Below is from Norris' Markov Chains. From the matrix decomposition of $P^n$, it is clear that $p_{11}^{(n)}=a+b(\frac{i}{2})^n+c(-\frac{i}{2})^n$ for some complex numbers $a,b,c$ that depends on the entries of $U, U^{-1}$. But how are we able to rewrite this in the form $\alpha + (\frac{1}{2})^n(\beta \cos\frac{n\pi}{2}+\gamma \sin\frac{n\pi}{2})$ for some real numbers $\alpha,\beta,\gamma$ because $p_{11}^{(n)}$ is real?
The disciplined approach here is to check this for 3 consecutive values, say $n= 0,1,2$. That is enough to uniquely specify and solve for $a$, $b$ and $c$ (using a Vandermonde matrix, shown at the end).
Based on limitting values you should be able to see $a =\alpha$. As far as the non-real eigenvalues and the real trig function approach, applying Euler's formula gives
$b\cdot \cos\frac{n\pi}{2}+ b\cdot i\cdot \sin\frac{n\pi}{2} + c\cdot \cos\frac{3n\pi}{2}+c\cdot i\cdot \sin\frac{3n\pi}{2} $
$= b\cdot \cos\frac{n\pi}{2}+ b\cdot i\cdot \sin\frac{n\pi}{2} + c\cdot \cos\frac{n\pi}{2}-c\cdot i\cdot \sin\frac{n\pi}{2}$
$= (b+c)\cdot \cos\frac{n\pi}{2}+ (b-c)\cdot i\cdot \sin\frac{n\pi}{2}$
where the middle equality makes use of the fact that $i$ and $-i$ (or $\lambda_2,\lambda_3$ on the unit circle) are conjugate pairs so they have the same real portions and 'opposite' imaginary portions given by cosine and sine respectively. Non-real eigenvalues necessarily come in conjugate pairs for matrices that have entirely real components.
so to make this very explicit in your particular problem, in the original formulation you have
$\left[\begin{matrix}1 & 1 & 1\\1 & \frac{i}{2} & - \frac{i}{2}\\1 & - \frac{1}{4} & - \frac{1}{4}\end{matrix}\right]\mathbf v = \mathbf e_1$
and your coefficient vector is
$\mathbf v=\left[\begin{matrix}a\\ b \\ c \end{matrix}\right] =\left[\begin{matrix}1 & 1 & 1\\1 & \frac{i}{2} & - \frac{i}{2}\\1 & - \frac{1}{4} & - \frac{1}{4}\end{matrix}\right]^{-1}\mathbf e_1 = \left[\begin{matrix}\frac{1}{5}\\\frac{2}{5} + \frac{i}{5}\\\frac{2}{5} - \frac{i}{5}\end{matrix}\right]$
and of course
$b+c = \frac{2}{5} + \frac{i}{5} + \frac{2}{5} - \frac{i}{5} = \frac{4}{5}$ and
$b-c = \frac{2}{5} + \frac{i}{5} -\big(\frac{2}{5} - \frac{i}{5}\big) = i\cdot \frac{2}{5}$
and when you plug this into
$ (b+c)\cdot \cos\frac{n\pi}{2}+ (b-c)\cdot i\cdot \sin\frac{n\pi}{2}\longrightarrow \frac{4}{5}\cdot \cos\frac{n\pi}{2}- \frac{2}{5}\cdot \sin\frac{n\pi}{2}$