Degree of a root of unity

The number in question is an 8 $^\text{th}$ root of unity . This means that it is a root of the equation $x^8-1=0.$ This polynomial can be factored:

$(x+1)(x-1)(x^2+1)(x^4+1)=0.$

The polynomial factor which the number in question is a root of is $(x^4+1).$ This cannot be factored any further (without using non-rational coefficients), so the degree of the number is $4.$

Consider

$\begin{aligned} \alpha & = \frac {\sqrt 2}2 + i \frac {\sqrt 2}2 \\ & = \frac 1{\sqrt 2}(1 + i) \\ \alpha^2 & = \frac 12 (2i) = i \\ \alpha^4 & = -1 \end{aligned}$

This implies that $\alpha = \frac {\sqrt 2}2 + i \frac {\sqrt 2}2$ is a root of $x^4+1$ . Therefore, its degree is $\boxed{4}$ .

I knew the number in question to be the (principal) square root of $i$ .

Now, the algebraic degree of $i$ is $2$ as its minimal polynomial is $x^2+1=0$ , and since the number in question is the square root of $i$ , it will therefore be a root of

$(x^2)^2+1=0$

or

$x^4+1=0$

This cannot be factored further so the number in question has algebraic degree $\boxed{4}$