Question

Find the number of complex numbers which satisfies both the equations $|z-1-i|=\sqrt{2}$ and $|z+1+i|=2$.

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

We have,
$|z-1-i|=\sqrt{2}$ and $|z+1+i|=2$
Now put $z=x+iy$
$\Rightarrow (x-1{)}^2+(y-1{)}^2=2\Rightarrow \left(1\right)$
and $(x+1{)}^2+(y+1{)}^2=4\Rightarrow \left(2\right)$
Now by $\left(2\right)-\left(1\right)$, we get, $x+y=\frac{1}{2}$
Thus by $\left(1\right)$, $(x-1{)}^2+(x+1/2{)}^2=2$
$\Rightarrow 2x^2-x-3/4=0$
Discriminant $>0\Rightarrow$ There will be two roots of above quadratic
Hence there will be two roots of given equations.