If p(x,y) denotes z = x + iy in Argand plane and `| (z - 1)/( z + 2i)|` = 1 the locus of P is a/an

Options

A. Hyperbola
B. Ellipse
C. Circle
D. Straight line

Answer

To solve the problem, we need to analyze the given equation involving the complex number \( z \) in the Argand plane. The equation is:

\[ \left| \frac{z - 1}{z + 2i} \right| = 1 \]

Here, \( z = x + iy \), where \( x \) and \( y \) are the real and imaginary parts of the complex number, respectively.

### Step 1: Rewrite the equation
We start by rewriting the modulus equation:

\[ \left| \frac{z - 1}{z + 2i} \right| = 1 \implies |z - 1| = |z + 2i| \]

This means the distance from the point \( z \) to the point \( 1 \) in the complex plane is equal to the distance from \( z \) to the point \( -2i \).

### Step 2: Express \( z \) in terms of \( x \) and \( y \)
Substituting \( z = x + iy \):

\[ |z - 1| = |(x + iy) - 1| = |(x - 1) + iy| = \sqrt{(x - 1)^2 + y^2} \]

\[ |z + 2i| = |(x + iy) + 2i| = |x + (y + 2)i| = \sqrt{x^2 + (y + 2)^2} \]

### Step 3: Set the two distances equal
Now we set the two expressions equal to each other:

\[ \sqrt{(x - 1)^2 + y^2} = \sqrt{x^2 + (y + 2)^2} \]

### Step 4: Square both sides
To eliminate the square roots, we square both sides:

\[ (x - 1)^2 + y^2 = x^2 + (y + 2)^2 \]

### Step 5: Expand both sides
Expanding both sides gives:

\[ (x^2 - 2x + 1 + y^2) = (x^2 + y^2 + 4y + 4) \]

### Step 6: Simplify the equation
Now, we simplify the equation by canceling \( x^2 \) and \( y^2 \) from both sides:

\[ -2x + 1 = 4y + 4 \]

### Step 7: Rearranging the equation
Rearranging gives:

\[ -2x - 4y + 1 - 4 = 0 \]

This simplifies to:

\[ -2x - 4y - 3 = 0 \]

### Step 8: Final form of the equation
We can divide the entire equation by -1 to make it more standard:

\[ 2x + 4y + 3 = 0 \]

### Conclusion
The locus of the point \( P(x, y) \) is a straight line, as it can be expressed in the form \( Ax + By + C = 0 \).