Question

Find complex numbers satisfying $|z-1|=|z-3|=|z-i|,$

• A. $-2-2i$
• B. $2+2i$
• C. $1+i$
• D. $-1-i$

Answer 1

The correct option is B $2+2i$

Let $z=x+iy$

Hence, $|z-1|=|z-3|$

$\Rightarrow (x-1{)}^2=(x-3{)}^2$

$\Rightarrow (2x-4)=0$

$\Rightarrow x=2$ -----(i)

and

$\Rightarrow |z-1|=|z-i|$

$\Rightarrow (x-1{)}^2+y^2=x^2+(y-1{)}^2$

Now substituting $x=2$, we get

$\Rightarrow 1+y^2=4+(y-1{)}^2$

$\Rightarrow (2y-1)=3$

$\Rightarrow 2y=4$

$\Rightarrow y=2$.

Hence, $z=x+iy$

$=2+2i$.