Trigonometric form of complex. Trigonometric form of a complex number. Converting a complex number from algebraic form to trigonometric form

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 on coordinate plane complex numbers:

3.3. Trigonometric form of a complex number

A complex number on the plane has the coordinates of a point M(x;y). In this case:

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. Actions with complex numbers in trigonometric form

· Addition and subtraction It’s more convenient to do 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 talk more about the trigonometric form of a complex number. Demonstrative form in practical tasks it occurs much less frequently. 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 this 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. Obviously. 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. Obviously. 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. Obviously. 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 is useful to remember the 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.

Introducing in complex form numbers and, the first and third numbers will be for independent decision.

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.

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 not defined.

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 is known, the square of the length of any vector equal to the sum 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 V in this case you can choose any angle: after all, at 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 is 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 n - 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°).