Question -10

Algebra Level 3

If $1,a_1,a_2,\ldots,a_{n-1}$ are the $n$ -th roots of unity, then find the value of $(2-a_1)\cdot(2-a_2)\cdot(2-a_3)\cdots(2-a_{n-1}).$

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2^n-1

2^n+1

2^n

1 solution

Anandhu Raj
Mar 1, 2015

We know that ,

$\displaystyle { z }^{ n }-1=\sum _{ r=1 }^{ n-1 }{ (z- } { a }_{ r })=(z-1)(z-{ { a }_{ 1 } })(z-{ a }_{ 2 }) \dots (z-{ a }_{ n-1 })$

Put $z=2$ to get,

$\displaystyle \Rightarrow { 2 }^{ n }-1=\sum _{ r=1 }^{ n-1 }{ (2- } { a }_{ r })=(2-1)(2-{ { a }_{ 1 } })(2-{ a }_{ 2 }) \dots (2-{ a }_{ n-1 })$

$\displaystyle \Rightarrow \boxed{ (2-{ { a }_{ 1 } })(2-{ a }_{ 2 }).............(2-{ a }_{ n-1 }) = { 2 }^{ n }-1}$

JEE MAIN METHOD:

let n=1 and 2 and check the options, (problem can be solved within a minute ) i m a bit laxy so im not writing d process

prajwal kavad - 6 years, 2 months ago

How do we reach up to the first relation?

Yogesh Verma - 6 years, 3 months ago

Sir, Please check the wikipedia page on "Root of unity".

Anandhu Raj - 6 years, 3 months ago

@Anandhu Raj and @Sandeep Bhardwaj if I take n=3, the answer of (2-a1) (2-a2) is coming 3, which should be 7 according to given answer, can you please help me with this.

Bhargav Upadhyay - 6 years, 3 months ago

If you take $n=3$ , It means $a_1,a_2$ are non-real cube roots of unity i.e. $\omega, \ and \ \omega^2$ .

So, $(2-\omega)(2-\omega^2)=4-2 \cdot(\omega+\omega^2)+\omega^3=4+2+1=7.$

Sandeep Bhardwaj - 6 years, 3 months ago

Now its clear, Thanks :-)

Bhargav Upadhyay - 6 years, 3 months ago

@Sandeep Bhardwaj https://brilliant.org/problems/find-the-areaonly-your-logic-can-help-you/?group=3UHxOzwinQpA

Yash Sharma - 6 years, 3 months ago

@Bhargav Upadhyay Now you get it,right?

Anandhu Raj - 6 years, 3 months ago

Yes, now it's clear!

Bhargav Upadhyay - 6 years, 3 months ago

Question -10

Looking for Top Rank in JEE? Alright, then go for this set : Expected JEE-Mains-2015-B .

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