Kompleks sonning ildizini ajratib olish. Murakkab sonning ildizini ajratib olish Ildizning barcha qiymatlarini qanday topish mumkin

ning ildizini aniq ajratib olish mumkin emas murakkab son, chunki u darajasiga teng bo'lgan bir qator qiymatlarga ega.

Murakkab sonlar trigonometrik shaklning kuchiga ko'tariladi, ular uchun Moyvard formulasi amal qiladi:

\(\ z^(k)=r^(k)(\cos k \varphi+i \sin k \varphi), \forall k \in N \)

Xuddi shunday, bu formuladan kompleks sonning k ildizini hisoblash uchun foydalaniladi (nolga teng emas):

\(\ z^(\frac(1)(k))=(r(\cos (\varphi+2 \pi n)+i \sin (\varphi+2 \pi n))))^(\frac( 1)(k))=r^(\frac(1)(k))\left(\cos \frac(\varphi+2 \pi n)(k)+i \sin \frac(\varphi+2 \) pi n)(k)\o'ng), \forall k>1, \forall n \in N \)

Agar kompleks son nolga teng bo'lmasa, u holda k darajali ildizlar har doim mavjud bo'lib, ular kompleks tekislikda ifodalanishi mumkin: ular koordinata boshi va radiusi \(\r) markazida joylashgan aylanaga chizilgan k-gonning uchlari bo'ladi. ^ (\ frac (1) (k)) \)

Muammoni hal qilishga misollar

Vazifa

\(\z=-1\) sonining uchinchi ildizini toping.

Yechim.

Avval \(\z=-1\) sonini ifodalaymiz trigonometrik shakl. \(\ z=-1 \) sonning haqiqiy qismi \(\ z=-1 \) son, xayoliy qismi \(\ y=\operatornomi(lm) \), \(\ z=). 0 \). Kompleks sonning trigonometrik shaklini topish uchun uning moduli va argumentini topish kerak.

Kompleks sonning moduli \(\z\) sondir:

\(\ r=\sqrt(x^(2)+y^(2))=\sqrt((-1)^(2)+0^(2))=\sqrt(1+0)=1 \ )

Argument quyidagi formula bo'yicha hisoblanadi:

\(\ \varphi=\arg z=\operatorname(arctg) \frac(y)(x)=\operatorname(arctg) \frac(0)(-1)=\operatorname(arctg) 0=\pi \)

Demak, kompleks sonning trigonometrik shakli quyidagicha: \(\z=1(\cos \pi+i \sin \pi)\)

Keyin uchinchi ildiz quyidagicha ko'rinadi:

\(\ =\cos \frac(\pi+2 \pi n)(3)+i \sin \frac(\pi+2 \pi n)(3) \), \(\ n=0,1, 2\ )

\(\ \omega_(1)=\cos \frac(\pi)(3)+i \sin \frac(\pi)(3)=\frac(1)(2)+i \frac(\sqrt( 3))(2)\)

\(\n=1\) uchun biz quyidagilarni olamiz:

\(\ \omega_(2)=\cos \pi+i \sin \pi=-1+i \cdot 0=-1 \)

\(\n=2\) uchun biz quyidagilarni olamiz:

\(\ \omega_(3)=\cos \frac(5 \pi)(3)+i \sin \frac(5 \pi)(3)=\frac(1)(2)+i \frac(- \sqrt(3))(2)=\frac(1)(2)-i \frac(\sqrt(3))(2) \)

Javob

\(\ \omega_(1)=\frac(1)(2)+i \frac(\sqrt(3))(2), \omega_(2)=-1, \omega_(3)=\frac( 1)(2)-i \frac(\sqrt(3))(2) \)

Vazifa

Raqamning ikkinchi ildizini chiqarish uchun \(\z=1-\sqrt(3)i\)

Yechim.

Boshlash uchun biz kompleks sonni trigonometrik shaklda ifodalaymiz.

Kompleks sonning haqiqiy qismi \(\ z=1-\sqrt(3) i \) sondir \(\ x=\operatorname(Re) z=1 \) , xayoliy qismi \(\ y=\ operator nomi (Im) z =-\sqrt(3) \) . Kompleks sonning trigonometrik shaklini topish uchun uning moduli va argumentini topish kerak.

Kompleks sonning moduli \(\r\) bu son:

\(\r=\sqrt(x^(2)+y^(2))=\sqrt(1^(2)+(-\sqrt(3))^(2))=\sqrt(1+3) )=2\)

Argument:

\(\ \varphi=\arg z=\operatorname(arctg) \frac(y)(x)=\operatorname(arctg) \frac(-\sqrt(3))(1)=\operatorname(arctg)(- \sqrt(3))=\frac(2 \pi)(3) \)

Shunday qilib, kompleks sonning trigonometrik shakli:

\(\ z=2\left(\cos \frac(2 \pi)(3)+i \sin \frac(2 \pi)(3)\o'ng) \)

2-darajali ildizni olish formulasini qo'llash orqali biz quyidagilarni olamiz:

\(\ z^(\frac(1)(2))=\left(2\left(\cos \frac(2 \pi)(3)+i \sin \frac(2 \pi)(3)\ o'ng)\o'ng)^(\frac(1)(2))=2^(\frac(1)(2))\left(\cos \frac(2 \pi)(3)+i \sin \frac (2 \pi)(3)\o'ng)^(\frac(1)(2))= \)

\(\ =\sqrt(2)\left(\cos \left(\frac(\pi)(3)+\pi n\right)+i \sin \left(\frac(\pi)(3)+ \pi n\o'ng)\o'ng), n=0,1 \)

\(\ \mathrm(n)=0 \) uchun biz quyidagilarni olamiz:

\(\ \omega_(1)=\sqrt(2)\left(\cos \left(\frac(\pi)(3)+0\right)+i \sin \left(\frac(\pi)( 3)+0\o'ng)\o'ng)=\sqrt(2)\left(\frac(1)(2)+i \frac(\sqrt(3))(2)\o'ng)=\frac(\sqrt (2))(2)+i \frac(\sqrt(6))(2) \)

\(\ \mathrm(n)=1 \) uchun biz quyidagilarni olamiz:

\(\ \omega_(2)=\sqrt(2)\left(\cos \left(\frac(\pi)(3)+\pi\right)+i \sin \left(\frac(\pi) (3)+\pi\o'ng)\o'ng)=\sqrt(2)\left(-\frac(1)(2)+i \frac(\sqrt(3))(2)\o'ng)=-\ frac(\sqrt(2))(2)+i \frac(\sqrt(6))(2) \)

Javob

\(\ \omega_(1)=\frac(\sqrt(2))(2)+i \frac(\sqrt(6))(2) ; \omega_(2)=-\frac(\sqrt(2) ))(2)+i \frac(\sqrt(6))(2) \)

trigonometrik shakldagi raqamlar.

Moivre formulasi

z 1 = r 1 (cos  1 + isin  1) va z 2 = r 2 (cos  2 + isin  2) bo‘lsin.

Kompleks sonni yozishning trigonometrik shakli ko'paytirish, bo'lish, butun son darajaga ko'tarish va n daraja ildizini olish amallarini bajarish uchun qulaydir.

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

Ikkita murakkab sonni ko'paytirishda trigonometrik shaklda ularning modullari ko'paytiriladi va ularning argumentlari qo'shiladi. Ajratish paytida ularning modullari bo'linadi va ularning argumentlari ayiriladi.

Kompleks sonni ko'paytirish qoidasining natijasi kompleks sonni darajaga ko'tarish qoidasidir.

z = r(cos  + i sin ).

z n = r n (cos n + isin n).

Bu nisbat deyiladi Moivre formulasi.

8.1-misol Raqamlarning hosilasi va qismini toping:

Yechim

z 1 ∙z 2
∙

=

;

8.2-misol Raqamni trigonometrik shaklda yozing

∙
–i) 7 .

Yechim

belgilaylik
va z 2 =
- men.

r 1 = |z 1 | = √ 1 2 + 1 2 = √ 2; ;

 1 = arg z 1 = arktan
;

z 1 =
;

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

z 2 = 2
) 5
z 1 5 = (

;
z 2 7 = 2 7
=

2 9

z = (

) 5 ·2 7§ 9 Kompleks sonning ildizini ajratib olishTa'rif. Ildiz n
kompleks sonning th darajasi
= 0.

z (belgilash

) w n = z bo'ladigan w kompleks sondir. Agar z = 0 bo'lsa, u holda

z  0, z = r(cos + isin) bo‘lsin. w = (cos + sin) ni belgilaymiz, keyin w n = z tenglamani quyidagi shaklda yozamiz.

 =

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

Demak,  n = r,

Shunday qilib, wk =

Bu qiymatlar orasida n ta farq bor.
Shuning uchun k = 0, 1, 2, …, n – 1.

Murakkab tekislikda bu nuqtalar radiusli aylana ichiga chizilgan muntazam n-burchakning uchlaridir.

markazi O nuqtada (12-rasm). 12-rasm
.

Yechim.

9.1-misol

Barcha qiymatlarni toping
Bu sonni trigonometrik shaklda ifodalaylik. Keling, uning moduli va argumentini topamiz.

w k =
.

, bu yerda k = 0, 1, 2, 3.
.

w 0 =
.

w 1 =
.

w 2 =
w 3 =

Murakkab tekislikda bu nuqtalar radiusli aylana ichiga chizilgan kvadratning uchlaridir

markazi koordinatali (13-rasm). 12-rasm
.

Yechim.

13-rasm 14-rasm

Barcha qiymatlarni toping
9.2-misol

w k =
z = – 64 = 64(cos +isin);
;

w 0 =
, bu yerda k = 0, 1, 2, 3, 4, 5.

;
w 1 =
.

w 3 =

w 4 =

;

belgilaylik
w 5 =
Murakkab tekislikda bu nuqtalar markazi O (0; 0) nuqtada bo'lgan radiusi 2 bo'lgan aylana ichiga chizilgan muntazam olti burchakli uchlaridir - 14-rasm. § 10 Kompleks sonning ko'rsatkichli shakli. .