Test Index

Complex Numbers Part 1

Section: Mathematics

© examsnet.com

Question : 42 of 100

Marks: +1, -0

If

z_{1}, z_{2}

are complex numbers such that

\Re(z_{1}) = |z_{1}-1|,

\Re(z_{2}) = |z_{2}-1|

\arg(z_{1}-z_{2}) = \frac{\pi}{6},

\Im(z_{1}+z_{2})

[3 Sep 2020 Shift 2]

$\frac{\sqrt{3}}{2}$
$\frac{2}{\sqrt{3}}$
$\frac{1}{\sqrt{3}}$
$2 \sqrt{3}$

Solution:

\Re(z) = |z-1|

\Rightarrow x = \sqrt{(x-1)^{2}+(y-0)^{2}} \;\; (x>0)

\Rightarrow y^{2}=2x-1

=4 \cdot \frac{1}{2}\left(x-\frac{1}{2}\right)

\Rightarrow

a parabola with focus (1,0)

\&

directrix as imaginary axis.

\therefore

Vertex

=\left(\frac{1}{2}, 0\right)

A(z_{1}) \& B(z_{2})

are two points on it such that slope of

AB = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}

\left(\arg(z_{1}-z_{2}) = \frac{\pi}{6}\right)

for

\;\;\; y^{2}=4 a x

Let

\;\; A(a t_{1}^{2}, 2 a t_{1}) \& B(a t_{2}^{2}, 2 a t_{2})

m_{AB} = \frac{2}{t_{1}+t_{2}}

= \frac{4 a}{y_{1}+y_{2}} = \frac{1}{\sqrt{3}}

(

Here

a = \frac{1}{2})

\Rightarrow y_{1}+y_{2}

=4 a \sqrt{3}=2 \sqrt{3}

© examsnet.com

Go to Question:

Prev Question

Next Question