MHT CET · Maths · Complex Number
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
- C 5
- D 2
Answer & Solution
Correct Answer
(C) 5
Step-by-step Solution
Detailed explanation
Z is a complex cube root of unity
\(\therefore \quad z^3=1...(i)\)
and \(1+z+z^2=0\)
\(\therefore z^2+1=-z\)...(ii)
Consider,
\(\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2 \)
\( =\left(1+\frac{1}{1}\right)^2+\left(z^3 \cdot z+\frac{1}{z^3 \cdot z}\right)^2 \)
\( =4+\left[z+\frac{1}{z}\right]^2...[\text{From (i)}]\)
\(=4+\left[\frac{z^2+1}{z}\right]^2 \)
\( =4+\left[\frac{-z}{z}\right]^2 ...[\text{From(ii)}]\)
\( =4+(-1)^2 =5\)
\(\therefore \quad z^3=1...(i)\)
and \(1+z+z^2=0\)
\(\therefore z^2+1=-z\)...(ii)
Consider,
\(\left(z^3+\frac{1}{z^3}\right)^2+\left(z^4+\frac{1}{z^4}\right)^2 \)
\( =\left(1+\frac{1}{1}\right)^2+\left(z^3 \cdot z+\frac{1}{z^3 \cdot z}\right)^2 \)
\( =4+\left[z+\frac{1}{z}\right]^2...[\text{From (i)}]\)
\(=4+\left[\frac{z^2+1}{z}\right]^2 \)
\( =4+\left[\frac{-z}{z}\right]^2 ...[\text{From(ii)}]\)
\( =4+(-1)^2 =5\)
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