If `Z=cos phi+isin phi(AA phi in ((pi)/(3),pi))`, then the value of `arg(Z^(2)-Z)` is equal to (where, `arg(Z)` represents the argument of the complex number Z lying in the interval `(-pi, pi] and i^(2)=-1`)

Options

A. `(3phi+pi)/(2)`
B. `(3phi)/(2)`
C. `(3)/(2)(phi-pi)`
D. `(3phi-pi)/(2)`

Answer

To solve the problem, we need to find the value of \( \arg(Z^2 - Z) \) given that \( Z = \cos \phi + i \sin \phi \) where \( \phi \in \left(\frac{\pi}{3}, \pi\right) \).

### Step 1: Express \( Z \) in exponential form
We can express \( Z \) using Euler's formula:
\[ Z = e^{i\phi} \]

### Step 2: Calculate \( Z^2 \)
Now, we calculate \( Z^2 \):
\[ Z^2 = (e^{i\phi})^2 = e^{2i\phi} \]

### Step 3: Calculate \( Z^2 - Z \)
Now we find \( Z^2 - Z \):
\[ Z^2 - Z = e^{2i\phi} - e^{i\phi} \]

### Step 4: Factor out \( e^{i\phi} \)
We can factor out \( e^{i\phi} \) from the expression:
\[ Z^2 - Z = e^{i\phi}(e^{i\phi} - 1) \]

### Step 5: Find the argument
To find \( \arg(Z^2 - Z) \), we can use the property of arguments:
\[ \arg(Z^2 - Z) = \arg(e^{i\phi}) + \arg(e^{i\phi} - 1) \]
Since \( \arg(e^{i\phi}) = \phi \), we have:
\[ \arg(Z^2 - Z) = \phi + \arg(e^{i\phi} - 1) \]

### Step 6: Calculate \( \arg(e^{i\phi} - 1) \)
To find \( \arg(e^{i\phi} - 1) \), we can express \( e^{i\phi} - 1 \) in terms of sine and cosine:
\[ e^{i\phi} - 1 = (\cos \phi - 1) + i \sin \phi \]
Now, we can use the formula for the argument of a complex number:
\[ \arg(a + bi) = \tan^{-1}\left(\frac{b}{a}\right) \]
Thus,
\[ \arg(e^{i\phi} - 1) = \tan^{-1}\left(\frac{\sin \phi}{\cos \phi - 1}\right) \]

### Step 7: Combine the arguments
Now we combine the arguments:
\[ \arg(Z^2 - Z) = \phi + \tan^{-1}\left(\frac{\sin \phi}{\cos \phi - 1}\right) \]

### Step 8: Simplify the expression
As \( \phi \) is in the interval \( \left(\frac{\pi}{3}, \pi\right) \), we can evaluate this further. However, we can note that \( \cos \phi - 1 \) is negative in this range, and \( \sin \phi \) is positive. Therefore, \( \tan^{-1}\left(\frac{\sin \phi}{\cos \phi - 1}\right) \) will yield a negative angle.

### Final Result
Thus, the final expression for \( \arg(Z^2 - Z) \) can be simplified further depending on the specific value of \( \phi \) chosen from the interval. However, generally, we can conclude that:
\[ \arg(Z^2 - Z) = \phi + \tan^{-1}\left(\frac{\sin \phi}{\cos \phi - 1}\right) \]