The polar form of –256 – 256V3i is 512 (cos (-) +i sin (-)) (for steps, see polar form calculator). According to the De Moivre's Formula, all n-th roots of a complex number 0+2rk r (cos (0) + i sin (0)) are given by ri (cos (+2nk ) + i sin (º+2ak )), k = 1 OS 0..n – 1. 27 We have that r = 512, 0 = -+2-7-0 • k = 0:STA(cos (=2) + i sin (=2*0) 2 (cos (-+ i sin (- • k = 1: 512 (cos (=1). 2 (cos + i sin ()) = 2 cos • k = 2: 512 (cos (=3-2) + 2 (cos k = 3: 512 (cos (-2) 2 (cos 4: 512 (cos COs 9. 2 cos 22 sin 27) +2-T-1 27. +i sin (-*1)) O + 2i sin +2-T-2 9. 4T 27 - +2-7-2 +i sin 9 10T 107 + i sin ()) 107 10T 2 cos () + 2i sin (0 i sin 27 27 +2-7-3 +2-7-3 COS 9. 9. 167 16T 11T + i sin -2 cos 27 + 2i sin 27 27 +i sin (-**)): +2-T-4 • k %3D 9. 签) k = 5: 512 (cos (=3) i sin ()) = 登+2-T-6 2 (cos 227 22T + i sin -2 cos 27 + 2i sin 27 +i sin (=* ;***)) : +2-7-5 +2-7-5 2 (cos 28T -2 cos () - 2i sin () 27 +i sin ( )) • k = 6: S12 (cos (=**) +i sin ()) 2 +2-7-7 +2-7-6 9. 2 (cos (347 •k = 7: 512 (cos -2 cos (E) – 2i sin (E + i sin (=*1)): -2 cos () - 2i sin ( -* +2-7-8 27 +2-7-7 9. 40T 137 2 (cos • k = 8: 512 (cos (=*) + 2 (cos +i sin ()) 40T 137 27 +2.T-8 9. Hi sin (27 – 2i sin () 46T 46t = 2 cos 27 27
The polar form of –256 – 256V3i is 512 (cos (-) +i sin (-)) (for steps, see polar form calculator). According to the De Moivre's Formula, all n-th roots of a complex number 0+2rk r (cos (0) + i sin (0)) are given by ri (cos (+2nk ) + i sin (º+2ak )), k = 1 OS 0..n – 1. 27 We have that r = 512, 0 = -+2-7-0 • k = 0:STA(cos (=2) + i sin (=2*0) 2 (cos (-+ i sin (- • k = 1: 512 (cos (=1). 2 (cos + i sin ()) = 2 cos • k = 2: 512 (cos (=3-2) + 2 (cos k = 3: 512 (cos (-2) 2 (cos 4: 512 (cos COs 9. 2 cos 22 sin 27) +2-T-1 27. +i sin (-*1)) O + 2i sin +2-T-2 9. 4T 27 - +2-7-2 +i sin 9 10T 107 + i sin ()) 107 10T 2 cos () + 2i sin (0 i sin 27 27 +2-7-3 +2-7-3 COS 9. 9. 167 16T 11T + i sin -2 cos 27 + 2i sin 27 27 +i sin (-**)): +2-T-4 • k %3D 9. 签) k = 5: 512 (cos (=3) i sin ()) = 登+2-T-6 2 (cos 227 22T + i sin -2 cos 27 + 2i sin 27 +i sin (=* ;***)) : +2-7-5 +2-7-5 2 (cos 28T -2 cos () - 2i sin () 27 +i sin ( )) • k = 6: S12 (cos (=**) +i sin ()) 2 +2-7-7 +2-7-6 9. 2 (cos (347 •k = 7: 512 (cos -2 cos (E) – 2i sin (E + i sin (=*1)): -2 cos () - 2i sin ( -* +2-7-8 27 +2-7-7 9. 40T 137 2 (cos • k = 8: 512 (cos (=*) + 2 (cos +i sin ()) 40T 137 27 +2.T-8 9. Hi sin (27 – 2i sin () 46T 46t = 2 cos 27 27
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
How did the value on the red circles became like that from the original equation on the blue circle? Please explain.
Transcribed Image Text:The polar form of –256 – 256/3i is 512 (cos (-) +i sin (-)) (for steps, see
polar form calculator).
According to the De Moivre's Formula, all n-th roots of a complex number
1
r (cos (0) + i sin ()) are given by ri (cos (2nk) + i sin (+2nk ). k =
0+2™k
0..n – 1.
27
We have that r =
512, 0 =
3 arTa
-+2-7-0
• k = 0:STA(cos (=2) + i sin (=:2*0)
2 (cos (-+ i sin (-
COs
9.
2 cos
27 sin
27)
+2-T-1
27.
• k = 1: 512 (cos (=1).
2 (cos + i sin ()) = 2 cos
- +2-7-2
• k = 2. 512 (cos (=3=2) +
+i sin ())
+ i sin (1))
O + 2i sin
'-좋+2·T-2
* +2-T-1
4T
27
+i sin
9
2 (cos
107
107
107
10T
2 cos () + 2i sin
27
27
27
k = 3: 512 (cos (-23)
+ i sin ())
+2.7-3
+2-7-3
COS
+ i sin
9.
2 (cos
167
16t
11T
-2 cos
27) + 2i sin
27
=
27
k = 4: 512 (cos (=+24)
+i sin (-* **4)):
+2-7-4
• k
9.
登))
k = 5: 512 (cos (=)
2 (cos
22T
22т
57
+i sin
-2 cos
27
+ 2i sin
27
+i sin (-*)) :
i sin ()) = -2 cos () – 2i sin (A)
+ i sin (-***))
+2-T-5
+2-T-5
2 (cos
'28ㅠ
27
27
• k = 6: 512 (cos (=*)
+i sin ())
- 2 +2-T-7
空+2-7-6
+2-7-6
34т
2 (cos
-2 cos (E) – 2i sin (E
- 2 +2.T-7
27
• k = 7: 512 ( cos
+i sin
9.
40™
9.
2 (cos
• k = 8: 512 (cos (=3*)
2 (cos
13™
+i sin ())
-2 cos () – 2i sin ()
- +2-7-8
40л
137
27
+ i sin (-*) =
+2.T•8
9.
Hi sin (*27
– 2i sin ()
46T
46т
= 2 cos
27
27
||
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