1- z 1- z 2- z 3- - z n -1 are the nth roots of unity- then the value of 1-3- z 1-1-3- z 2-1-3- z n -1 is equal toA- n - 3 n -1-3 n -1-1B- n - 3n-1-3n-1-1C- n - 3 n -1-3 n -1-1-2D- n - 3n-1-3n-1-1-2

The correct option is C n.3^n 1/3^n 1 1/2z^n 1 = (z 1)(z z_1)(z z_2) …… (z z_n 1) Differentiating w.r.t. X nz^n 1/z^n 1 = 1/z 1 + 1/z z_1 + …… + 1/z z_n 1 ...

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Byju's Answer

Standard XII

Mathematics

AM,GM,HM Inequality

1, z 1, z 2, ...

Question

$1, z_{1}, z_{2}, z_{3}, \dots \dots, z_{n - 1}$ are the nth roots of unity, then the value of $\frac{1}{3 - z_{1}} + \frac{1}{3 - z_{2}} + \dots \dots + \frac{1}{3 - z_{n - 1}}$ is equal to

\frac{n {.3}^{n - 1}}{3^{n - 1}} + \frac{1}{2}

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\frac{n {.3}^{n - 1}}{3^{n - 1}} - 1

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\frac{n {.3}^{n - 1}}{3^{n - 1}} + 1

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\frac{n {.3}^{n - 1}}{3^{n - 1}} - \frac{1}{2}

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Solution

The correct option is D $\frac{n {.3}^{n - 1}}{3^{n - 1}} - \frac{1}{2}$
$z^{n} - 1 = (z - 1) (z - z_{1}) (z - z_{2}) \dots \dots (z - z_{n - 1})$
Differentiating w.r.t. X
$\frac{n z^{n - 1}}{z^{n} - 1} = \frac{1}{z - 1} + \frac{1}{z - z_{1}} + \dots \dots + \frac{1}{z - z_{n - 1}}$

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\frac{1}{3 - z_{1}} + \frac{1}{3 - z_{2}} + \dots + \frac{1}{3 - z_{n - 1}}

is equal to :

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1,z1,z2,z3,……,zn−1 are the nth roots of unity, then the value of 13−z1+13−z2+……+13−zn−1 is equal to

The correct option is D n.3n−13n−1−12zn−1=(z−1)(z−z1)(z−z2)……(z−zn−1) Differentiating w.r.t. X nzn−1zn−1=1z−1+1z−z1+……+1z−zn−1

$1, z_{1}, z_{2}, z_{3}, \dots \dots, z_{n - 1}$ are the nth roots of unity, then the value of $\frac{1}{3 - z_{1}} + \frac{1}{3 - z_{2}} + \dots \dots + \frac{1}{3 - z_{n - 1}}$ is equal to

The correct option is D $\frac{n {.3}^{n - 1}}{3^{n - 1}} - \frac{1}{2}$
$z^{n} - 1 = (z - 1) (z - z_{1}) (z - z_{2}) \dots \dots (z - z_{n - 1})$
Differentiating w.r.t. X
$\frac{n z^{n - 1}}{z^{n} - 1} = \frac{1}{z - 1} + \frac{1}{z - z_{1}} + \dots \dots + \frac{1}{z - z_{n - 1}}$