Solution: Note that \[ \sum_{n\in \mathbb{N} } 2^{-n}z^{2n} = \sum_{n\in \mathbb{N} } 2^{-n} \left( z^2 \right) ^n. \] We know that the series $\displaystyle \sum_{n\in \mathbb{N} } a_nz^n$ converges if $\vert z \vert \lt R$, where $R$ is the radius of convergence given by \[ R = \lim_{n \to \infty} \frac{a_n}{a_{n+1}}. \]

In the given series, the radius of convergence will be \begin{align*} R & = \lim_{n \to \infty} \frac{a_n}{a_{n+1}} \\ & = \lim_{n \to \infty} \frac{2^{-n}}{2^{-(n+1)}} \\ & = \lim_{n \to \infty} \frac{1}{2^{-1}} \\ & = 2. \end{align*} Therefore, the power series will converge if \[ \vert w \vert \lt 2 \implies \left\vert z^2 \right\vert \lt 2 \implies \vert z \vert \lt \sqrt{2}. \] Now we need to check whether the series converges at the boundary points, that is, at $z=\pm \sqrt{2} $. If $z=\pm \sqrt{2} $, then the series becomes \[ \sum_{n\in \mathbb{N} } 2^{-n} \left( \pm\sqrt{2} \right) ^{2n} = \sum_{n\in \mathbb{N} } 2^{-n}2^{n} = \sum_{n\in \mathbb{N} } 1, \] which is a divergent series. Hence the region of convergence of the given power series will be $\vert z \vert \lt \sqrt{2} $.