Each term gives equal value?

Note that $z^3 = 1$ . Hence, $\begin{aligned} \sum_{i=1}^{21} (z^i + \frac{1}{z^i})^2 &= \sum_{i=1}^{21} z^{2i} + 2 + z^{-2i} \\ &= 7(z^2 + z + 1) + 42 + 7(z + z^2 + 1) \\ &= 0 + 42 + 0 \\ &= 42. \end{aligned}$

Notice that roots of given quadratic are $\omega$ and $\omega^{2}$ and that $\omega\omega^2$ =1 and $\omega+\omega^2$ =-1.

So terms where powers are not integer multiple of 3 (14 such terms) becomes 1 and remaining 7 terms become $(1+1)^{2}$ =4

Hence the total sum will be 14×1+7×4= $\boxed{42}$

Comparing the given equation with itself when multiplied by $z$ , we can conclude that $z^3 = 1$

Dividing the equation by $z$ and replacing $1$ with $z^3$ , we get:

$z^3+z+\frac{1}{z} = 0$ (I)

Since we can multiply or divide (I) by any exponent of $z$ to reach the desired exponents, it follows that $z^{k+2}+z^k+z^{k-2} = 0 \forall k\in \mathbb{Z}$

$\implies \left(z+\dfrac{1}{z}\right)^2+\left(z^2+\dfrac{1}{z^2}\right)^2+\left(z^3+\dfrac{1}{z^3}\right)^2+\ldots+\left(z^{21}+\dfrac{1}{z^{21}}\right)^2$

$= \underbrace{z^2+z^4+z^6}_{0}+\underbrace{\ldots+z^{42}}_{0}+\underbrace{\dfrac{1}{z^2}+\dfrac{1}{z^4}+\dfrac{1}{z^6}}_{0}+\underbrace{\ldots+\dfrac{1}{z^{42}}}_{0}+42 = \boxed{42}$

Note that $(z-1)(z^2+z+1)=0 \Rightarrow z^3=1$ and we know that $z \neq 1$ . So Let $S$ for the sum :

$S= (z+\frac{1}{z})^2+(z+\frac{1}{z})^2+(z^3+\frac{1}{z^3})^2+...(z^{21}+\frac{1}{z^{21}})^2$

$S= 7[(z+\frac{1}{z})^2+(z^2+\frac{1}{z^2})^2+(z^3+\frac{1}{z^3})^2]$

$S=7[(z^2+2+\frac{1}{z^2})+(z^4+2+\frac{1}{z^4})+4]$

$S=7[(z^2+2+\frac{1}{z^2})+(z+2+\frac{1}{z})+4]$

$S=7[(z^2+\frac{1}{z^2})+(z+\frac{1}{z})+8]$

Using the relation $z^2+z+1=0$ :

$S=7[(\frac{1}{z^2})+(\frac{1}{z})+7]$

Observe that: $z^2+z+1=0 \Rightarrow \frac{1}{z}+\frac{1}{z^2}=\frac{z+1}{z^2}=-1$

Finally :

$S=7 *6 = 42$