If a point P denoting the complex number z moves on the complex plane such tha `"Re"(z^(2))="Re"(z+bar(z))`, then the locus of P (z) is

Options

A. A circle
B. Rectangular hyperbola
C. Ellipse
D. pair of straight lines

Answer

To find the locus of the point \( P \) denoting the complex number \( z \) such that \( \text{Re}(z^2) = \text{Re}(z + \bar{z}) \), we will follow these steps:

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

### Step 2: Calculate \( z^2 \)
We calculate \( z^2 \):
\[ z^2 = (x + iy)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + 2xyi \]
The real part of \( z^2 \) is \( x^2 - y^2 \).

### Step 3: Calculate \( z + \bar{z} \)
Next, we calculate \( z + \bar{z} \):
\[ z + \bar{z} = (x + iy) + (x - iy) = 2x \]
The real part of \( z + \bar{z} \) is \( 2x \).

### Step 4: Set up the equation
According to the problem, we have:
\[ \text{Re}(z^2) = \text{Re}(z + \bar{z}) \]
This gives us the equation:
\[ x^2 - y^2 = 2x \]

### Step 5: Rearranging the equation
Rearranging the equation, we get:
\[ x^2 - 2x - y^2 = 0 \]
We can complete the square for the \( x \) terms:
\[ (x - 1)^2 - 1 - y^2 = 0 \]
This simplifies to:
\[ (x - 1)^2 - y^2 = 1 \]

### Step 6: Identify the locus
The equation \( (x - 1)^2 - y^2 = 1 \) represents a hyperbola centered at \( (1, 0) \) with the transverse axis along the x-axis.

### Final Result
Thus, the locus of the point \( P \) is a hyperbola.

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