Let z1 and z2 be the nt h roots of unity which are ends of a line segment that subtend a right angle at the origin. Then, n must be of the form

Question

Let z1 and z2 be the nt h roots of unity which are ends of a line segment that subtend a right angle at the origin. Then, n must be of the form (A) 4k+1 (B) 4k+2 (C) 4k+3 (D) 4k

Tardigrade Admin · Accepted Answer

We know that 1(1/n)=cos (2 r π /n)+isin â¡ (2 r π /n),r=0,1, ldots ldots ,n-1 Let, z1=cos (2 r1 π /n)+ isin â¡ (2 r1 π /n) And, z2=cos (2 r2 π /n)+isin â¡ (2 r2 π /n) . Then, angle z1oz2=amp((z1/z2))=amp(z1)-amp(z2) ⇒ amp(z1)-amp(z2)=(2 (r1 - r2) π /n)=(π /2) ∴ n=4 (r1 - r2)=4× integer. So, n is of the form 4k .

Q. Let $z_{1}$ and $z_{2}$ be the $n^{t h}$ roots of unity which are ends of a line segment that subtend a right angle at the origin. Then, $n$ must be of the form

$4 k + 1$

$4 k + 2$

$4 k + 3$

$4 k$

Q. Let z1​ and z2​ be the nth roots of unity which are ends of a line segment that subtend a right angle at the origin. Then, n must be of the form

4k+1

4k+2

4k+3

4k

Q. Let $z_{1}$ and $z_{2}$ be the $n^{t h}$ roots of unity which are ends of a line segment that subtend a right angle at the origin. Then, $n$ must be of the form

$4 k + 1$

$4 k + 2$

$4 k + 3$

$4 k$