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) $A_1{A}_r=|1-e^{i2(r-1)\frac{\pi }{5}}|=|1-\cos \frac{2(r-1)\pi }{5}-i\sin \frac{2(r-1)\pi }{5}|$
$=|2{\sin }^2\frac{(r-1)\pi }{5}+2i\sin \frac{(r-1)\pi }{5}\cos \frac{(r-1)\pi }{5}|$
$=|-2i\sin \frac{(r-1)\pi }{5}(\cos \frac{(r-1)\pi }{5}+i\sin \frac{(r-1)\pi }{5})|=2\sin \frac{(r-1)\pi }{5}$
$\therefore A_1{A}_2^2+A_1{A}_3^2+A_1{A}_4^2+A_1{A}_5^2=4{\sin }^2\frac{\pi }{5}+4{\sin }^2\frac{2\pi }{5}+4{\sin }^2\frac{3\pi }{5}+4{\sin }^2\frac{4\pi }{5}$
$=2[(1-\cos \frac{2\pi }{5})+(1-\cos \frac{4\pi }{5})+(1-\cos \frac{6\pi }{5})+(1-\cos \frac{8\pi }{5})]$
$=2[4-\frac{\cos \pi \sin \frac{4\pi }{5}}{\sin \frac{\pi }{5}}]=10$
(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
$\Rightarrow (-2)(-1-{\alpha }_1)(-1-{\alpha }_2)(-1-{\alpha }_3)(-1-{\alpha }_4)=(-1{)}^5-1$
$\Rightarrow (1+{\alpha }_1)(1+{\alpha }_2)(1+{\alpha }_3)(1+{\alpha }_4)=1$