If `omega=1` is the complex cube root of unity and matrix `H=|{:(,omega,0),(,0,omega):}|`, then `H^(70)` is equal to:

Options

A. 0
B. `-H`
C. `H^(2)`
D. H

Answer

To solve the problem, we need to find \( H^{70} \) where \( H \) is defined as:

\[ H = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} \]

and \( \omega \) is a complex cube root of unity, specifically \( \omega = e^{2\pi i / 3} \).

### Step 1: Understanding the properties of \( \omega \)

The complex cube roots of unity satisfy the equation:

\[ \omega^3 = 1 \]

This implies that:

\[ \omega^0 = 1, \quad \omega^1 = \omega, \quad \omega^2 = \omega^2, \quad \omega^3 = 1, \quad \omega^4 = \omega, \quad \omega^5 = \omega^2, \quad \text{and so on.} \]

### Step 2: Finding \( H^n \)

The matrix \( H \) is a diagonal matrix, and the powers of a diagonal matrix can be computed by raising each diagonal entry to the power \( n \):

\[ H^n = \begin{pmatrix} \omega^n & 0 \\ 0 & \omega^n \end{pmatrix} \]

### Step 3: Calculate \( H^{70} \)

Now, we need to find \( H^{70} \):

\[ H^{70} = \begin{pmatrix} \omega^{70} & 0 \\ 0 & \omega^{70} \end{pmatrix} \]

### Step 4: Simplifying \( \omega^{70} \)

To simplify \( \omega^{70} \), we can use the property of \( \omega \):

\[ \omega^{70} = \omega^{3 \cdot 23 + 1} = (\omega^3)^{23} \cdot \omega^1 = 1^{23} \cdot \omega = \omega \]

### Step 5: Substitute back into \( H^{70} \)

Thus, we have:

\[ H^{70} = \begin{pmatrix} \omega & 0 \\ 0 & \omega \end{pmatrix} = H \]

### Conclusion

Therefore, the final result is:

\[ H^{70} = H \]