Q98 / 150 · MHT CET 2024 (03 May Shift 2) · Mathematics

If $z^2+z+1=0$ then $\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2=$ where $\mathrm{z}=\mathrm{w}=$ complex cube root of unity

  • A
    4
  • B
    1
  • Correct answer
    C
    5
  • D
    2

Solution

Z is a complex cube root of unity $\therefore \quad z^3=1...(i)$ and $1+z+z^2=0$ $\therefore \quad z^2+1=-z$...(ii) Consider, $\begin{aligned} & \left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2 \\ & =\left(1+\frac{1}{1}\right)^2+\left(z^3 \cdot z+\frac{1}{z^3 \cdot z}\right)^2 \\ & =4+\left[z+\frac{1}{z}\right]^2 \end{aligned}$...[From (i)] $\begin{aligned} & =4+\left[\frac{z^2+1}{z}\right]^2 \\ & =4+\left[\frac{-z}{z}\right]^2 ...[From(ii)]\\ & =4+(-1)^2 =5 \end{aligned}$

Hence, the correct answer is option: 'C'.

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