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 C n 3^n 1/3^n 1 1/2 (z^n 1)=(z 1)(z z_1)(z z_2).....(z z_n 1) (1) Differentiating w.r.t. x, and then dividing by (1), we have ...

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Byju's Answer

Standard XII

Mathematics

Modulus of a Complex Number

1, z1, z2, z3...

Question

$1, z_{1}, z_{2}, z_{3}, . . . . ., z_{n - 1}$ are the nth roots of unity, then the value of $\frac{1}{(3 - z_{1})} + \frac{1}{(3 - z_{2})} + . . . . . + \frac{1}{(3 - z_{n - 1})}$ 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 C $\frac{n 3^{n - 1}}{3^{n} - 1} - \frac{1}{2}$
$(z^{n} - 1) = (z - 1) (z - z_{1}) (z - z_{2}) . . . . . (z - z_{n - 1}) (1)$
Differentiating w.r.t. x, and then dividing by (1), we have
$\frac{n z^{n - 1}}{z^{n} - 1} = \frac{1}{z - 1} + \frac{1}{z - z_{1}} + \frac{1}{z - z_{2}} + . . . . . + \frac{1}{z - z_{n - 1}}$
Putting $z = 3$ , we get
$\frac{n 3^{n - 1}}{3^{n} - 1} = \frac{1}{2} + \frac{1}{3 - z_{1}} + \frac{1}{3 - z_{2}} + . . . . . + \frac{1}{3 - z_{n - 1}}$
$\Rightarrow \frac{1}{3 - z_{1}} + \frac{1}{3 - z_{2}} + . . . . + \frac{1}{3 - z_{n - 1}} = \frac{n 3^{n - 1}}{3^{n} - 1} - \frac{1}{2}$

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1, z_{1}, z_{2}, . . . z_{n - 1}

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1,z1,z2,z3,.....,zn−1 are the nth roots of unity, then the value of 1(3−z1)+1(3−z2)+.....+1(3−zn−1) is equal to

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