Statement-1 : The locus of z , if ` arg((z-1)/(z+1)) = pi/2` is a circle.
and
Statement -2 : ` |(z-2)/(z+2)| = pi/2`, then the locus of z is a circle.

Options

A. Statement -1 is True, Statement -2 is True, Statement -2 is a correct explanation for statement -2
B. Statement -1 is True, Statement -2 is True , Statement -2 is NOT a correct explanation for Statement -2
C. Statement -1 is True, Statement -2 is False
D. Statement -1 is Flase, Statement -2 is True

Answer

To solve the problem step by step, we will analyze both statements regarding the locus of the complex number \( z \).

### Step 1: Analyze Statement 1
We need to determine the locus of \( z \) such that:
\[ \arg\left(\frac{z-1}{z+1}\right) = \frac{\pi}{2} \]

### Step 2: Substitute \( z \)
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then we can express \( z - 1 \) and \( z + 1 \):
\[ z - 1 = (x - 1) + iy \]
\[ z + 1 = (x + 1) + iy \]

### Step 3: Write the Expression
Now, we can write the expression:
\[ \frac{z-1}{z+1} = \frac{(x-1) + iy}{(x+1) + iy} \]

### Step 4: Rationalize the Denominator
To simplify, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{(x-1) + iy}{(x+1) + iy} \cdot \frac{(x+1) - iy}{(x+1) - iy} = \frac{((x-1)(x+1) + y^2) + i(y(x+1) - y(x-1))}{(x+1)^2 + y^2} \]

### Step 5: Simplify the Numerator
The numerator simplifies to:
\[ (x^2 - 1 + y^2) + i(2y) \]
Thus, we have:
\[ \frac{(x^2 + y^2 - 1) + i(2y)}{(x+1)^2 + y^2} \]

### Step 6: Find the Argument
The argument of a complex number \( a + ib \) is given by \( \tan^{-1}\left(\frac{b}{a}\right) \). Here:
\[ \arg\left(\frac{(x^2 + y^2 - 1) + i(2y)}{(x+1)^2 + y^2}\right) = \tan^{-1}\left(\frac{2y}{x^2 + y^2 - 1}\right) \]
Setting this equal to \( \frac{\pi}{2} \) implies:
\[ \tan^{-1}\left(\frac{2y}{x^2 + y^2 - 1}\right) = \frac{\pi}{2} \]

### Step 7: Determine Conditions for Infinity
The tangent function approaches infinity when its denominator is zero:
\[ x^2 + y^2 - 1 = 0 \implies x^2 + y^2 = 1 \]

### Step 8: Conclusion for Statement 1
The equation \( x^2 + y^2 = 1 \) represents a circle centered at the origin with a radius of 1. Therefore, Statement 1 is **true**.

### Step 9: Analyze Statement 2
Now we analyze Statement 2:
\[ \left|\frac{z-2}{z+2}\right| = \frac{\pi}{2} \]
This is not correctly stated, as it should be an equality involving a modulus. We will assume it is meant to be an argument condition similar to Statement 1.

### Step 10: Substitute \( z \) Again
Let \( z = x + iy \):
\[ \frac{z-2}{z+2} = \frac{(x-2) + iy}{(x+2) + iy} \]

### Step 11: Rationalize the Denominator Again
Following similar steps as before, we rationalize:
\[ \frac{(x-2) + iy}{(x+2) + iy} \cdot \frac{(x+2) - iy}{(x+2) - iy} \]

### Step 12: Simplify the Expression
This leads to:
\[ \frac{(x^2 - 4 + y^2) + i(2y)}{(x+2)^2 + y^2} \]

### Step 13: Find the Argument
Setting the argument to \( \frac{\pi}{2} \) gives:
\[ \frac{2y}{x^2 + y^2 - 4} = \infty \implies x^2 + y^2 - 4 = 0 \implies x^2 + y^2 = 4 \]

### Step 14: Conclusion for Statement 2
The equation \( x^2 + y^2 = 4 \) represents a circle centered at the origin with a radius of 2. Therefore, Statement 2 is also **true**.

### Final Conclusion
Both statements are true, confirming that the locus described in each case is indeed a circle.

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