Given, $$\mathrm{Z}_k=\cos \frac{2 k \pi}{10}+i \sin \frac{2 k \pi}{10}, k=1,2,3, \ldots, 9$$

$$\begin{aligned} & \text { (P) } Z_k \cdot Z_j=1 \\ & \Rightarrow\left(\cos \frac{2 k \pi}{10}+i \sin \frac{2 k \pi}{10}\right) \\ & \left(\cos \frac{2 j \pi}{10}+i \sin \frac{2 j \pi}{10}\right)=1 \end{aligned}$$

$$\begin{aligned} & \Rightarrow \cos \left(\frac{2 \pi(k+j)}{10}\right)+i \sin \frac{2 \pi(k+j)}{10}=1 \\ & \Rightarrow \cos \frac{2 \pi(k+j)}{10}=1 \text { and } \sin \frac{2 \pi(k+j)}{10}=0 \\ & \Rightarrow \quad \frac{2 \pi(k+j)}{10}=2 \lambda \pi \text { where } \lambda \in \text { Integer } \\ & \Rightarrow \quad k+j=10 \lambda \\ & \Rightarrow \quad j=10 \lambda-k \\ \end{aligned}$$

For every value of $$k$$ there is a possible value of $$j$$

Hence, $$P$$ match with 1.

(Q) Given $$Z_1 \cdot Z=Z_k$$

$$\begin{aligned} & \left(\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}\right) \cdot z=\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10} \\ & z=\frac{\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10}}{\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}} \\ & z=\cos \frac{2 \pi(k-1)}{10}+i \sin \frac{2 \pi(k-1)}{10} \end{aligned}$$

For every value of $$k$$, there is a unique value of $$z$$ is possible.

$$\therefore$$ The statement $$z_1 \cdot z=z_k$$ has no solution z in the set of complex numbers is false.

Hence, Q match with 2.

(R) Given, $$z_k=\cos \frac{2 \pi k}{10}+i \sin \frac{2 \pi k}{10}, k=1,2,3, \ldots 9$$

Here, $$z_1, z_2, z_3, \ldots z_9$$ are the roots of $$z^{10}-1=0$$

We know $$z^{10}-1=(z-1)$$

$$\begin{aligned} & \left(z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1\right) \\ & =(z-1)\left(z-z_1\right)\left(z-z_2\right) \ldots\left(z-z_9\right) \\ \Rightarrow & 1+z+z^2+\ldots+z^9=\left(z-z_1\right)\left(z-z_2\right) \\ & \left(z-z_3\right) \ldots\left(z-z_9\right) \\ \Rightarrow & \left|1+z+z^2+\cdots+z^9\right| \\ & =\left|\left(z-z_1\right)\left(z-z_2\right)\left(z-z_3\right) \ldots\left(z-z_9\right)\right| \\ \Rightarrow & \left|z-z_1\right| \cdot\left|z-z_2\right|\left|z-z_3\right| \ldots\left|z-z_9\right| \\ & =\left|1+z+z^2+\ldots+z^9\right| \end{aligned}$$

Put $$z=1$$

$$\begin{aligned} & \Rightarrow\left|1-z_1\right| \cdot\left|1-z_2\right|\left|1-z_3\right| \ldots\left|1-z_9\right|=10 \\ & \Rightarrow \frac{\left|1-z_1\right|\left|1-z_2\right|\left|1-z_3\right| \ldots\left|1-z_9\right|}{10}=1 \end{aligned}$$

Hence, R match with 3.

(S) We know the sum of all roots of $$z^{10}-1=0$$ is zero.

$$\begin{aligned} & \Rightarrow 1+z_1+z_2+z_3+\ldots+z_9=0 \\ & \Rightarrow 1+\left(\cos \frac{2 \pi}{10}+i \sin \frac{2 \pi}{10}\right) \\ & +\left(\cos \left(\frac{4 \pi}{10}\right)+i \sin \left(\frac{4 \pi}{10}\right)\right) \\ & +\left(\cos \frac{6 \pi}{10}+i \sin \frac{6 \pi}{10}\right) \\ & +\ldots+\cos \frac{18 \pi}{10}+i \sin \frac{18 \pi}{10}=0 \\ \end{aligned}$$

$$\begin{aligned} & \begin{array}{l} \Rightarrow\left(1+\cos \frac{2 \pi}{10}+\cos \frac{4 \pi}{10}+\cdots+\cos \frac{18 \pi}{10}\right) \\ +i\left(\sin \frac{2 \pi}{10}+\sin \frac{4 \pi}{10}+\ldots+\sin \frac{18 \pi}{10}\right) \\ =0+i 0 \end{array} \\ & \Rightarrow 1+\cos \frac{2 \pi}{10}+\cos \frac{4 \pi}{10}+\cos \frac{6 \pi}{10}+\ldots+\cos \frac{18 \pi}{10}=0 \\ & \Rightarrow \sum_{k=1}^9 \cos \frac{2 k \pi}{10}=-1 \\ & \Rightarrow 1-\sum_{k=1}^9 \cos \frac{2 k \pi}{10}=1-(-1)=2 \end{aligned}$$

Hint:

(i) Recall $$(\cos \theta+i \sin \theta)(\cos \phi+i \sin \phi)=\cos (\theta+\phi)+i \sin (\theta+\phi)$$

(ii) $$\cos \theta=0$$ when $$\theta=2 n \pi$$, where $$n$$ is an integer. and $$\sin \theta=0$$ when $$\theta=n \pi$$, where $$n$$ is an integer.

(iii) $$1, z_1, z_2, z_3, \ldots z_9$$ are the roots of $$z^{10}-1=0$$.

(iv) $$z^{10}-1=(z-1)\left(1+z+z^2+\ldots+z^9\right)$$

$$=(z-1)\left(z-z_1\right)\left(z-z_2\right) \ldots\left(z-z_9\right)$$