If `alpha+ibeta=tan^(-1) (z), z=x+iy` and `alpha` is constant, the locus of 'z' is

Options

A. `x^(2)+y^(2)+2xcot 2alpha=1`
B. `cot 2alpha(x^(2)+y^(2))=1+x`
C. `x^(2)+y^(2)+2ytanalpha=1`
D. `x^(2)+y^(2)+2xsinx2alpha=1`

Answer

To find the locus of \( z = x + iy \) given that \( \alpha + i\beta = \tan^{-1}(z) \) and \( \alpha \) is a constant, we can follow these steps:

### Step 1: Set up the equation
We start with the equation:
\[ \alpha + i\beta = \tan^{-1}(z) = \tan^{-1}(x + iy) \]
This gives us our first equation:
\[ \alpha + i\beta = \tan^{-1}(x + iy) \]

### Step 2: Find the conjugate
Next, we find the conjugate of the equation:
\[ \alpha - i\beta = \tan^{-1}(x - iy) \]
This gives us our second equation:
\[ \alpha - i\beta = \tan^{-1}(x - iy) \]

### Step 3: Add the two equations
Now we add the first and second equations:
\[ (\alpha + i\beta) + (\alpha - i\beta) = \tan^{-1}(x + iy) + \tan^{-1}(x - iy) \]
This simplifies to:
\[ 2\alpha = \tan^{-1}(x + iy) + \tan^{-1}(x - iy) \]

### Step 4: Use the tangent addition formula
Using the formula for the sum of inverse tangents:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]
we can substitute \( a = x + iy \) and \( b = x - iy \):
\[ \tan^{-1}(x + iy) + \tan^{-1}(x - iy) = \tan^{-1}\left(\frac{(x + iy) + (x - iy)}{1 - (x + iy)(x - iy)}\right) \]

### Step 5: Simplify the right-hand side
Calculating the numerator:
\[ (x + iy) + (x - iy) = 2x \]
Calculating the denominator:
\[ 1 - (x + iy)(x - iy) = 1 - (x^2 + y^2) = 1 - x^2 - y^2 \]
Thus, we have:
\[ 2\alpha = \tan^{-1}\left(\frac{2x}{1 - (x^2 + y^2)}\right) \]

### Step 6: Take the tangent of both sides
Taking the tangent of both sides, we get:
\[ \tan(2\alpha) = \frac{2x}{1 - (x^2 + y^2)} \]

### Step 7: Rearranging the equation
Rearranging gives us:
\[ \tan(2\alpha)(1 - (x^2 + y^2)) = 2x \]
Expanding this:
\[ \tan(2\alpha) - \tan(2\alpha)(x^2 + y^2) = 2x \]
Rearranging further:
\[ \tan(2\alpha)(x^2 + y^2) + 2x - \tan(2\alpha) = 0 \]

### Step 8: Final form of the locus
Dividing through by \(\tan(2\alpha)\):
\[ x^2 + y^2 + \frac{2x}{\tan(2\alpha)} = 1 \]
Letting \( k = \cot(2\alpha) \) (since \(\cot(2\alpha) = \frac{1}{\tan(2\alpha)}\)), we can rewrite this as:
\[ x^2 + y^2 + 2x \cdot k = 1 \]
This represents a circle.

### Conclusion
The locus of \( z \) is given by:
\[ x^2 + y^2 + 2x \cot(2\alpha) = 1 \]