Show that `sqrt([-1sqrt({-1-sqrt(-1+ ..."to"oo)})]) = omega, or omega^(2)`

Answer

To solve the equation \( \sqrt{-1 \sqrt{-1 - \sqrt{-1 + \ldots}}} = \omega \) or \( \omega^2 \), where \( \omega \) and \( \omega^2 \) are the cube roots of unity, we can follow these steps:

### Step 1: Define the Expression
Let \( y = \sqrt{-1 \sqrt{-1 - \sqrt{-1 + \ldots}}} \).

### Step 2: Rewrite the Expression
From the definition, we can express \( y \) as:
\[ y = \sqrt{-1 - y} \]
This is because the expression inside the square root continues indefinitely.

### Step 3: Square Both Sides
Now, we square both sides to eliminate the square root:
\[ y^2 = -1 - y \]

### Step 4: Rearrange the Equation
Rearranging gives us:
\[ y^2 + y + 1 = 0 \]

### Step 5: Solve the Quadratic Equation
We can solve this quadratic equation using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = 1 \).

Substituting these values in:
\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ y = \frac{-1 \pm \sqrt{1 - 4}}{2} \]
\[ y = \frac{-1 \pm \sqrt{-3}}{2} \]

### Step 6: Simplify the Roots
Since \( \sqrt{-3} = i\sqrt{3} \), we can write:
\[ y = \frac{-1 \pm i\sqrt{3}}{2} \]

### Step 7: Identify the Roots
The two roots are:
\[ y_1 = \frac{-1 + i\sqrt{3}}{2}, \quad y_2 = \frac{-1 - i\sqrt{3}}{2} \]

### Step 8: Relate to Cube Roots of Unity
The roots \( y_1 \) and \( y_2 \) can be identified as the cube roots of unity:
\[ \omega = \frac{-1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{-1 - i\sqrt{3}}{2} \]

### Conclusion
Thus, we have shown that:
\[ \sqrt{-1 \sqrt{-1 - \sqrt{-1 + \ldots}}} = \omega \text{ or } \omega^2 \]

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