Complex number

Rearranging, $x = \dfrac {1}{\sqrt 2} + i \dfrac {1}{\sqrt 2} = \cos \begin{pmatrix} \dfrac π4 \end{pmatrix} + i \sin \begin{pmatrix} \dfrac π4 \end{pmatrix}$

In Euler Notation, $x = e^{i\frac π4}$

Now, $S= x^4 + 2 = (e^{i\frac π4})^4 + 2 = e^{iπ} + 2 = (\cos (π) + i \sin (π)) + 2 = (-1 + 0 ) + 2$

$S = \boxed{1}$

Given that

$x\sqrt{2} = 1 + \sqrt{-1},$

$\Rightarrow x = \dfrac{1+i}{\sqrt2}$ $~~~~~~~~~~~~~~~~~~~~~~~$ $\left[ i = \sqrt{-1}\right]$

$\Rightarrow x^2 = \dfrac{1+2.1.i+i^2}{2}$

$\Rightarrow x^2 = \dfrac{1 - 1 +2i}{2}$

$x^2 =i$

Now,

$x^4+2$

$=(x^2)^2+2$

$=(i)^2+2$

$= -1+2$

$=1$