Question

If $z_1$ and $z_2$ both satisfy the relation $z + \bar{z} = 2|z - 1|$ and $arg (z_1 - z_2) = \pi/4$, then $Im(z_1 + z_2)$ equals

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

Let $z = a + ib$
$\therefore \bar{z} = a - ib$
$\Rightarrow a = \frac{ z + \bar{z}}{2} = $ Real part of $z$
$\Rightarrow 2a = 2|a -1 + ib| \,\, (\because z + \bar{z} = 2| z - 1|)$
$\Rightarrow 2a = 1 + b^2$
$\Rightarrow 2(a_1 - a_2) = b_1^2 - b_2^2 $
$= (b_1 + b_2)(b_1 - b_2) \,...(i)$
But $arg(z_1 - z_2) = \pi/4$
$\therefore tan^{-1}\left(\frac{b_1 - b_2}{a_1 - a_2}\right) = \pi/4$
$\Rightarrow b_1 - b_2 = a_1 - a_2 \,...(ii)$
Using (ii) in (i) we get,
$b_1 + b_2 = 2$
$\Rightarrow Im (z_ 1 + z_2) = 2$