Question

If $|z-2|=min\{|z-1|,|z-5|\}$, where $z$ is a complex number, then possible value(s) of $Re\left(z\right)$ is/are

• A. $\frac{3}{2}$
• B. $\frac{5}{2}$
• C. $\frac{7}{2}$
• D. $\frac{9}{2}$

Answer 1

Answer: (A), (C)

$\frac{7}{2}$

Given : $|z-2|=min\{|z-1|,|z-5|\}$
Let $z=x+iy$
When $|z-1|<|z-5|$
$\Rightarrow (x-1{)}^2+y^2<(x-5{)}^2+y^2\Rightarrow -2x+1<-10x+25\Rightarrow x<3$
Now,
$|z-2|=|z-1|$
$z$ is perpendicular bisector of line segment joining points $(2,0)$ and $(1,0)$
$x=\frac{3}{2}$

When $|z-1|\ge |z-5|$
$\Rightarrow x\ge 3$
Now,
$|z-2|=|z-5|$
$z$ is perpendicular bisector of line segment joining the points $(2,0)$ and $(5,0)$
$x=\frac{7}{2}$

Hence, the possible values of $Re\left(z\right)$ are $\frac{3}{2}$ and $\frac{7}{2}.$