How to take the root of a complex number. Degree with arbitrary rational exponent

Complex number Z called rootn– c, If Z n = c.

Find all root values n– oh degree of complex number With. Let c=| c|·(cos Arg c+ i· sin ArgWith), A Z = | Z|·(withos Arg Z + i· sin Arg Z) , Where Z root n- th degree from a complex number With. Then it must be = c = | c|·(cos Arg c+ i· sin ArgWith). Hence it follows that
And n· Arg Z = ArgWith
Arg Z =
(k=0,1,…) . Hence, Z =
(cos
+ i· sin
), (k=0,1,…) . It is easy to see that any of the values
, (k=0,1,…) different from one of the corresponding values
,(k = 0,1,…, n-1) to a multiple 2π. That's why , (k = 0,1,…, n-1) .

Example.

Calculate the root of (-1).

, obviously |-1| = 1, arg (-1) = π

-1 = 1 (cos π + i· sin π )

, (k = 0, 1).

= i

Degree with arbitrary rational exponent

Take an arbitrary complex number With. If n natural number, then With n = | c| n ·(Withos nArgwith +i· sin nArgWith)(6). This formula is also true in the case n = 0 (c≠0)
. Let n < 0 And n Z And c ≠ 0, Then

With n =
(cos nArgWith+i sin nArgWith) = (cos nArgWith+ i sin nArgWith) . Thus, formula (6) is valid for any n.

Let's take a rational number , Where q natural number, and R is an integer.

Then under degree c r let's understand the number
.

We get that ,

(k = 0, 1, …, q-1). These values q pieces, if the fraction is not reduced.

Lecture №3 The limit of a sequence of complex numbers

A complex-valued function of a natural argument is called sequence of complex numbers and denoted (With n ) or With 1 , With 2 , ..., With n . With n = a n + b n · i (n = 1,2, ...) complex numbers.

With 1 , With 2 , … - members of the sequence; With n - common member

Complex number With = a+ b· i called limit of a sequence of complex numbers (c n ) , Where With n = a n + b n · i (n = 1, 2, …) , where for any

, that for all n > N the inequality
. A sequence that has a finite limit is called converging sequence.

Theorem.

In order for a sequence of complex numbers (with n ) (With n = a n + b n · i) converged to a number with = a+ b· i, is necessary and sufficient for the equalitylim a n = a, lim b n = b.

Proof.

We will prove the theorem based on the following obvious double inequality

, Where Z = x + y· i (2)

Necessity. Let lim(With n ) = with. Let us show that the equalities lim a n = a And lim b n = b (3).

Obviously (4)

Because
, When n → ∞ , then it follows from the left side of inequality (4) that
And
, When n → ∞ . therefore equalities (3) hold. The need has been proven.

Adequacy. Now let equalities (3) hold. It follows from equality (3) that
And
, When n → ∞ , therefore, due to the right side of inequality (4), it will be
, When n→∞ , Means lim(With n )=s. Sufficiency has been proven.

So, the question of the convergence of a sequence of complex numbers is equivalent to the convergence of two real number sequences, therefore, all the basic properties of the limits of real number sequences apply to sequences of complex numbers.

For example, for sequences of complex numbers, the Cauchy criterion is valid: in order for a sequence of complex numbers (with n ) converged, it is necessary and sufficient that for any

, that for anyn, m > Nthe inequality
.

Theorem.

Let a sequence of complex numbers (with n ) And (z n ) converge respectively to with andz, then the equalitylim(With n z n ) = c z, lim(With n · z n ) = c· z. If it is known for certain thatzis not equal to 0, then the equality
.

numbers in trigonometric form.

De Moivre formula

Let z 1 = r 1 (cos  1 + isin  1) and z 2 = r 2 (cos  2 + isin  2).

The trigonometric form of a complex number is convenient to use to perform the operations of multiplication, division, raising to an integer power, and extracting a root of degree n.

z 1 ∙ z 2 = r 1 ∙ r 2 (cos ( 1 +  2) + i sin( 1 +  2)).

When multiplying two complex numbers in trigonometric form, their moduli are multiplied and their arguments are added. When dividing their moduli are divided and their arguments subtracted.

A consequence of the rule for multiplying a complex number is the rule for raising a complex number to a power.

z = r(cos  + i sin ).

z n \u003d r n (cos n + isin n).

This ratio is called De Moivre's formula.

Example 8.1 Find the product and quotient of numbers:

And

Solution

z1∙z2
∙

=

;

Example 8.2 Write a number in trigonometric form

∙
-i) 7 .

Solution

Denote
and z 2 =
– i.

r 1 = |z 1 | = √ 1 2 + 1 2 = √ 2;  1 = argz 1 = arctg ;

z1 =
;

r 2 = |z 2 | = √(√ 3) 2 + (– 1) 2 = 2;  2 = arg z 2 = arctg
;

z2 = 2
;

z 1 5 = (
) 5
; z 2 7 = 2 7

z = (
) 5 2 7
=

2 9

§ 9 Extracting the root of a complex number

Definition. rootnth power of a complex number z (denoting
) is a complex number w such that w n = z. If z = 0, then
= 0.

Let z  0, z = r(cos + isin). Denote w = (cos + sin), then we write the equation w n = z in the following form

 n (cos(n ) + isin(n )) = r(cos + isin).

Hence  n = r,

 =

Thus w k =
·
.

There are exactly n distinct values among these values.

Therefore, k = 0, 1, 2, …, n – 1.

On the complex plane, these points are the vertices of a regular n-gon inscribed in a circle with a radius
centered at point O (Figure 12).

Figure 12

Example 9.1 Find all values
.

Solution.

Let's represent this number in trigonometric form. Find its modulus and argument.

w k =
, where k = 0, 1, 2, 3.

w 0 =
.

w 1 =
.

w 2 =
.

w 3 =
.

On the complex plane, these points are the vertices of a square inscribed in a circle with radius
centered at the origin (Figure 13).

Figure 13 Figure 14

Example 9.2 Find all values
.

Solution.

z = - 64 = 64(cos + isin);

w k =
, where k = 0, 1, 2, 3, 4, 5.

w 0 =
; w 1 =
;

w 2 =
w 3 =

w4 =
; w 5 =
.

On the complex plane, these points are the vertices of a regular hexagon inscribed in a circle with a radius of 2 centered at the point O (0; 0) - Figure 14.

§ 10 The exponential form of a complex number.

Euler formula

Denote
= cos  + isin  and
= cos  - isin  . These ratios are called Euler formulas .

Function
has the usual properties of an exponential function:

Let the complex number z be written in the trigonometric form z = r(cos + isin).

Using the Euler formula, we can write:

z = r
.

This entry is called indicative form complex number. Using it, we get the rules for multiplication, division, exponentiation and root extraction.

If z 1 = r 1
and z 2 = r 2
?That

z 1 z 2 = r 1 r 2
;

z n = r n

, where k = 0, 1, … , n – 1.

Example 10.1 Write a number in algebraic form

z=
.

Solution.

Example 10.2 Solve the equation z 2 + (4 - 3i)z + 4 - 6i = 0.

Solution.

For any complex coefficients, this equation has two roots z 1 and z 1 (possibly coinciding). These roots can be found using the same formula as in the real case. Because
takes on two values that differ only in sign, then this formula has the form: