For two unimobular complex numbers `z_(1)` and `z_(2)`, find `[(bar(z)_(1),-z_(2)),(bar(z)_(2),z_(1))]^(-1) [(z_(1),z_(2)),(-bar(z)_(2),bar(z)_(1))]^(-1)`

Explanation

To solve the problem, we need to find the value of the expression:

\[ \left[ \begin{pmatrix} \bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1 \end{pmatrix} \right]^{-1} \left[ \begin{pmatrix} z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1 \end{pmatrix} \right]^{-1} \]

where \( z_1 \) and \( z_2 \) are unimodular complex numbers (i.e., \( |z_1| = 1 \) and \( |z_2| = 1 \)).

### Step 1: Define Matrices A and B
Let:
\[ A = \begin{pmatrix} \bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1 \end{pmatrix} \]
\[ B = \begin{pmatrix} z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1 \end{pmatrix} \]

### Step 2: Use the Inverse Property of Matrices
Using the property of inverses, we have:
\[ A^{-1} B^{-1} = (BA)^{-1} \]

### Step 3: Compute the Product \( BA \)
Now, we compute the product \( BA \):
\[ BA = \begin{pmatrix} z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1 \end{pmatrix} \begin{pmatrix} \bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1 \end{pmatrix} \]

Calculating this product:
1. The element at (1,1):
\[ z_1 \bar{z}_1 + z_2 \bar{z}_2 = |z_1|^2 + |z_2|^2 = 1 + 1 = 2 \]
2. The element at (1,2):
\[ z_1 (-z_2) + z_2 z_1 = -z_1 z_2 + z_2 z_1 = 0 \]
3. The element at (2,1):
\[ -\bar{z}_2 \bar{z}_1 + \bar{z}_1 z_2 = -\bar{z}_2 \bar{z}_1 + \bar{z}_1 z_2 = 0 \]
4. The element at (2,2):
\[ -\bar{z}_2 (-z_2) + \bar{z}_1 z_1 = |z_2|^2 + |z_1|^2 = 1 + 1 = 2 \]

Thus, we have:
\[ BA = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \]

### Step 4: Find the Inverse of \( BA \)
Now, we find the inverse of \( BA \):
\[ (BA)^{-1} = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \]

### Final Result
Thus, the final answer is:
\[ \left[ \begin{pmatrix} \bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1 \end{pmatrix} \right]^{-1} \left[ \begin{pmatrix} z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1 \end{pmatrix} \right]^{-1} = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix} \]