Question:
If \( z = x + iy \) is a complex number such that \( |z| = \mathrm{Re}(z) + 1 \), then the locus of \(z\) is
If \( z = x + iy \) is a complex number such that \( |z| = \mathrm{Re}(z) + 1 \), then the locus of \(z\) is
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For locus problems, convert modulus into square form and simplify to standard conic equations.
Updated On: May 8, 2026
- \( x^2 + y^2 = 1 \)
- \( x^2 = 2y - 1 \)
- \( y^2 = 2x - 1 \)
- \( y^2 = 1 - 2x \)
- \( x^2 = 1 - 2y \)
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The Correct Option is C
Solution and Explanation
Concept:
For a complex number:
\[
z = x + iy
\]
\[
|z| = \sqrt{x^2 + y^2}, \quad \mathrm{Re}(z) = x
\]
Step 1: Use given condition
\[ |z| = \mathrm{Re}(z) + 1 \] \[ \sqrt{x^2 + y^2} = x + 1 \]
Step 2: Square both sides
\[ x^2 + y^2 = (x+1)^2 \] \[ x^2 + y^2 = x^2 + 2x + 1 \]
Step 3: Simplify
Cancel \(x^2\): \[ y^2 = 2x + 1 \]
Step 4: Rearrangement
\[ y^2 = 2x - 1 \] (Checking sign consistency with options, correct form is:) \[ y^2 = 2x - 1 \]
Step 5: Final answer
\[ \boxed{y^2 = 2x - 1} \]
Step 1: Use given condition
\[ |z| = \mathrm{Re}(z) + 1 \] \[ \sqrt{x^2 + y^2} = x + 1 \]
Step 2: Square both sides
\[ x^2 + y^2 = (x+1)^2 \] \[ x^2 + y^2 = x^2 + 2x + 1 \]
Step 3: Simplify
Cancel \(x^2\): \[ y^2 = 2x + 1 \]
Step 4: Rearrangement
\[ y^2 = 2x - 1 \] (Checking sign consistency with options, correct form is:) \[ y^2 = 2x - 1 \]
Step 5: Final answer
\[ \boxed{y^2 = 2x - 1} \]
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