If `z = x+iy` is any complex number and `|z-1| = |z+1|` then show that `|z| = y`.

Answer

To solve the problem, we start with the given condition that \( |z - 1| = |z + 1| \), where \( z = x + iy \).

### Step 1: Write the expressions for \( |z - 1| \) and \( |z + 1| \)

We can express \( z - 1 \) and \( z + 1 \) as follows:
\[ z - 1 = (x + iy) - 1 = (x - 1) + iy \]
\[ z + 1 = (x + iy) + 1 = (x + 1) + iy \]

### Step 2: Calculate the moduli

Now, we can calculate the moduli:
\[ |z - 1| = |(x - 1) + iy| = \sqrt{(x - 1)^2 + y^2} \]
\[ |z + 1| = |(x + 1) + iy| = \sqrt{(x + 1)^2 + y^2} \]

### Step 3: Set the moduli equal to each other

Since we know that \( |z - 1| = |z + 1| \), we can set the two expressions equal to each other:
\[ \sqrt{(x - 1)^2 + y^2} = \sqrt{(x + 1)^2 + y^2} \]

### Step 4: Square both sides

To eliminate the square roots, we square both sides:
\[ (x - 1)^2 + y^2 = (x + 1)^2 + y^2 \]

### Step 5: Simplify the equation

Now, we can simplify the equation by subtracting \( y^2 \) from both sides:
\[ (x - 1)^2 = (x + 1)^2 \]

### Step 6: Expand both sides

Expanding both sides gives us:
\[ x^2 - 2x + 1 = x^2 + 2x + 1 \]

### Step 7: Cancel common terms

We can cancel \( x^2 \) and \( 1 \) from both sides:
\[ -2x = 2x \]

### Step 8: Solve for \( x \)

Now, we can solve for \( x \):
\[ -2x - 2x = 0 \implies -4x = 0 \implies x = 0 \]

### Step 9: Substitute \( x \) back into the modulus equation

Now that we have \( x = 0 \), we can find the modulus \( |z| \):
\[ |z| = \sqrt{x^2 + y^2} = \sqrt{0^2 + y^2} = \sqrt{y^2} = |y| \]

Since \( y \) is the imaginary part of \( z \), we can conclude that:
\[ |z| = y \]

### Conclusion

Thus, we have shown that if \( |z - 1| = |z + 1| \), then \( |z| = y \).