For a complex number Z, if the argument of `3+3i and (Z-2) (bar(Z)-1)` are equal, then the maximum distance of Z from the x - axis is equal to (where, `i^(2)=-1`)

Options

A. `((1+sqrt2))/(2)" units"`
B. `2" units"`
C. `(3)/(2)" units"`
D. `((sqrt2+2))/(2)" units"`

Answer

To solve the problem, we need to find the maximum distance of the complex number \( Z \) from the x-axis, given that the arguments of \( 3 + 3i \) and \( (Z - 2)(\bar{Z} - 1) \) are equal.

### Step 1: Define the complex number
Let \( Z = x + iy \), where \( x \) and \( y \) are real numbers.

### Step 2: Calculate \( Z - 2 \) and \( \bar{Z} - 1 \)
We have:
\[ Z - 2 = (x - 2) + iy \]
\[ \bar{Z} = x - iy \quad \text{so} \quad \bar{Z} - 1 = (x - 1) - iy \]

### Step 3: Calculate the product \( (Z - 2)(\bar{Z} - 1) \)
Now, we compute:
\[ (Z - 2)(\bar{Z} - 1) = [(x - 2) + iy][(x - 1) - iy] \]
Using the distributive property:
\[ = (x - 2)(x - 1) - (x - 2)iy + iy(x - 1) + y^2 \]
\[ = (x^2 - 3x + 2 + y^2) + i(y(x - 1) - (x - 2)) \]
\[ = (x^2 - 3x + 2 + y^2) + i(yx - y - x + 2) \]

### Step 4: Find the argument of \( 3 + 3i \)
The argument of \( 3 + 3i \) is given by:
\[ \text{arg}(3 + 3i) = \tan^{-1}\left(\frac{3}{3}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]

### Step 5: Set the arguments equal
For the arguments to be equal:
\[ \frac{yx - y - x + 2}{x^2 - 3x + 2 + y^2} = 1 \]
This implies:
\[ yx - y - x + 2 = x^2 - 3x + 2 + y^2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ yx - y - x + 2 - (x^2 - 3x + 2 + y^2) = 0 \]
\[ yx - y - x - x^2 + 3x - y^2 = 0 \]
\[ -x^2 + (y + 3)x - y - y^2 = 0 \]

### Step 7: Identify the circle
This is a quadratic equation in \( x \). The discriminant must be non-negative for real solutions:
\[ D = (y + 3)^2 - 4(-1)(-y - y^2) \geq 0 \]
\[ D = (y + 3)^2 - 4y - 4y^2 \geq 0 \]
\[ D = -3y^2 + 2y + 9 \geq 0 \]

### Step 8: Find the maximum distance from the x-axis
The maximum distance of \( Z \) from the x-axis is the maximum value of \( y \). To find this, we need to solve the quadratic inequality:
\[ -3y^2 + 2y + 9 \geq 0 \]
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{(2)^2 - 4(-3)(9)}}{2(-3)} \]
\[ = \frac{-2 \pm \sqrt{4 + 108}}{-6} = \frac{-2 \pm \sqrt{112}}{-6} = \frac{-2 \pm 4\sqrt{7}}{-6} \]
\[ = \frac{2 \mp 4\sqrt{7}}{6} = \frac{1 \mp 2\sqrt{7}}{3} \]
The maximum value occurs at:
\[ y = \frac{1 + 2\sqrt{7}}{3} \]

### Final Answer
Thus, the maximum distance of \( Z \) from the x-axis is:
\[ \frac{1 + 2\sqrt{7}}{3} \]