Question

Match the elements of List 1 with the elements of List 2:

Answer 1

Element A:
Expression ${(x^n+\frac{1}{x^n})}^2=(x^2{)}^n+\frac{1}{(x^2{)}^n}+2$,........... {1}
$x$ is the root of the quadratic equation: $x^2=x+1=0$
$\therefore x=\frac{1\pm i\sqrt{3}}{2}$
Now, $x^2=\frac{-1\pm i\sqrt{3}}{2}$, which is same as cube roots of unity, i.e. $\omega$ and ${\omega }^2$
Substituting $x^2=\omega$ in {1}, we get
${\omega }^n+{\left(\frac{1}{\omega }\right)}^n+2={\omega }^n+({\omega }^2{)}^n+2$,........... $(\because \frac{1}{\omega }={\omega }^2)$
$\sum _{n=1}^5{(x^n+\frac{1}{x^n})}^2=\sum _{n=1}^5({\omega }^n+({\omega }^2{)}^n+2)=8$........$(\because 1+\omega +{\omega }^2=0,$ and ${\omega }^3=1)$
Element B:
Expression, ${\left[\frac{1+\cos \theta +i\sin \theta }{\sin \theta +i(1+\cos \theta )}\right]}^4={\left[\frac{2\sin \theta (1+\cos \theta )+i({\sin }^2\theta -(1+\cos \theta {)}^2)}{2(1+\cos \theta )}\right]}^4$
$\Rightarrow {[\sin \theta +i\frac{({\sin }^2\theta -(1+\cos \theta {)}^2)}{2(1+\cos \theta )}]}^4=(\sin \theta -i\cos \theta {)}^4$
$\Rightarrow \cos 4\theta +i\sin 4\theta =\cos n\theta +i\sin n\theta$
$\therefore n=4$
Elecment C:
Let $z=x+iy$, then $\overline{z}=x-iy$
$\frac{Im\left(z\right)}{Re\left(z\right)}=\frac{y}{x}=tan\theta =\sqrt{2}-1,$..... $(\because ta{n}^{-1}\left(\frac{Im\left(z\right)}{Re\left(z\right)}\right)=\theta )$
$\therefore \theta ={22.5}^o$
But $z$ and $\overline{z}$ are adjacent vertices of a regular polygon, so angle between two adjacent vertices is ${45}^o$
$\therefore$ Total number of sides, $n=8$
$\frac{n}{4}=2$
Element D:
Expression $(r-\omega )(r-{\omega }^2)=r^2-r(\omega +{\omega }^2)-{\omega }^3$
$\Rightarrow r^2+r-1$........$(\because 1+\omega +{\omega }^2=0,$ and ${\omega }^3=1)$
$\therefore \frac{1}{50}\left\{\sum _{r=1}^{10}(r-\omega )(r-{\omega }^2)\right\}=\frac{1}{50}\left\{{\sum }_{r=1}^{10}\right(r^2+r-1\left)\right\}=\frac{450}{50}=9$