Solution: Note that \[ \frac{1}{\sqrt{2}} + \frac{\iota}{\sqrt{2} } = \cos \frac{\pi}{4} + \iota \sin \frac{\pi}{4}. \] We also recall the \emph{De moire's Theorem:} For any complex number $z=r(\cos \theta +\iota \sin \theta )$ and for any $n\in \mathbb{Z}$, we have \[ z^ = r^n \left( \cos n \theta +\sin n \theta \right) . \] Thus, \begin{align*} \left( \frac{1}{\sqrt{2}} + \frac{\iota}{\sqrt{2} } \right) ^{2023} & = \cos \left( \frac{2023\pi}{4} \right) + \iota \sin \left( \frac{2023\pi}{4} \right)\\ & = \cos \left( 505\pi+\frac{3\pi}{4} \right) + \iota \sin \left( 505\pi+\frac{3\pi}{4} \right) \\ & = -\cos \frac{3\pi}{4} + \iota \left( -\sin \frac{3\pi}{4} \right) \\ & = -\left( -\frac{1}{\sqrt{2} } \right) +\iota \left( -\frac{1}{\sqrt{2} } \right) \\ & = \frac{1}{\sqrt{2} } -\frac{\iota}{\sqrt{2}}\\ & = \frac{1-\iota}{\sqrt{2} }. \end{align*} Therefore, \[ \textcolor{blue}{\boxed{ \left( \frac{1+\iota}{\sqrt{2}} \right)^{2023}=\frac{1-\iota}{\sqrt{2} }}} \]