Question

can anybody solve this : if 1,a1,a2....a(n-1) are nth roots of unity,then the value of (1-a1) (1-a2)(1-a3)...(1-a(n-)) is equal to ?

Answer 1

Hello, @Reeti1 :)

Answer 2

(:

Answer 3

HI

Answer 4

What's up?

Answer 5

NM ! YOU SAY ?
your asl ?

Answer 6

14
I'm a boy
England

Now, let's get started on this question :)

Answer 7

ahh this is 12th grade mathematics dude ;)

Answer 8

Well, if you say so :3

Answer 9

can you solve it ?

Answer 10

Of course I can :P

Answer 11

plaese move then

Answer 12

Moving...

Answer 13

move on*

Answer 14

Okay, all the nth roots of unity are roots of this polynomial...
\[\huge x^n - 1 = 0\]

Answer 15

OKAY

Answer 16

Consequently, this polynomial is equal to...
\[\LARGE (x-1)(x-a_1)(x-a_2)...(x-a_{n-1})=0\]

Answer 17

But also
\[\LARGE x^n - 1 = (x-1)(x^{n-1}+x^{n-2}+...x+1)\]

Answer 18

Sorry
\[\Large (x-a_1)(x-a_2)...(x-a_{n-1})=x^{n-1}+x^{n-2}+...x+1\]

Answer 19

Is everything clear? @Reeti1

Answer 20

the choices are :
a. 1
b. 1/2
c. n

Answer 21

Yeah yeah... but everything is clear to you so far?
How we arrived at this...
\[\Large (x-a_1)(x-a_2)...(x-a_{n-1})=x^{n-1}+x^{n-2}+...x+1\]

Answer 22

answer is n i guess

Answer 23

YES!
See?
Just replace x with 1
\[\Large (1-a_1)(1-a_2)...(1-a_{n-1})=1^{n-1}+1^{n-2}+...1+1\]

Answer 24

And you get n 1's that you'll add to get... n
haha
did I solve it or what? ^.^

Answer 25

you did it ! but wow how ?
you're just 14

Answer 26

I'm actually well over 200?
^.^
@dmezzullo already said on chat... age doesn't matter XD

Answer 27

thank you :)

Answer 28

;)

Answer 29

ok one more

Answer 30

Let's go :)

Answer 31

cube root of i is equal to ?

Answer 32

cube root of i?
:(
\[\huge \sqrt[3]i\]

Answer 33

yep

Answer 34

is that even possible? :/

Answer 35

haha of course it is :D

Answer 36

First, put i in polar form ^.^

Answer 37

okay done