Find the square root of a complex number.

02.12.2022

With and natural number n 2 .

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

Let's find all the values of the root n– oh power of a 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- oh power of a complex number With. Then it must be = c = | c|·(cos Arg c+ i· sin ArgWith). 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,…) differs from one of the corresponding values
,(k = 0,1,…, n-1) by multiple 2π. That's why , (k = 0,1,…, n-1) .

Example.

Let's calculate the root of (-1).

, obviously |-1| = 1, arg (-1) = π

-1 = 1·(cos π + i· sin π )

, (k = 0, 1).

= i

Power with an arbitrary rational exponent

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

Then under degree c r we will understand the number
.

We get that ,

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

Lecture No. 3 The limit of a sequence of complex numbers

A complex-valued function of a natural argument is called sequence of complex numbers and is designated (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 in front of everyone n > N inequality holds
. A sequence having a finite limit is called convergent 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 equality to holdlim 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 ) = s. Let us show that the equalities are true lim a n = a And lim b n = b (3).

Obviously (4)

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

Adequacy. Let now equalities (3) be satisfied. From equality (3) it follows that
And
, When n → ∞ , therefore, due to the right side of inequality (4), it will be
, When n→∞ , Means lim(With n )=c. 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 ) converges, it is necessary and sufficient that for any

, that for anyn, m > Ninequality holds
.

Theorem.

Let a sequence of complex numbers (with n ) And (z n ) converge to c and respectivelyz, then the equalities are truelim(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 is true
.

numbers in trigonometric form.

Moivre's formula

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

The trigonometric form of writing a complex number is convenient to use to perform the operations of multiplication, division, raising to an integer power, and extracting the 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 modules are multiplied and their arguments are added. When dividing their modules are divided and their arguments are subtracted.

A corollary 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 = r n (cos n + isin n).

This ratio is called Moivre's formula.

Example 8.1 Find the product and quotient of numbers:

And

Solution

z 1 ∙z 2
∙

=

;

Example 8.2 Write a number in trigonometric form

∙
–i) 7 .

Solution

Let's denote
and z 2 =
–i.

r 1 = |z 1 | = √ 1 2 + 1 2 = √ 2; ;

 1 = arg z 1 = arctan
;

z 1 =
;

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

z 2 = 2
) 5
z 1 5 = (

;
z 2 7 = 2 7
=

2 9

z = (

) 5 ·2 7n§ 9 Extracting the root of a complex number Definition. Root
th power of a complex number
= 0.

z (denote

) is a complex number w such that w n = z. If z = 0, then

Let z  0, z = r(cos + isin). Let us 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 wk =

Among these values there are exactly n different ones.
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 of radius

with center at point O (Figure 12). Figure 12
.

Example 9.1

Find all values

Solution.
Let's represent this number in trigonometric form. Let's 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 of radius

with the center at the origin (Figure 13). Figure 12
.

Example 9.1

Figure 13 Figure 14

Solution.
Example 9.2

w k =
z = – 64 = 64(cos +isin);
;

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

;
w 1 =
.

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

§ 10 Exponential form of a complex number.

Euler's formula

Let's denote
= cos  + isin  and
= cos  - isin  . These relations are called .

Euler's formulas
Function

has the usual properties of an exponential function:

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

Using Euler's formula, we can write:
.

z = r This entry is called exponential form

complex number. Using it, we obtain 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
.

Example 9.1

z = Example 10.2

Example 9.1

Solve the equation z 2 + (4 – 3i)z + 4 – 6i = 0.
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