Hello,

$\displaystyle A-\lambda I_2=\begin{pmatrix} \cos \theta-\lambda & \sin \theta \\ -\sin \theta & \cos \theta-\lambda \end{pmatrix}$

The determinant is :

$\displaystyle p(\lambda)=(\cos \theta-\lambda)^2+\sin^2 \theta$

Find $\displaystyle \lambda$ such that $\displaystyle (\cos \theta-\lambda)^2+\sin^2 \theta=0$

Now, you can see that this is true if and only if $\displaystyle \sin^2 \theta=-(\cos \theta-\lambda)^2=(i [\cos \theta-\lambda])^2$

Hence $\displaystyle \sin \theta=\pm i (\cos \theta-\lambda)$

$\displaystyle \pm i \sin \theta=\cos \theta-\lambda$

$\displaystyle \lambda=\cos \theta \pm i \sin \theta$

which gives the values you want, thanks to the identity $\displaystyle \cos \theta+i \sin \theta=e^{i \theta}$ ($\displaystyle \cos \theta-i \sin \theta=\cos(-\theta)+i \sin (-\theta)=e^{-i\theta}$, because cos is an even function and sin is an odd function)