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Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+ i \sin \left(\frac{2 k \pi}{10}\right) ; k =1,2, \ldots 9$.
| 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 | $4.$ $2$ |
Codes: $ \quad P \quad Q \quad R \quad S$
$\quad 1 \quad 2 \quad 4 \quad 3 $
$\quad 2 \quad 1 \quad 3 \quad 4 $
$\quad 1 \quad 2 \quad 3 \quad 4 $
$\quad 2 \quad 1 \quad 4 \quad 3 $
Solution
$(P)$ $z_k z_j=1 \quad \Rightarrow \quad z_1=z_{10-k}$
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
$\frac{z^{10}-1}{z-1}=1+z+\ldots+z^9=\left(z-z_1\right)\left(z-z_2\right) \ldots\left(z-z_9\right)$
Substituting $z=1$, we get $\frac{\left(1-z_1\right)\left(1-z_2\right) \ldots\left(1-z_9\right)}{10}=\frac{10}{10}=1$
$(S)$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)=1-\left\{\right.$ sum of real parts of roots of $z^{10}=1$ except 1$\}$ $=1-(-1)=2$
$\left(\text { as } 1+z_1+z_2+\ldots+z_9=0\right) \Rightarrow \quad \sum \operatorname{Re}\left(z_k\right)+1=0$
