If `y_(1)=max||z-omega|-|z-omega^(2)||`, where `|z|=2` and `y_(2)=max||z-omega|-|z-omega^(2)||`, where `|z|=(1)/(2)` and `omega` and `omega^(2)` are complex cube roots of unity, then

Options

A. `y_(1)=sqrt(3)`, `y_(2)=sqrt(3)`
B. `y_(1) lt sqrt(3)`, `y_(2)=sqrt(3)`
C. `y_(1)=sqrt(3)`, `y_(2) lt sqrt(3)`
D. `y_(1) gt 3`, `y_(2) lt sqrt(3)`

Explanation

To solve the problem, we need to find the maximum values of the expressions given for \( y_1 \) and \( y_2 \) under the specified conditions.

### Step 1: Understanding the complex cube roots of unity
The complex cube roots of unity are given by:
\[ \omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i, \quad \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \]
These roots satisfy the equation \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

### Step 2: Finding \( y_1 \)
We need to calculate:
\[ y_1 = \max ||z - \omega| - |z - \omega^2|| \]
where \( |z| = 2 \).

Using the triangle inequality:
\[ ||z - \omega| - |z - \omega^2|| \leq |(\omega^2 - \omega)| \]
Calculating \( |\omega^2 - \omega| \):
\[ \omega^2 - \omega = \left(-\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) - \left(-\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) = -\sqrt{3} i \]
Thus,
\[ |\omega^2 - \omega| = \sqrt{3} \]
So, we have:
\[ ||z - \omega| - |z - \omega^2|| \leq \sqrt{3} \]
To check if equality can be achieved, we can position \( z \) such that it is on the line connecting \( \omega \) and \( \omega^2 \). Therefore, \( y_1 = \sqrt{3} \).

### Step 3: Finding \( y_2 \)
Now we calculate:
\[ y_2 = \max ||z - \omega| - |z - \omega^2|| \]
where \( |z| = \frac{1}{2} \).

Using the triangle inequality again:
\[ ||z - \omega| - |z - \omega^2|| \leq |\omega^2 - \omega| = \sqrt{3} \]
However, since \( |z| = \frac{1}{2} \), we need to check if we can achieve this maximum. The maximum distance from \( z \) to \( \omega \) and \( \omega^2 \) will not reach \( \sqrt{3} \) because \( z \) is constrained to a smaller circle.

Thus, we conclude:
\[ y_2 < \sqrt{3} \]

### Final Conclusion
From our calculations, we have:
\[ y_1 = \sqrt{3} \quad \text{and} \quad y_2 < \sqrt{3} \]