If the complex numbers $z_1=a+i,$ $z_2=1+ib,$ $z_3=0$  form an equilateral triangle $0<a;b<1,$ then:

Let $A=z_1=a+i,$ $B=z_2=1+ib,O=z_3=0$

$\Delta OAB$ is equilateral so$OA=OB=AB$

$\Rightarrow$    $1+a^2=1+b^2={\left(a-1\right)}^2+{\left(1-b\right)}^2$

$\therefore$     $a^2=b^2$

Since both $aandb$  are greater than 0, We have $a=b$

$\therefore$  $1+a^2=2{\left(a-1\right)}^2=2a^2-4a+2$

$\Rightarrow$   $a^2-4a+1=0$

$\therefore$   $a=\frac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3}$

Since $a<1$  we have $a=2-\sqrt{3}=b.$