If z=x+iy is a complex number satisfying `|z+i/2|^2=|z-i/2|^2` , then the locus of z is

Answer

To solve the problem, we need to analyze the equation given:

\[ |z + \frac{i}{2}|^2 = |z - \frac{i}{2}|^2 \]

where \( z = x + iy \).

### Step 1: Substitute \( z \) in the equation
Substituting \( z = x + iy \) into the equation gives:

\[ | (x + iy) + \frac{i}{2} |^2 = | (x + iy) - \frac{i}{2} |^2 \]

This simplifies to:

\[ | x + i(y + \frac{1}{2}) |^2 = | x + i(y - \frac{1}{2}) |^2 \]

### Step 2: Calculate the modulus squared
The modulus squared of a complex number \( a + ib \) is given by \( a^2 + b^2 \). Thus, we have:

\[ |x|^2 + |y + \frac{1}{2}|^2 = |x|^2 + |y - \frac{1}{2}|^2 \]

This leads to:

\[ x^2 + \left(y + \frac{1}{2}\right)^2 = x^2 + \left(y - \frac{1}{2}\right)^2 \]

### Step 3: Expand both sides
Expanding both sides, we get:

\[ x^2 + \left(y^2 + y + \frac{1}{4}\right) = x^2 + \left(y^2 - y + \frac{1}{4}\right) \]

### Step 4: Simplify the equation
Cancelling \( x^2 \) and \( \frac{1}{4} \) from both sides gives:

\[ y^2 + y = y^2 - y \]

### Step 5: Rearranging the equation
Rearranging the equation results in:

\[ y + y = 0 \implies 2y = 0 \]

### Step 6: Solve for \( y \)
From \( 2y = 0 \), we find:

\[ y = 0 \]

### Conclusion: Identify the locus
Since \( y = 0 \), this means that the locus of \( z \) is along the x-axis.

Thus, the required locus of \( z \) is the x-axis.

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