Đặt $z=x+yi$
($x,y \in R$)
$\Leftrightarrow \left| {z + 1 - 2i} \right| = \left| {\bar z + 3 + 4i} \right|$
$ \Leftrightarrow \left| {(x + 1) + (y - 2)i} \right| = \left| {(x + 3) + ( - y + 4)i} \right|$
$ \Leftrightarrow {(x + 1)^2} + {(y - 2)^2} = {(x + 3)^2} + {( - y + 4)^2}$
$ \Leftrightarrow x - y + 5 = 0$ (1)
$*\frac{{z - 2i}}{{\bar z + i}} = \frac{{x + (y - 2)i}}{{x + ( - y + 1)i}} = \frac{{\left[ {x + (y - 2)i} \right]\left[ {x - ( - y + 1)i} \right]}}{{{x^2} + {{( - y + 1)}^2}}} = \frac{{{x^2} - {y^2} + 3y - 2}}{{{x^2} + {{( - y + 1)}^2}}} + \frac{{2xy - 3x}}{{{x^2} + {{( - y + 1)}^2}}}i$
Để $\frac{z-2i}{\overline{z}+i}$ là số ảo
thì ${x^2} - {y^2} + 3y - 2 = 0$ (2)
* (1), (2) $\Rightarrow x = - \frac{{12}}{7},y = \frac{{23}}{7}$
Vậy $z = - \frac{{12}}{7} + \frac{{23}}{7}i$