Complex Numbers

Concept:The roots of $x^2+x+1=0$ are $\omega$ and $\omega^2$, the complex cube roots of unity, satisfying $\omega^3=1$, $1+\omega+\omega^2=0$, and $\omega^2=\overline{\omega}$.Similarly, $x^2-x+1=0$ has roots $-\omega$ and $-\omega^2$.For $x^2+x-1=0$, use the relation $x+1 = \frac{1}{x}$ derived from the equation.Explanation:(P) Let $\alpha=\omega$, $\beta=\omega^2$. Then $\alpha+1=1+\omega=-\omega^2$ and $\beta+1=1+\omega^2=-\omega$.$(\alpha+1)^{2026}=(-\omega^2)^{2026}=\omega^{4052}=(\omega^3)^{1350}\cdot\omega^2=1\cdot\omega^2=\omega^2$. So $\frac{1}{(\alpha+1)^{2026}}=\frac{1}{\omega^2}=\omega$.$(\beta+1)^{2026}=(-\omega)^{2026}=\omega^{2026}=(\omega^3)^{675}\cdot\omega=1\cdot\omega=\omega$. So $\frac{1}{(\beta+1)^{2026}}=\frac{1}{\omega}=\omega^2$.The roots $\omega$ and $\omega^2$ satisfy $x^2+x+1=0$. Hence (P) matches (1).(Q) Using the previous result: $\frac{1}{(\alpha+1)^{2027}}=\frac{1}{(\alpha+1)^{2026}\cdot(\alpha+1)}=\frac{1}{\omega^2\cdot(-\omega^2)}=-\frac{1}{\omega^4}=-\frac{1}{\omega}=-\omega^2$.Similarly, $\frac{1}{(\beta+1)^{2027}}=\frac{1}{\omega\cdot(-\omega)}=-\frac{1}{\omega^2}=-\omega$.The roots $-\omega$ and $-\omega^2$ satisfy $x^2-x+1=0$. Hence (Q) matches (2).(R) For $x^2-x+1=0$, roots are $\gamma=-\omega$, $\delta=-\omega^2$.$\gamma-1=-\omega-1=-(\omega+1)=\omega^2$ and $\delta-1=-\omega^2-1=-(\omega^2+1)=\omega$.$(\gamma-1)^{2026}=(\omega^2)^{2026}=\omega^{4052}=\omega^2$, so $\frac{1}{(\gamma-1)^{2026}}=\frac{1}{\omega^2}=\omega$.$(\delta-1)^{2026}=\omega^{2026}=\omega$, so $\frac{1}{(\delta-1)^{2026}}=\frac{1}{\omega}=\omega^2$.Sum = $\omega+\omega^2=-1$. Hence (R) matches (4).(S) For $x^2+x-1=0$, roots $p$ and $r$ satisfy $p+r=-1$, $pr=-1$.From the equation, $x(x+1)=1$, so $x+1=\frac{1}{x}$. Then $\frac{1}{(p+1)^3}=p^3$ and $\frac{1}{(r+1)^3}=r^3$.Sum $p^3+r^3=(p+r)^3-3pr(p+r)=(-1)^3-3(-1)(-1)=-1-3=-4$.Hence (S) matches (5).Thus matches: P→1, Q→2, R→4, S→5.Answer:The correct option is C: (P)→(1), (Q)→(2), (R)→(4), (S)→(5).