Complex Numbers Ques 94
Let $z _k=\cos (\frac{2 k \pi}{10})+i \sin (\frac{2 k \pi}{10}) ; k=1,2, \ldots 9$.
| Column I | Column II | |
|---|---|---|
| P. $\quad$ For each $z _k$, there exists a $z _j$ such that $z _k \cdot z _j=1$ |
(i) True | |
| Q. $\quad$ There exists a $k \in{1,2, \ldots, 9}$ such that $z _1 \cdot z=z _k$ has no solution $z$ in the set of complex numbers |
(ii) False | |
| R. $\quad \frac{\left|1-z _1\right|\left|1-z _2\right| \ldots\left|1-z _9\right|}{10}$ equal | (iii) 1 | |
| S. $\quad 1-\sum _{k=1}^{9} \cos \frac{2 k \pi}{10}$ equals | (iv) 2 |
Codes
| P | Q | R | S | |
|---|---|---|---|---|
| (a) | (i) | (ii) | (iv) | (iii) |
| (b) | (ii) | (i) | (iii) | (iv) |
| (c) | (i) | (ii) | (iii) | (iv) |
| (d) | (ii) | (i) | (iv) | (iii) |
Show Answer
Answer:
Correct Answer: 94.(c)
Solution:
Formula:
Steps to determine nth roots of a complex number:
- (P) PLAN $e^{i \theta} \cdot e^{i \alpha}=e^{i(\theta+\alpha)}$
Given $\quad z _k=e^{i \frac{2 k \pi}{10}} \Rightarrow z _k \cdot z _j=e^{i (\frac{2 \pi}{10})(k+j)}$
$z _k$ is 10 th root of unity.
$\Rightarrow \bar{z} _k$ will also be 10th root of unity.
Taking, $z _j$ as $\bar{z} _k$, we have $z _k \cdot z _j=1$ (True)
(Q) PLAN
$\begin{aligned} \frac{e^{i \theta}}{e^{i \alpha}} & =e^{i(\theta-\alpha)} \\ z & =z _k / z _1=e^{i (\frac{2 k \pi}{10}-\frac{2 \pi}{10})}=e^{i \frac{\pi}{5}(k-1)} \end{aligned} $
For $k=2 ; z=e^{i \frac{\pi}{5}}$ which is in the given set (False)
(R) PLAN
(i) $1-\cos 2 \theta=2 \sin ^{2} \theta$
(ii) $\sin 2 \theta=2 \sin \theta \cos \theta$ and
$ \text { (i) } \cos 36^{\circ}=\frac{\sqrt{5}-1}{4} $
(ii) $\cos 108^{\circ}=\frac{\sqrt{5}+1}{4} \frac{\left|1-z _1\right|\left|1-z _2\right| \ldots\left|1-z _9\right|}{10}$
NOTE $\left|1-z _k\right|=|1-\cos \frac{2 \pi k}{10}-i \sin \frac{2 \pi k}{10}|$
$ =|2 \sin \frac{\pi k}{10}| \quad |\sin \frac{\pi k}{10}-i \cos \frac{\pi k}{10}|=2\left|\sin \frac{\pi k}{10}\right| $
Now, required product is
$ \begin{aligned} & \frac{2^{9} \sin \frac{\pi}{10} \cdot \sin \frac{2 \pi}{10} \cdot \sin \frac{3 \pi}{10} \ldots \sin \frac{8 \pi}{10} \cdot \sin \frac{9 \pi}{10}}{10} \\ & =\frac{2^{9} (\sin \frac{\pi}{10} \sin \frac{2 \pi}{10} \sin \frac{3 \pi}{10} \sin \frac{4 \pi}{10}){ }^{2} \sin \frac{5 \pi}{10}}{10} \\ & =\frac{2^{9} (\sin \frac{\pi}{10} \cos \frac{\pi}{10} \cdot \sin \frac{2 \pi}{10} \cos \frac{2 \pi}{10}){ }^{2} \cdot 1}{10} \end{aligned} $
$ \begin{aligned} & =\frac{2^{9} (\frac{1}{2} \sin \frac{\pi}{5} \cdot \frac{1}{2} \sin \frac{2 \pi}{5}){ }^{2}}{10} \\ & =\frac{2^{5}\left(\sin 36^{\circ} \cdot \sin 72^{\circ}\right)^{2}}{10} \\ & =\frac{2^{5}}{2^{2} \times 10}\left(2 \sin 36^{\circ} \sin 72^{\circ}\right)^{2} \\ & =\frac{2^{2}}{5}\left(\cos 36^{\circ}-\cos 108^{\circ}\right)^{2} \\ & =\frac{2^{2}}{5} \quad [(\frac{\sqrt{5}-1}{4})+(\frac{\sqrt{5}+1}{4})]^2=\frac{2^{2}}{5} \cdot \frac{5}{4}=1 \end{aligned} $
(S) Sum of $n$th roots of unity $=0$
$ \begin{aligned} 1+\alpha+\alpha^{2}+\alpha^{3}+\ldots+\alpha^{9} & =0 \\ 1+\sum _{k=1}^{9} \alpha^{k} & =0 \\ 1+\sum _{k=1}^{9} \cos \frac{2 k \pi}{10}+i \sin \frac{2 k \pi}{10} & =0 \\ 1+\sum _{k=1}^{9} \cos \frac{2 k \pi}{10} & =0 \\94-\sum _{k=1}^{9} \cos \frac{2 k \pi}{10} & =2 \end{aligned} $
$(P) \rightarrow (i), (Q) \rightarrow (ii),(R) \rightarrow (iii),(S) \rightarrow (iv)$
BETA
