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let z be a complex number satisfying z 16 4 z 1 th
Question:
Let \( z \) be a complex number satisfying \( |z + 16| = 4|z + 1| \). Then:
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When squaring both sides of an equation involving complex numbers expressed in modulus form, ensure to expand and simplify carefully to find the modulus \( |z| \).
KEAM - 2024
KEAM
Updated On:
Mar 10, 2025
\( |z| = 2 \)
\( |z| = 4 \)
\( |z| = 8 \)
\( |z| = 10 \)
\( |z| = 16 \)
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The Correct Option is
B
Solution and Explanation
We start with the given condition: \[ |z + 16| = 4|z + 1| \] Assume \( z = x + iy \), where \( x \) and \( y \) are real numbers. This leads to: \[ \sqrt{(x + 16)^2 + y^2} = 4\sqrt{(x + 1)^2 + y^2} \] Squaring both sides to eliminate the square roots: \[ (x + 16)^2 + y^2 = 16((x + 1)^2 + y^2) \] Expanding and simplifying: \[ x^2 + 32x + 256 + y^2 = 16(x^2 + 2x + 1 + y^2) \] \[ x^2 + 32x + 256 + y^2 = 16x^2 + 32x + 16 + 16y^2 \] Bringing all terms to one side and simplifying: \[ 15x^2 + 15y^2 = 240 \] \[ x^2 + y^2 = 16 \] Thus, \( |z| = \sqrt{x^2 + y^2} = \sqrt{16} = 4 \).
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