Question

If $z^2+z+1=0$, where $z$ is a complex number, then the value of ${\left(z+\frac{1}{z}\right)}^2+{\left(z^2+\frac{1}{z^2}\right)}^2+{\left(z^3+\frac{1}{z^3}\right)}^2+...+{\left(z^6+\frac{1}{z^6}\right)}^2$

• A. $6$
• B. $12$
• C. $18$
• D. $24$

Answer 1

The correct option is B

$12$

Explanation for the correct option.

Step 1. Find the roots of the equation.

The roots of the equation $z^2+z+1=0$ are complex cube roots of unity. So $z=\omega$ or ${\omega }^2$.

The sum of roots is $\omega +{\omega }^2=-1$ and the product of roots is ${\omega }^3=1$.

Step 2. Write the given expression in terms of $\omega$ and find its value.

In the given espression ${\left(z+\frac{1}{z}\right)}^2+{\left(z^2+\frac{1}{z^2}\right)}^2+{\left(z^3+\frac{1}{z^3}\right)}^2+...+{\left(z^6+\frac{1}{z^6}\right)}^2$ substitute $\omega$ for $z$.

${\left(\omega +\frac{1}{\omega }\right)}^2+{\left({\omega }^2+\frac{1}{{\omega }^2}\right)}^2+{\left({\omega }^3+\frac{1}{{\omega }^3}\right)}^2+{\left({\omega }^4+\frac{1}{{\omega }^4}\right)}^2+{\left({\omega }^5+\frac{1}{{\omega }^5}\right)}^2+{\left({\omega }^6+\frac{1}{{\omega }^6}\right)}^2$

Now, as ${\omega }^3=1$ so the equation can be written as:

${\left(\omega +\frac{{\omega }^3}{\omega }\right)}^2+{\left({\omega }^2+\frac{{\omega }^3}{{\omega }^2}\right)}^2+{\left(1+\frac{1}{1}\right)}^2+{\left({\omega }^3\omega +\frac{{\omega }^3}{{\omega }^3\omega }\right)}^2+{\left({\omega }^3{\omega }^2+\frac{{\omega }^3}{{\omega }^3{\omega }^2}\right)}^2+{\left(1+\frac{1}{1}\right)}^2$

Simplifying by using laws of exponent as:

${\left(\omega +{\omega }^2\right)}^2+{\left({\omega }^2+\omega \right)}^2+2^2+{\left(\omega +{\omega }^2\right)}^2+{\left({\omega }^2+\omega \right)}^2+2^2$

Now as $\omega +{\omega }^2=-1$ so the value can be found as:

$\begin{array}{rcl}{\left(\omega +{\omega }^2\right)}^2+{\left({\omega }^2+\omega \right)}^2+2^2+{\left(\omega +{\omega }^2\right)}^2+{\left({\omega }^2+\omega \right)}^2+2^2& =& {\left(-1\right)}^2+{\left(-1\right)}^2+4+{\left(-1\right)}^2+{\left(-1\right)}^2+4\\ & =& 1+1+4+1+1+4\\ & =& 12\end{array}$

So the value of the expression ${\left(z+\frac{1}{z}\right)}^2+{\left(z^2+\frac{1}{z^2}\right)}^2+{\left(z^3+\frac{1}{z^3}\right)}^2+...+{\left(z^6+\frac{1}{z^6}\right)}^2$ is $12$.

Hence, the correct option is B.