If 1- z 1- z 2- - z n -1 are the roots of z n -1-0- then 1-3- z 1-1-3- z 2-1-3- z 3-1-3- z n -1 is equal toA- -r-1n-1 r - 3r-1-r-1n 3r-1B- -r-1n r - 3r-1-r-1n 3r-1n - 3n-1C- n - 3 r - n -1- n -3 n -1-1-2D- n - 3n1-3n-1-1-2

1,z_1,z_2,...,z_n 1 are the roots of z^n 1=0 ⇒(z 1)(z z_1)(z z_2)...(z z_n 1)=0 ⇒1+z+z^2+...+z^n 1=0 Replace 13 z=y⇒ z=3y 1y 1+3y 1y+(3y 1y)^2+...+(3y 1y)^n ...

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Standard XII

Mathematics

Properties of nth Root of a Complex Number

If 1, z 1, z ...

Question

If $1, z_{1}, z_{2}, . . ., z_{n - 1}$ are the roots of $z^{n} - 1 = 0,$ then $\frac{1}{3 - z_{1}} + \frac{1}{3 - z_{2}} + \frac{1}{3 - z_{3}} + . . . + \frac{1}{3 - z_{n - 1}}$ is equal to

\frac{n - 1 \sum r = 1 r \cdot 3^{r - 1}}{n \sum r = 1 3^{r - 1}}

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\frac{n \sum r = 1 r \cdot 3^{r - 1}}{n \sum r = 1 3^{r - 1}}

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

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

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Solution

The correct options are
A $\frac{n - 1 \sum r = 1 r \cdot 3^{r - 1}}{n \sum r = 1 3^{r - 1}}$
C $\frac{n \cdot 3^{n - 1}}{3^{n} - 1} - \frac{1}{2}$
$1, z_{1}, z_{2}, . . ., z_{n - 1}$ are the roots of $z^{n} - 1 = 0$
$\Rightarrow (z - 1) (z - z_{1}) (z - z_{2}) . . . (z - z_{n - 1}) = 0$

$\Rightarrow 1 + z + z^{2} + . . . + z^{n - 1} = 0$

Replace $\frac{1}{3 - z} = y \Rightarrow z = \frac{3 y - 1}{y}$

$1 + \frac{3 y - 1}{y} + {(\frac{3 y - 1}{y})}^{2} + . . . + {(\frac{3 y - 1}{y})}^{n - 1} = 0$

Multiply by $y^{n - 1}$
$\Rightarrow y^{n - 1} + y^{n - 2} (3 y - 1) + y^{n - 3} (3 y - 1)^{2} + . . . + (3 y - 1)^{n - 1} = 0$

$\Rightarrow y^{n - 1} (1 + 3 + 3^{2} + . . . + 3^{n - 1}) - y^{n - 2} (1 + 2 \cdot 3 + 3 \cdot 3^{2} + 4 \cdot 3^{3} + . . . + (n - 1) 3^{n - 2}) + . . . + (- 1)^{n - 1} = 0$

We know that, sum of roots $= - \frac{b}{a}$
$∴ \frac{1}{3 - z_{1}} + \frac{1}{3 - z_{2}} + \frac{1}{3 - z_{3}} + . . . + \frac{1}{3 - z_{n - 1}} = \frac{1 + 2 \cdot 3 + 3 \cdot 3^{2} + 4 \cdot 3^{3} + . . . + (n - 1) 3^{n - 2}}{1 + 3 + 3^{2} + . . . + 3^{n - 1}}$
$= \frac{n - 1 \sum r = 1 r \cdot 3^{r - 1}}{n \sum r = 1 3^{r - 1}}$

Let $S = n - 1 \sum r = 1 r \cdot 3^{r - 1}$
$S = 1 + 2 \cdot 3 + 3 \cdot 3^{2} + 4 \cdot 3^{3} + . . . + (n - 1) 3^{n - 2}$
$3 S = 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + . . . + (n - 2) 3^{n - 2} + (n - 1) 3^{n - 1}$
Subtracting above equations, we get
$- 2 S = \frac{3^{n - 1} - 1}{2} - (n - 1) 3^{n - 1}$
$\Rightarrow S = \frac{n \cdot 3^{n - 1}}{2} - \frac{3^{n} - 1}{4}$

$S^{'} = n \sum r = 1 3^{r - 1} = \frac{3^{n} - 1}{2}$

Required Answer=
$= \frac{n \cdot 3^{n - 1}}{3^{n} - 1} - \frac{1}{2}$

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Similar questions

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

Q. If

1, z_{1}, z_{2}, z_{3}, \dots z_{n - 1}

are

n

roots of unity then the value of

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

is equal to :

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

are the n 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

1, z_{1}, z_{2}, z_{3}, . . ., z_{n - 1}

are the

n

th 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

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

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If 1,z1,z2,...,zn−1 are the roots of zn−1=0, then 13−z1+13−z2+13−z3+...+13−zn−1 is equal to

If $1, z_{1}, z_{2}, . . ., z_{n - 1}$ are the roots of $z^{n} - 1 = 0,$ then $\frac{1}{3 - z_{1}} + \frac{1}{3 - z_{2}} + \frac{1}{3 - z_{3}} + . . . + \frac{1}{3 - z_{n - 1}}$ is equal to