n
n
si
i
n
s
(co
r
z
n
n
ϕ
ϕ
+
=
Muavr formulasi o’rinli.
Muavr formulasi kompleks sondan italgan darajali ildiz chiqarish
masalasini hal qilishga imkon beradi. w kompleks son z kompleks sonning n-
darajali
(
)
N
∈
n
ildizi deyiladi, agar
z
w
n
=
bo’lsa. z ning n-darajali ildizini
n
z
orqali belgilaymiz.
Berilgan z kompleks sonning n-darajali ildizi bir nechta qiymatlarga ega,
shuning uchun
n
z = w yozuv w son shu qiymatlardan biri ekanligini bildiradi.
Bitta mulohaza davomida
n
z ifoda z kompleks son n-darajali ildizining faqat
bitta qiymatini bildiradi.
Agar tekstdan ildizning aynnan shu qiymati haqida gapirilayotganligi
ma’lum bo’lsa, bu haqida alohida eslatilmaydi, masalan, kompleks son modulini
hisoblashda.
Agar
0
≠
+
=
)
n
isi
s
r(co
z
ϕ
ϕ
bo’lsa, z ning n-darajali ildizi uchun
)
2
2
(
n
n
isi
n
s
co
r
w
z
n
k
n
κπ
ϕ
κπ
ϕ
+
+
+
=
=
,
bu yerda
1
...,
1
,
0
−
=
n
k
, formula o’rinli. Bu formula z kompleks son n-darajali
ildizining n ta har xil qiymatilarini beradi.
1-m i s o l. Hisoblang:
20
1
3
1
−
+
i
i
.
Yechish.
)
р
n
isi
р
s
(co
i
3
3
2
3
1
+
=
+
,
р)
n
isi
р
s
(co
i
4
7
4
7
2
1
+
=
−
,
bo’lganligi sababli
16
=
+
=
+
=
=
+
+
+
=
−
+
−
+
=
−
+
π
π
π
π
12
140
12
140
2
12
7
12
7
2
4
3
4
3
2
4
4
2
3
3
2
1
3
1
10
20
20
20
20
n
isi
s
co
n
isi
s
co
р
р
n
isi
р
р
s
co
)
р
(
n
isi
)
р
(
s
(co
)
р
n
isi
р
s
(co
i
i
( )
.
1
2
2
2
2
2
2
2
4
7
4
7
2
9
10
10
i
i
n
isi
s
co
−
=
−
=
+
=
π
π
■
2-m i s o l. Hisoblang:
n
б)
n
isi
б
s
co
(
+
+
1
.
Yechish.
≤
≤
+
+
+
−
≤
≤
+
=
+
+
булса
агар
n
isi
сos
сos
булса
агар
n
isi
сos
сos
n
isi
сos
π
α
π
π
α
π
α
α
π
α
α
α
α
α
α
2
,
2
2
2
2
2
2
0
,
2
2
2
2
1
,
(28 ( r) mashqqa qarang) bo’lganligi uchun
π
α
<
≤
0
bo’lganda
+
=
+
+
2
2
2
2
)
1
(
n
n
isi
n
s
co
s
co
n
isi
s
co
n
n
n
α
α
α
α
α
,
π
α
π
2
≤
≤
bo’lganda esa
(
)
+
+
+
−
=
+
+
π
α
π
α
α
α
α
n
n
n
isi
n
n
s
co
s
co
n
isi
s
co
n
n
n
2
2
2
)
2
(
1
. ■
3-m i s o l.
4
16
−
ildizning barcha qiymatlarini toping.
Yechish.
16
−
=
z
ni trigonometrik shaklga keltiramiz:
(
)
π
π
n
isi
s
co
z
+
=
−
=
16
16
.
U holda ildiz chiqarish formulasiga ko’ra
+
+
+
=
4
2
4
2
2
рк
р
n
isi
рк
р
s
co
w
к
,
3
,
2
,
1
,
0
=
k
.
Natijada,
2
2
4
4
2
0
i
р
n
isi
р
s
co
w
+
=
+
=
,
2
2
4
3
4
3
2
1
i
n
isi
s
co
w
+
−
=
+
=
π
π
,
2
2
4
5
4
5
2
2
i
n
isi
s
co
w
−
−
=
+
=
π
π
,
2
2
4
7
4
7
2
3
i
n
isi
s
co
w
−
=
+
=
π
π
. ■
4-m i s o l.
4
3
1
1
i
i
−
+
−
to’plam elementlarining trigonometrik shaklini
yozing.
17
Yechish.
+
=
+
−
π
π
4
3
4
3
2
1
n
isi
s
co
i
va
−
+
−
=
−
3
3
2
3
1
π
π
n
isi
s
co
i
bo’lganligi uchun
+
=
+
+
+
=
−
+
−
π
π
π
π
π
π
12
13
12
13
2
2
3
4
3
3
4
3
2
2
3
1
1
n
isi
s
co
n
isi
s
co
i
i
.
Natijada,
+
+
+
=
−
+
−
48
24
13
48
24
13
2
1
3
1
1
8
4
κπ
π
κπ
π
n
isi
s
co
i
i
,
.
,
,
,
3
2
1
0
=
κ
■
Muavr formulasi ba’zi trigonometrik ifodalarni almashtirishda qulayliklar
yaratadi.
5-m i s o l.
ϕ
5
tg
ni
ϕ
tg
orqali ifodalang.
Yechish. Darajaga ko’tarish formulasiga ko’ra
(
)
5
sin
cos
5
sin
5
cos
ϕ
ϕ
ϕ
ϕ
i
i
+
=
+
.
Nyuton binom formulasini qo’llab, quyidagini hosil qilamiz:
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
5
4
3
2
2
3
4
5
5
10
10
5
5
5
in
s
i
in
s
s
co
in
s
s
ico
in
s
s
co
in
s
s
ico
s
co
in
s
i
s
co
+
+
−
−
−
+
=
+
chunki
i
, i
-i, i
, i
-
i
=
=
=
=
5
4
3
2
1
1
. Mos ravishda haqiqiy va mavhum
qismlarini tenglashtirib,
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
5
2
2
4
4
2
3
5
10
5
5
,
5
10
5
in
s
in
s
s
co
in
s
s
co
in
s
in
s
s
co
in
s
s
co
s
co
s
co
+
−
=
=
+
−
=
munosabatlarni hosil qilamiz. Bulardan
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
4
2
5
3
4
2
3
5
5
2
2
4
5
10
1
10
5
5
10
10
5
5
tg
tg
tg
tg
tg
in
s
s
co
in
s
s
co
s
co
in
s
in
s
s
co
in
s
s
co
tg
+
−
+
−
=
+
−
+
−
=
.
Bu yerda biz kasrning surat va maxrajini
ϕ
5
s
co
ga bo’ldik. ■
6-m i s o l.
ϕ
5
in
s
ni kaðrali argumentlarning trigonometrik funksiyalari
orqali chiziqli ifodalang.
Yechish.
ϕ
ϕ
in
s
i
s
co
z
+
=
bo’lsin, u holda
ϕ
ϕ
sin
i
cos
z
-1
−
=
,
ϕ
ϕ
k
isin
k
cos
z
k
+
=
,
ϕ
ϕ
k
isin
-
k
cos
z
k
=
−
,
2
-1
z
z
cos
+
=
ϕ
,
i
z
-
z
sin
-1
2
=
ϕ
,
2
-k
k
z
z
k
cos
+
=
ϕ
,
i
z
z
k
sin
-k
k
2
−
=
ϕ
.
Bularga ko’ra
(
) (
) (
)
=
+
−
=
−
+
−
−
−
=
=
−
+
−
+
−
=
−
=
−
−
−
−
−
−
−
i
in
s
i
in
s
i
in
s
i
i
z
z
z
z
z
z
i
z
z
z
z
z
z
i
z
z
in
s
32
20
3
10
5
2
32
10
5
32
5
10
10
5
2
1
3
3
5
5
5
3
1
3
5
5
1
5
ϕ
ϕ
ϕ
ϕ
18
=
.
16
10
3
5
5
ϕ
ϕ
ϕ
in
s
in
s
in
s
+
−
■
Xuddi shunga o’xshash yo’l Bilan istalgan
ϕ
ϕ
κ
m
in
s
s
co
ifodani karrali
argumentning trigonometrik funksiyalari orqali chiziqli ifodalash mumkin.
M A S H Q L A R
32. Hisoblang: a)
20
1
3
1
−
+
i
i
; b)
24
2
3
1
−
−
i
;
c)
(
)
( )
(
)
( )
20
15
20
15
1
3
1
1
3
1
i
i
i
i
+
−
−
+
−
+
−
; d)
( )
( )
N
∈
−
+
−
+
n
i
i
n
n
,
1
1
1
2
1
2
; e)
(
)
4
1 i
tg
z
−
=
;
f)
(
)
4
2
i
tg
−
; g)
5
5
6
1
5
6
+
+
π
π
s
co
i
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