Question

Let $z_1=1+i$ and $Z_2=-1-i$. Find $z_3ϵC$ such that triangle $z_1{z}_2{z}_3$ is equilateral.

Answer 1

Let us have point $z_3$ as $x+\iota y$

Now, for $△z_1{z}_2{z}_3$ to be equilateral, $z_1{z}_2=z_2{z}_3=z_1{z}_3$

$\therefore z_1{z}_2=\sqrt{(1-(-1){)}^2+(1-(-1){)}^2}=2\sqrt{2}$ ...(1)

$\therefore z_2{z}_3=\sqrt{(x-(-1){)}^2+(y-(-1){)}^2}$ ...(2)

$\therefore z_1{z}_3=\sqrt{(x+1{)}^2+(y+1{)}^2}$ ...(3)

equating (1) and (2),

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

squaring both sides,

$8=x^2+2x+y^2+2y+2$

$\therefore x^2+2x+y^2+2y-6=0$ ...(4)

equating (3) and (1),

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

squaring both sides,

$8=x^2-2x+y^2-2y+2$

$\therefore x^2-2x+y^2-2y-6=0$ ...(5)

subtracting (5) from (4), we have

$4x+4y=0⟹x=-y$

Hence, substituting in (4), we have

$x^2+2x+x^2-2x-6=0$

$\therefore x^2=3$

$\therefore x=\pm \sqrt{3}$

Hence, $z_3$ can be $(\sqrt{3},-\sqrt{3})$ or $(-\sqrt{3},\sqrt{3})$.