This problem is just a matter of obtaining the reduced row echelon form of the matrix. This is a standard technique for solving for the kernel of a matrix. Finding the kernel of $det(A - \lambda I)$ gives you the eigenvector associated with the eigenvalue $\lambda$. Since the eigenvector is complex, the other complex eigenvector is the complex conjugate of the first. Therefore, we only need to solve for one.
Using $\lambda = 1 + i$,
\begin{bmatrix}
1-i & 1 & \frac{1}{\sqrt{2}} \\
1 & 1-i & -\frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & -i \\
\end{bmatrix}
Add the second row to the first, and multiply the bottom row by $-\sqrt{2}$:
\begin{bmatrix}
2-i & 2-i & 0 \\
1 & 1-i & -\frac{1}{\sqrt{2}} \\
1 & -1 & i\sqrt{2} \\
\end{bmatrix}
multiply the top row by $\frac{1}{5}(2+i)$
\begin{bmatrix}
1 & 1 & 0 \\
1 & 1-i & -\frac{1}{\sqrt{2}} \\
1 & -1 & i\sqrt{2} \\
\end{bmatrix}
subtract the first row from the other two rows:
\begin{bmatrix}
1 & 1 & 0 \\
0 & -i & -\frac{1}{\sqrt{2}} \\
0 & -2 & i\sqrt{2} \\
\end{bmatrix}
multiply second row by $i$:
\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & -\frac{i}{\sqrt{2}} \\
0 & -2 & i\sqrt{2} \\
\end{bmatrix}
subtract the second row from the first, and add 2 times the second row to the third:
\begin{bmatrix}
1 & 0 & \frac{i}{\sqrt{2}} \\
0 & 1 & -\frac{i}{\sqrt{2}} \\
0 & 0 & 0 \\
\end{bmatrix}
This is the reduced row echelon form. Therefore, the eigenvector associated with $\lambda = 1 + i$ is:
\begin{bmatrix}
1 \\
-1 \\
i \sqrt{2} \\
\end{bmatrix}
and so the eigenvector associated with $\lambda = 1-i$ is:
\begin{bmatrix}
1 \\
-1 \\
-i \sqrt{2} \\
\end{bmatrix}
Best Answer
In general, when talking about "largest" eigenvalue, we are usually talking about largest in absolute value (or magnitude,) where $|a+bi|=\sqrt{a^2+b^2}$.
This means sometimes that there isn't one eigenvalue that is "largest", because two different eigenvalues can have the same absolute value.
As mentioned by others, complex numbers are not themselves ordered.
As mentioned in the comments below, if you know a matrix has only real eigenvalues, then the question of "largest" and "smallest" eigenvalues will depend on the context.
The "largest" eigenvalue for a matrix $A$ is often interesting, particularly when it is unique, because then for large $n$, $A^n$ is dominated by the action on the eigenvectors for those values. This is useful for putting bounds on $A^n\mathbf v$.