Minimal Polynomial

For the sake of algebra, let $\alpha = x$

$x = i - \sqrt{2}$

$x+ \sqrt{2} = i$ (Transposition)

$x^2+2\sqrt{2}x+2 = -1$ (Squaring both sides)

$x^2+3 = -2\sqrt{2}x$ (Transposition)

$x^4+6x^2+9 = 8x$ (Squaring both sides)

$x^4-2x^2+9 = 0$ (Transposition)

Oh my goodness, we magically acquired an polynomial function in $x$ for which $f(x)$ is $0$ !

So, $f(x)=x^4-2x^2+9$ , $f(1) = 1^4 -2(1)^2 +9 = \boxed{8}$

$\begin{aligned} \alpha & = -\sqrt{2}+i \\ \implies \alpha^0 & = 1 \\ \alpha^2 & = (-\sqrt{2}+i)^2 = 2-1-2\sqrt{2}i = 1 - 2 \sqrt{2}i \\ \alpha^4 & = (1-2\sqrt{2}i)^2 = 1-8 -4\sqrt{2}i = -7 -4\sqrt{2}i \end{aligned}$

$\implies \begin{cases} \alpha^4 = -7 -4\sqrt{2}i \\ \alpha^2 = 1 -2\sqrt{2}i \\ \alpha^0 = 1 \end{cases}$

To eliminate the imaginary part, we do:

$\begin{aligned} \alpha^4 - 2\alpha^2 & = -9 \\ & = - 9 \alpha^0 \\ \implies \alpha^4 - 2\alpha^2 + 9 \alpha^0 & = 0 \\ \implies f(x) & = x^4 - 2x^2 + 9 \\ f(1) & = 1-2+9 = \boxed{8} \end{aligned}$

$\alpha = i-\sqrt{2} \\ \alpha^{2} = i^2 +{\sqrt{2}^{2} - 2\cdot{i}\cdot{\sqrt{2}}} \\ \alpha^2 - 1 = -2 \cdot {\sqrt{2}} \cdot I \\ \alpha^4 - 2\alpha^2 + 1 = -8 \\ f(\alpha)=\alpha^4 - 2\alpha^2 + 9=0 \\ f(1)= 1 - 2 + 9 \\ f(1) =8$