometric form is:
Examination:
4) And the fourth interesting case. It's obvious that. Formal calculation according to 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 see
which is the same angle.
Thus, the entry becomes: 
Attention! In no case should you use the evenness of the cosine, the oddness of the sine and carry out further "simplification" of the record:
By the way, it is useful to remember the appearance and properties of trigonometric and inverse trigonometric functions, reference materials are in the last paragraphs of the page Graphs and properties of basic elementary functions. And complex numbers are much easier to learn!
In the design of the simplest examples, this should be written : "Obviously the modulus is... obviously the argument is...". This is really obvious and easily solved verbally.
Let's move on to 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 (the 1st and 4th coordinate quarters, or the right half-plane), then the argument must be found by the formula.
2) If (2nd coordinate quarter), then the argument must be found by the formula 
.
3) If (3rd coordinate quarter), then the argument must be found by the formula 
.
Example 8
Express complex numbers in trigonometric form: ,,,.
As soon as there are ready-made formulas, then the drawing is not necessary. But there is one point: when you are asked to present a number in trigonometric form, then drawing is better to do anyway. The fact is that teachers often reject a solution without a drawing, the absence of a drawing is a serious reason for a minus and a failure.
We represent the numbers and in complex form, the first and third numbers will be for an independent solution.
Let's represent the number in trigonometric form. Find its modulus and argument.
Since (case 2), then
- here you need to use the oddness of the arc tangent. Unfortunately, there is no value in the table, 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. Find its modulus and argument.
Since (case 1), then (minus 60 degrees).
In this way:
is a number in trigonometric form.
And here, as already noted, the cons do not touch.
In addition to the funny graphical verification method, there is also an analytical verification, which has already been carried out in Example 7. We use table of values of trigonometric functions, while taking into account that the angle is exactly the tabular angle (or 300 degrees): - numbers in the original algebraic form.
Numbers and represent in trigonometric form yourself. 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 the complex number and is the argument of the complex number.
What needs to be done to represent a complex number in exponential form? Almost the same: execute the drawing, find the module and argument. And write the number as .
For example, for the number of the previous example, we found the modulus and argument:,. Then this number in exponential form will be written as follows:
The number in exponential form would look like this:
Number 
- So:
The only advice is do not touch the indicator exponents, there is no need to rearrange factors, open brackets, etc. A complex number in exponential form is written strictly in form.
Actions on complex numbers written in algebraic form
The algebraic form of the 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 of addition of 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 performed according to the usual rule for multiplication of 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 of the divisor.
The 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 private z from division z 1 \u003d 3 - 2 z 2 = 3 – i.
z= .
4. Solve the equation:, x And y Î R.
(2x+y) + (x+y)i = 2 + 3i.
By virtue of 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 is called a plane with Cartesian coordinates ( x, y), if each point with coordinates ( a, b) is assigned a complex number z = a + bi. In this case, the abscissa axis is called real axis, and the y-axis is imaginary. Then every complex number a+bi geometrically represented on a plane as a point A (a, b) or vector .
Therefore, the position of the point BUT(and hence the complex number z) can be set 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 a vector is called complex number modulus and is denoted by | z|=r, and the angle j called complex number argument and denoted j = argz.
It is clear that | z| ³ 0 and | z | = 0 Û z= 0.
From fig. 2 shows that .
The argument of a complex number is defined ambiguously, and up to 2 pk, kÎ Z.
From fig. 2 also shows that if z=a+bi And j=argz, then
cos j =, sin j =, tg j = .
If zОR And z > 0 then argz = 0 +2pk;
if z ОR And z< 0 then argz = p + 2pk;
if z= 0,argz not determined.
The main value of the argument is determined on the interval 0 £argz£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. Determine on the complex plane the areas specified 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 is the equation of a circle with radius 5 and centered at the origin.
2) Circle with radius 6 centered at the origin.
3) Circle with radius 3 centered at a point z0 = 2 + i.
4) A ring bounded by circles with radii 6 and 7 centered at a point z 0 = i.
3. Find the module and argument of numbers: 1) ; 2).
1) ; but = 1, b = Þ 
,
Þ j 1 = 
.
2) z 2 = –2 – 2i; a =–2, b=-2 Þ 
,
.
Note: When defining the main argument, use the complex plane.
In this way: z 1 = .
2) 
, r 2 =
1, j 2 = , 
.
3) 
, r 3 = 1, j 3 = , 
.
4) , r 4 = 1, j4 = , 
.
Trigonometric 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 of the plane according to the following rule: a + bi = M ( a ; b ) (Fig. 1).
Picture 1
b) A complex number can be represented as a vector that starts at the pointABOUT and end at a given point (Fig. 2).
Figure 2
Example 7. Plot 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 set using the radius - vector with coordinates( a ; b ) (Fig. 4).
Figure 4
Definition . Vector length representing the 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 value 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 denotedBUT rg z orφ .
Complex number argumentz = 0 not determined. Complex number argumentz≠ 0 is a multi-valued quantity and is determined up to the term2πk (k = 0; - 1; 1; - 2; 2; ...): Arg z = arg z + 2πk , wherearg z - the main value of the argument, enclosed in the interval(-π; π] , i.e-π < arg z ≤ π (sometimes the value belonging to the interval is taken as the main value of the argument .
This formula forr =1 often referred to as De 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 extracting the square root of a complex numbera + bi we have two cases:
ifb
> about
, then 
;