Question

Let $z=x+iy,$ where $x,y \in \mathbb{R}$ and $i^{2}=-1$. If $|z-i|=|z-1|$, then $y=$ ________.

• A. $-x$
• B. $x+1$
• C. $-x-1$
• D. $x+2$
• E. $x$

Hint:

$|z-a|=|z-b|$ represents the perpendicular bisector of the line joining $a$ and $b$.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
The equation \(|z - z_1| = |z - z_2|\) describes the set of all points \(z\) in the complex plane that are equidistant from two fixed points \(z_1\) and \(z_2\). Geometrically, this locus is the perpendicular bisector of the line segment connecting \(z_1\) and \(z_2\). We can solve this algebraically by substituting \(z = x + iy\) and using the definition of the modulus.
Step 2: Key Formula or Approach
The modulus of a complex number \(a + bi\) is given by \(|a + bi| = \sqrt{a^2 + b^2}\). We will substitute \(z = x + iy\) into the given equation and solve for \(y\).
\[ |(x+iy) - i| = |(x+iy) - 1| \] Step 3: Detailed Explanation
1. Substitute \(z = x + iy\) into the equation.
\[ |x + iy - i| = |x + iy - 1| \] 2. Group the real and imaginary parts inside the moduli.
\[ |x + i(y-1)| = |(x-1) + iy| \] 3. Apply the modulus formula to both sides.
\[ \sqrt{x^2 + (y-1)^2} = \sqrt{(x-1)^2 + y^2} \] 4. Square both sides to eliminate the square roots.
\[ x^2 + (y-1)^2 = (x-1)^2 + y^2 \] 5. Expand the squared binomials.
\[ x^2 + (y^2 - 2y + 1) = (x^2 - 2x + 1) + y^2 \] 6. Simplify the equation by canceling terms that appear on both sides.
The \(x^2\), \(y^2\), and \(+1\) terms cancel out.
\[ \cancel{x^2} + \cancel{y^2} - 2y + \cancel{1} = \cancel{x^2} - 2x + \cancel{1} + \cancel{y^2} \] We are left with:
\[ -2y = -2x \] 7. Solve for y.
Divide both sides by -2:
\[ y = x \] Step 4: Final Answer
The relation between y and x is \(y = x\).