Eigenvalues for Characteristic polynomial $λ^4=-1$

Question

I supposed that the eigenvalues $λ_1=λ_2=i$ and $λ_3=λ_4=-i$. However looking at the literature, the actual eigenvalues are \begin{aligned} λ_1&=\sqrt 2/2+i\sqrt 2/2, \\ λ_2&=\sqrt 2/2-i\sqrt 2/2, \\ λ_3&=-\sqrt 2/2+i\sqrt 2/2, \\ λ_4&=-\sqrt 2/2-i\sqrt 2/2 \end{aligned} Can someone tell me where did I go wrong and why is the answer above is the correct one? Thank you

Answer 1

$$i^4 = i\cdot i \cdot i \cdot i = i^2 \cdot i^2 = (-1)(-1)=1$$

So $i,-i$ are solutions for $\lambda^4 = 1$ rather than $\lambda^4 = -1$.

You can do the same with the other solutions and see that $\lambda^4$ is indeed $-1$.

Answer 2

$$ \lambda ^4=-1 = e^{i\pi}$$

$$\lambda = e^{i(\pi /4 +2k\pi/4)}, k=0,1,2,3 $$ Which give you the eigenvalues.