Question:
If \(z_1,\ z_2\) are complex numbers such that
\(\mathrm{Re}(z_1)=|z_1-1|\), \(\mathrm{Re}(z_2)=|z_2-1|\), and
\(\arg(z_1-z_2)=\frac{\pi}{6}\), then \(\mathrm{Im}(z_1+z_2)\) is equal to :
If \(z_1,\ z_2\) are complex numbers such that
\(\mathrm{Re}(z_1)=|z_1-1|\), \(\mathrm{Re}(z_2)=|z_2-1|\), and
\(\arg(z_1-z_2)=\frac{\pi}{6}\), then \(\mathrm{Im}(z_1+z_2)\) is equal to :
\(\mathrm{Re}(z_1)=|z_1-1|\), \(\mathrm{Re}(z_2)=|z_2-1|\), and
\(\arg(z_1-z_2)=\frac{\pi}{6}\), then \(\mathrm{Im}(z_1+z_2)\) is equal to :
Show Hint
\(\mathrm{Re}(z)=|z-1|\) describes the parabola \(y^2=2x-1\); the argument of \(z_1-z_2\) gives the slope of the chord joining the two points.
Updated On: Nov 5, 2026
- \(\frac{\sqrt3}{2}\)
- \(\frac{1}{\sqrt3}\)
- \(\frac{2}{\sqrt3}\)
- \(2\sqrt3\)
Show Solution
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The Correct Option is D
Solution and Explanation
Let \(z=x+iy\). \(\mathrm{Re}(z)=|z-1|\Rightarrow x^2=(x-1)^2+y^2\Rightarrow y^2=2x-1\) (a parabola). Both \(z_1,z_2\) lie on it.
\(\arg(z_1-z_2)=\frac\pi6\Rightarrow\) slope of chord \(=\tan\frac\pi6=\frac1{\sqrt3}\).
Slope \(=\frac{y_1-y_2}{x_1-x_2}=\frac{y_1-y_2}{\frac{y_1^2-y_2^2}{2}}=\frac{2}{y_1+y_2}\).
So \(\frac{2}{y_1+y_2}=\frac1{\sqrt3}\Rightarrow y_1+y_2=2\sqrt3\), i.e. \(\mathrm{Im}(z_1+z_2)=2\sqrt3\).
Answer: (4) \(2\sqrt3\)
\(\arg(z_1-z_2)=\frac\pi6\Rightarrow\) slope of chord \(=\tan\frac\pi6=\frac1{\sqrt3}\).
Slope \(=\frac{y_1-y_2}{x_1-x_2}=\frac{y_1-y_2}{\frac{y_1^2-y_2^2}{2}}=\frac{2}{y_1+y_2}\).
So \(\frac{2}{y_1+y_2}=\frac1{\sqrt3}\Rightarrow y_1+y_2=2\sqrt3\), i.e. \(\mathrm{Im}(z_1+z_2)=2\sqrt3\).
Answer: (4) \(2\sqrt3\)
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