Question

Prove that ω^248+ω^154+5i^400=4
Note:
(i=√-1 and ω is the complex cube root of unity)

Answer 1

whats w^248? can u calculate that?? knowing w^3=1???

Answer 2

i think it will break into parts (ω^248)

Answer 3

....tell me whats w^6?

Answer 4

sorry,i did'nt get you

Answer 5

okk,as w^3 = 1
w^6=w^3*w^3=1*1=1
so any power of w which is multiple of 3 must equal 1 !!!
so,w^246=1
so w^248=w^246*w^2=w^2
got this???

Answer 6

there is an example on breaking in my book:
ω^32 = (ω^3)^10.ω^2
can i do this by above example

Answer 7

yes,thats what i did,i break w^248 as (w^3)^82.w^2 to get w^2 only
u got that ???

Answer 8

ok

Answer 9

so,can u do same thing for w^154??

Answer 10

yes

Answer 11

then tell me what w^154 equals?

Answer 12

ω^154=ω^153.ω
(ω^3)^51.ω
ω

Answer 13

sorry i am slow in typing

Answer 14

correct,so now your 1st 2 terms are w+w^2
to find that from 1+w+w^2=0 is easy....

Answer 15

but how will 5i^400 be solved

Answer 16

i^400 = (i^4)^100 = 1

Answer 17

ω^248+ω^154+5i^400
= ω + ω^2 + 5(1)
= ω + ω^2 + 1 + 4
= 1+ω + ω^2 + 4
= ...

Answer 18

5i^400= 5(-1)^400 = 5
so,
w^2+w+5
-1+5=4
am i right

Answer 19

yup :)

Answer 20

5i^400 = 5(√-1)^400

Answer 21

how will 5(√-1)^400 be solved

Answer 22

i=swrt(-1)
so i^2=1
so i^400=(i^2)^200=1^200=1

Answer 23

i= \(\sqrt{-1}\)
\(i^2 = (\sqrt{-1})^2 = -1\)
\(i^3 = (\sqrt{-1})^3= -1i = -i\)
\(i^4 = (\sqrt{-1})^4=(-1)^2 = 1\)

Answer 24

i^2 is -1....

Answer 25

sorry,
i=sqrt(-1)
so i^2=-1
so i^4=1
so i^400=(i^4)^100=1^100=1

Answer 26

what??
w^2+w+1=0
we have to prove that it equals 4

Answer 27

we already did,1st 2 terms = w+w^2 =-1
3rd term =5 i^400=5*(i^4)^100=5*(1^100)=5
5-1=4

Answer 28

oh i forgot about 5 :D thanks

Answer 29

5*(i^4)^100=5*(1^100)
how did i^4 become 1

Answer 30

I did explain above.

i^4 = (i^2)^2 = (-1)^2 = 1

i^2 = \(\sqrt{-1} \times \sqrt{-1} = -1\)