Mht Cet
Complex Numbers
MHT CET 2024 3rd May Evening Shift
MCQ+2 / -02024
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 $z=w=$ complex cube root of unity
Mht Cet
MHT CET 2024 3rd May Evening Shift
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 $z=w=$ complex cube root of unity