Question

Complex Numbers problem

Answer 1

Click the attachment

Answer 2

x+1/x = sqrt 3

square it,
what do u get?

Answer 3

\[x^2+1/x^2 = 1\] @ hartnn

Answer 4

thats correct :)
now square it again

Answer 5

\[x^4+1/x^4=-1,x^8+x^8=-3\]

Answer 6

-sqrt(3)

Answer 7

x+1/x = sqrt 3
cube it
what do u get?

Answer 8

sry \[x^8+1/x^8=-3\]

Answer 9

b.) -sqrt(3)

Answer 10

http://www.wolframalpha.com/input/?i=%281%2F2+%28sqrt%283%29%2Bi%29%29%5E7%2B1%2F%281%2F2+%28sqrt%283%29%2Bi%29%29%5E7

Answer 11

lets see whether x^8+1/x^8 is required

but x^4+1/x^4 was required
and so is x^3+1/x^3

Answer 12

klied, aren't you enthusiastic on how we get the answer? :)

Answer 13

^_^. i am.

Answer 14

very good :)
lets see what vijay gets for
x^3+1/x^3

Answer 15

x^3+1/x^3 =3sqrt(3)-6

Answer 16

how?

Answer 17

x^3+x+1/x+1/x^3 = 3 sqrt 3

x^3+sqrt 3 +1/x^3 = 3 sqrt 3

so,
x^3+1/x^3 = 2 sqrt 3

how else have you done?

Answer 18

Sry i made a mistake]

Answer 19

no prob :)
lets move ahead.

we have x^4+1/x^4 = -1
and x^3+a/x^3 = 2 sqrt 3

and we want x^7+1/x^7

what do you suggest we should do? :)

Answer 20

***1/x^3

Answer 21

i dont have an idea how to proceed further

Answer 22

ok, just multiply these 2 equations together!

x^4+1/x^4 = -1
x^3+1/x^3 = 2 sqrt 3

what do u get?

Answer 23

i think i made a mistake in finding x^3+1/x^3 :P

Answer 24

I got it but x^3+1/x^3 = 1 not 2sqrt3

Answer 25

x^3+3x+3/x+1/x^3 = 3 sqrt 3

x^3+3sqrt 3 +1/x^3 = 3 sqrt 3

oh, how did u get 1 ?

Answer 26

and you got it for x^7+1/x^7 ?

Answer 27

yea can u help in this question

Answer 28

bdw, x^3+1/x^3 = 0
and let me try the next question :)

Answer 29

the long way :
let
z1= x1 + i y1
z2 = x2 +i y2
z3 = x3 +i y3

so, x1 + x2 + x3 = 0 , y1+y2+y3 = 0
x1^2 +y1^2 = x2^2 +y2^2 = x3^2 +y3^2 = 1/3

then try to evaluate each term
|z2 -z3| = |x2 -x3 + i (y2 -y3)|
= sqrt [ (x2-x3)^2 + (y2-y3)^2 ]

i hope theres a shorter way

Answer 30

Ok i will try it out can u guess how to proceed the next question

Answer 31

\(\Large \sqrt{1+i\sqrt3 }=x+iy \)
then square both sides...

Answer 32

1+i sqrt3 =x^2-y^2+2xyi

Answer 33

or converting it to trigonometric form is way easier,

1+ i sqrt 3 = 2 [1/2 + i sqrt 3/2] = 2 [1 <60 ]

Answer 34

and then taking its sqrt

sqrt 2 [1 <30]

note : while taking square root of complex number r < theta
we get
sqrt r < theta/2

Answer 35

bring back 1<30 in x+iy form :)

Answer 36

[1 <30] what is this i cant get

Answer 37

1< 30 = cos 30 + i sin 30

Answer 38

< = "angle"

Answer 39

ok continue

Answer 40

\(\Large 1 \angle 30 = \cos 30 + i\sin 30 = \sqrt 3/2 + i/2\)

Answer 41

multiply that by sqrt 2, and you're done :)

Answer 42

thnx can u tell how to proceed in this last problem

Answer 43

that tests how well you know your trigonometry

arg (x+iy ) = arctan (y/x)

arg (1-sina+icos a) = arctan (cos a/(1-sin a))

Answer 44

use double angle formulas for

cos a and 1-sin a

Answer 45

can u proceed forward

Answer 46

cos 2x = cos^2 x - sin^2 x = (cos x + sin x)(cos x -sin x)

1-sin2x = cos^2x -2 sin x cos x + sin^2 x = (cos x - sin x)^2

so,
cos 2x /(1-sin 2x) = (cos x + sin x) / (cos x - sin x)

here, a =2x, so x = a/2

Answer 47

should i still proceed further or you can take it from here?

Answer 48

how to get the answer ?

Answer 49

by simplifying further and using tan (x+pi/4 ) formula

Answer 50

can u proceed a little further

Answer 51

sure,

(cos x + sin x) / (cos x - sin x) = (1+ tan x)/ (1- tan x)

also,
tan (x+pi/4) = (1+tan x tan pi/4)/(1- tan x tan pi/4) = (1+tan x)/(1-tan x)

Answer 52

I got it Thanks a lot hartnn. I wont forget you

Answer 53

welcome vijay ^_^
always happy to help :)