Question

Let $1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$, be the $5^{\text {th }}$ roots of unity. If $A_1, A_2, A_3, A_4, A_5$ are vertices of a polygon formed by $1, \alpha_1, \alpha_2, \alpha_3, \alpha_4$, on complex plane then which of following statement(s) is/are Correct? (where $A_i A_j$ is length of line segment joining $A_i$ and $A_j$ )

• A. Product of distances $\left(A_1 A_2\right)\left(A_1 A_3\right)\left(A_1 A_4\right)\left(A_1 A_5\right)\left(A_2 A_3\right)\left(A_2 A_4\right)\left(A_2 A_5\right)\left(A_3 A_4\right)\left(A_3 A_5\right)\left(A_4 A_5\right)$ is equal to $(\sqrt{5})^5$
• B. $\left(A_1 A_2\right)^2+\left(A_1 A_3\right)^2+\left(A_1 A_4\right)^2+\left(A_1 A_5\right)^2=10$
• C. Area of the circle inscribed in the polygon is $\frac{\pi}{2}\left(1+\cos \frac{2 \pi}{5}\right)$
• D. $\left(1+\alpha_1\right)\left(1+\alpha_2\right)\left(1+\alpha_3\right)\left(1+\alpha_4\right)=0$

Answer 1

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

$\alpha^5=1$
Roots $1, \alpha_1, \alpha_2, \alpha_3, \alpha_5$ can be taken $1, \alpha, \alpha^2, \alpha^3, \alpha^4$

(A) $(A_1 A_2)(A_1 A_3)(A_1 A_4)(A_1 A_5)(A_2 A_3)(A_2 A_4)(A_2 A_5)(A_3 A_4)(A_3 A_5)(A_4 A_5)$

$=(|1-\alpha||1-\alpha^2||1-\alpha^3||1-\alpha^4|)(|\alpha-\alpha^2||\alpha-\alpha^3||\alpha-\alpha^4|)(|\alpha^2-\alpha^3||\alpha^2-\alpha^4|)(|\alpha^3-\alpha^4|) $
$=(|1-\alpha||1-\alpha^2|)^5=(\sqrt{5})^5$
(B) $A_1 A_r=|1-e^{i 2(r-1) \frac{\pi}{5}}|=|1-\cos \frac{2(r-1) \pi}{5}-i \sin \frac{2(r-1) \pi}{5}| $
$ =|2 \sin ^2 \frac{(r-1) \pi}{5}+2 i \sin \frac{(r-1) \pi}{5} \cos \frac{(r-1) \pi}{5}| $
$ =|-2 i \sin \frac{(r-1) \pi}{5}(\cos \frac{(r-1) \pi}{5}+i \sin \frac{(r-1) \pi}{5})|=2 \sin \frac{(r-1) \pi}{5}$
$\therefore A_1 A_2^2+A_1 A_3^2+A_1 A_4^2+A_1 A_5^2=4 \sin ^2 \frac{\pi}{5}+4 \sin ^2 \frac{2 \pi}{5}+4 \sin ^2 \frac{3 \pi}{5}+4 \sin ^2 \frac{4 \pi}{5}$
$=2[(1-\cos \frac{2 \pi}{5})+(1-\cos \frac{4 \pi}{5})+(1-\cos \frac{6 \pi}{5})+(1-\cos \frac{8 \pi}{5})]$
$=2[4-\frac{\cos \pi \sin \frac{4 \pi}{5}}{\sin \frac{\pi}{5}}]=10$
(C) $r=\cos \frac{\pi}{5}$
Area $=\pi \cos ^2 \frac{\pi}{5}=\frac{\pi}{2}[1+\cos \frac{2 \pi}{5}]$

(D) $x^5=1 \Rightarrow(x-1)\left(x-\alpha_1\right)\left(x-\alpha_2\right)\left(x-\alpha_3\right)\left(x-\alpha_4\right)=x^5-1$
Put $x=-1$ in above equation
$\Rightarrow(-2)(-1-\alpha_1)(-1-\alpha_2)(-1-\alpha_3)(-1-\alpha_4)=(-1)^5-1 $
$\Rightarrow(1+\alpha_1)(1+\alpha_2)(1+\alpha_3)(1+\alpha_4)=1$