Let z_k=cos left(frac{2 k pi}{10}right)+ i si | vedclass

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Question

$(P)$ $z_k z_j=1 \quad \Rightarrow \quad z_1=z_{10-k}$

Hence for each $k \in\{1,2,3, \ldots, 9\}$ there exists $z_1$ such that $z_k$. $z_1=1$ True

Vedclass · Accepted Answer

$\quad 1 \quad 2 \quad 3 \quad 4 $

Vedclass · Answer

$(P)$ $z_k z_j=1 \quad \Rightarrow \quad z_1=z_{10-k}$

Hence for each $k \in\{1,2,3, \ldots, 9\}$ there exists $z_1$ such that $z_k$. $z_1=1$ True

$(Q)$ $z_1 \cdot z=z_k \quad \Rightarrow \quad z=z_{k-1}$ for $k=2,3,4, \ldots, 9$ 

$z =1$ for $k =1$ False

$(R)$ $z_1, z_2, \ldots, z_9$ are roots of the equation $z^{10}=1$ other then unity, hence

$\frac{z^{10}-1}{z-1}=1+z+\ldots+z^9=\left(z-z_1ight)\left(z-z_2ight) \ldots\left(z-z_9ight)$

Substituting $z=1$, we get $\frac{\left(1-z_1ight)\left(1-z_2ight) \ldots\left(1-z_9ight)}{10}=\frac{10}{10}=1$

$(S)$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}ight)=1-\left\{ight.$ sum of real parts of roots of $z^{10}=1$ except 1$\}$ $=1-(-1)=2$

$\left(	ext { as } 1+z_1+z_2+\ldots+z_9=0ight) \Rightarrow \quad \sum \operatorname{Re}\left(z_kight)+1=0$

List $I$	List $II$
$P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$	$1.$ True
$Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers.	$2.$ False
$R.$ $\frac{\left\|1-z_1\right\|\left\|1-z_2\right\| \ldots . .\left\|1-z_9\right\|}{10}$ equals	$3.$ $1$
$S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals	$4.$ $2$

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