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Đặt $z=x+yi$ ($x,y \in R)$*\left| {\frac{{z + 1 - 2i}}{{\bar z + 3 + 4i}}} \right| = 1 $$\Leftrightarrow \left| {z + 1 - 2i} \right| = \left| {\bar z + 3 + 4i} \right|$$ \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$

Đặt $z=x+yi$ ($x,y \in R$)$*\left| {\frac{{z + 1 - 2i}}{{\bar z + 3 + 4i}}} \right| = 1$$\Leftrightarrow \left| {z + 1 - 2i} \right| = \left| {\bar z + 3 + 4i} \right|$$\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 $z = - \frac{{12}}{7} + \frac{{23}}{7}i$