It is impossible to unambiguously extract the root of a complex number, since it has a number of values equal to its power.
Complex numbers are raised to the power of trigonometric form, for which Moyward's formula is valid:
\(\ z^(k)=r^(k)(\cos k \varphi+i \sin k \varphi), \forall k \in N \)
Similarly, this formula is used to calculate the kth root of a complex number (not equal to zero):
\(\ z^(\frac(1)(k))=(r(\cos (\varphi+2 \pi n)+i \sin (\varphi+2 \pi n)))^(\frac( 1)(k))=r^(\frac(1)(k))\left(\cos \frac(\varphi+2 \pi n)(k)+i \sin \frac(\varphi+2 \ pi n)(k)\right), \forall k>1, \forall n \in N \)
If the complex number is not zero, then roots of degree k always exist, and they can be represented in the complex plane: they will be the vertices of a k-gon inscribed in a circle centered at the origin and radius \(\r^(\frac(1) (k))\)
Examples of problem solving
Find the third root of the number \(\z=-1\).
First we express the number \(\z=-1\) in trigonometric form. The real part of the number \(\ z=-1 \) is the number \(\ z=-1 \), the imaginary part is \(\ y=\operatorname(lm) \), \(\ z=0 \). To find the trigonometric form of a complex number, you need to find its modulus and argument.
The modulus of a complex number \(\z\) is the number:
\(\ r=\sqrt(x^(2)+y^(2))=\sqrt((-1)^(2)+0^(2))=\sqrt(1+0)=1 \ )
The argument is calculated using the formula:
\(\ \varphi=\arg z=\operatorname(arctg) \frac(y)(x)=\operatorname(arctg) \frac(0)(-1)=\operatorname(arctg) 0=\pi \)
Therefore, the trigonometric form of a complex number is: \(\z=1(\cos \pi+i \sin \pi)\)
Then the 3rd root looks like this:
\(\ =\cos \frac(\pi+2 \pi n)(3)+i \sin \frac(\pi+2 \pi n)(3) \), \(\ n=0.1, 2\)
\(\ \omega_(1)=\cos \frac(\pi)(3)+i \sin \frac(\pi)(3)=\frac(1)(2)+i \frac(\sqrt( 3))(2)\)
For \(\n=1\) we get:
\(\ \omega_(2)=\cos \pi+i \sin \pi=-1+i \cdot 0=-1 \)
For \(\n=2\) we get:
\(\ \omega_(3)=\cos \frac(5 \pi)(3)+i \sin \frac(5 \pi)(3)=\frac(1)(2)+i \frac(- \sqrt(3))(2)=\frac(1)(2)-i \frac(\sqrt(3))(2) \)
\(\ \omega_(1)=\frac(1)(2)+i \frac(\sqrt(3))(2), \omega_(2)=-1, \omega_(3)=\frac( 1)(2)-i \frac(\sqrt(3))(2) \)
To extract the 2nd root of a number \(\z=1-\sqrt(3)i\)
To begin with, we express a complex number in trigonometric form.
The real part of a complex number \(\ z=1-\sqrt(3) i \) is the number \(\ x=\operatorname(Re) z=1 \) , the imaginary part \(\ y=\operatorname(Im) z =-\sqrt(3) \) . To find the trigonometric form of a complex number, you need to find its modulus and argument.
The modulus of a complex number \(\r\) is the number:
\(\r=\sqrt(x^(2)+y^(2))=\sqrt(1^(2)+(-\sqrt(3))^(2))=\sqrt(1+3 )=2\)
Argument:
\(\ \varphi=\arg z=\operatorname(arctg) \frac(y)(x)=\operatorname(arctg) \frac(-\sqrt(3))(1)=\operatorname(arctg)(- \sqrt(3))=\frac(2 \pi)(3) \)
Therefore, the trigonometric form of a complex number is:
\(\ z=2\left(\cos \frac(2 \pi)(3)+i \sin \frac(2 \pi)(3)\right) \)
Applying the formula for extracting the 2nd degree root, we get:
\(\ z^(\frac(1)(2))=\left(2\left(\cos \frac(2 \pi)(3)+i \sin \frac(2 \pi)(3)\ right)\right)^(\frac(1)(2))=2^(\frac(1)(2))\left(\cos \frac(2 \pi)(3)+i \sin \frac (2 \pi)(3)\right)^(\frac(1)(2))= \)
\(\ =\sqrt(2)\left(\cos \left(\frac(\pi)(3)+\pi n\right)+i \sin \left(\frac(\pi)(3)+ \pi n\right)\right), n=0,1 \)
For \(\ \mathrm(n)=0 \) we get:
\(\ \omega_(1)=\sqrt(2)\left(\cos \left(\frac(\pi)(3)+0\right)+i \sin \left(\frac(\pi)( 3)+0\right)\right)=\sqrt(2)\left(\frac(1)(2)+i \frac(\sqrt(3))(2)\right)=\frac(\sqrt (2))(2)+i \frac(\sqrt(6))(2) \)
For \(\ \mathrm(n)=1 \) we get:
\(\ \omega_(2)=\sqrt(2)\left(\cos \left(\frac(\pi)(3)+\pi\right)+i \sin \left(\frac(\pi) (3)+\pi\right)\right)=\sqrt(2)\left(-\frac(1)(2)+i \frac(\sqrt(3))(2)\right)=-\ frac(\sqrt(2))(2)+i \frac(\sqrt(6))(2) \)
\(\ \omega_(1)=\frac(\sqrt(2))(2)+i \frac(\sqrt(6))(2) ; \omega_(2)=-\frac(\sqrt(2 ))(2)+i \frac(\sqrt(6))(2) \)
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 
;
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 (denote 
) is a complex number w such that w n = z. If z = 0, then 
= 0.
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 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 = 
w 4 = 
; w 5 = 
.
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
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 two values that differ only in sign, then this formula looks like:
Since –9 = 9 e i, then the values 
there will be numbers:
Then 
And 
.
|
Example 10.3 Solve the equations z 3 +1 = 0; z 3 = – 1. |
Solution.
The required roots of the equation will be the values 
.
For z = –1 we have r = 1, arg(–1) = .
w k = 
, k = 0, 1, 2.
Exercises
9 Present numbers in exponential form:
|
b) |
G) |
+i;
.10 Write numbers in exponential and algebraic forms:
|
A) |
V) |
|
b) |
d) 7(cos0 + isin0). |
11 Write the numbers in algebraic and geometric forms:
|
A) |
b) |
V) |
G) |
12 Numbers are given 
Presenting them in exponential form, find 
.
13 Using the exponential form of a complex number, perform the following steps:
A) 
b) 
V) 
G) 
|
d) |
|
|
|
|
.