Question

Consider a complex number $z =\left(\cos \frac{\pi}{5}+i \sin \frac{\pi}{5}\right)$ and a set $A =\left\{ z _{ k } \mid z _{ k }= z ^{ k }, k =0,1,2,3, \ldots \ldots 9\right\}$, then

• A. all elements of A lie on a unit circle centred at origin.
• B. $\arg \left(z_p\right)-\arg \left(z_q\right)=\frac{2 \pi}{5}$ for exactly 8 ordered pairs of $\left(z_p, z_q\right)$ where $z_p, z_q \in A$ and $\arg z \in[0,2 \pi)$.
• C. number of elements of $A$ for which $\operatorname{Re}\left( z _{ k }\right)>\sin 18^{\circ}$ is 3 .
• D. number of elements of A for which $\operatorname{Im}\left(z_k\right)+\sin 36^{\circ}>0$ is 6 .

Answer 1

Answer: (A), (B), (C), (D)

(A) (A) is obivious
(B) $\left( z _2, z _0\right)\left( z _3, z _1\right)\left( z _4, z _2\right)\left( z _5, z _3\right),\left( z _6, z _4\right)\left( z _7, z _5\right)\left( z _8, z _6\right)\left( z _9, z _7\right)$ 8 ordered pairs

(C) $z _0, z _1, z _9$ are 3 elements of $A$ for which $\operatorname{Re}( z )>\sin 18^{\circ}$.
(D) $\operatorname{Im}\left( z _{ k }\right)>-\sin 36^{\circ} \Rightarrow z _0, z _1, z _2, z _3, z _4, z _5$ are required elements of