Question

I. $\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$ if $\frac{z_1}{z_2}$ is purely imaginary
II. If z is purely real then $z=\overline{z}$

• A. Only II is true
• B. Only I is true
• C. Both I and II are true
• D. I is false but II is true

Answer 1

Answer: (C)

Both I and II are true

$\left|\frac{z_1-z_2}{z_1-z_2}\right|=1$
Dividing by $z_2$
$\left|\frac{z_1/z_2-1}{z_1/z_2+1}\right|=1$
Let $z_1/z_2$ be a complex number $u=x+iy$
$\Rightarrow \left|\frac{u-1}{u+1}\right|=1$
$\Rightarrow {|u-1|}^2={|u+1|}^2$
$\Rightarrow (x-1{)}^2+y^2=(x+1{)}^2+y^2$
$\therefore (x-1{)}^2=(x+1{)}^2$
Which is possible only if x=0
$\therefore u=iy$ ( purely imaginary)
ii) If $z$ is purely real, then $z=x$ (imaginary part zero)
Hence $\overline{z}$ is also equal to $x$.
$\Rightarrow z=\overline{z}$