Z₁ and Z₂ be complex numbers such that Z₁$\ne$Z₂ and $\left|{\mathrm{Z}}_1\right|=\left|{\mathrm{Z}}_2\right|$. If Z₁ has positive real  part and Z₂ has negative imaginary part, then $\frac{{\mathrm{Z}}_1+{\mathrm{Z}}_2}{{\mathrm{Z}}_1-{\mathrm{Z}}_2}$ may be

$$\begin{gathered} {\mathrm{Z}}_1=\mathrm{a}+\mathrm{ib},\mathrm{a}>\mathrm{o},\mathrm{b}\in \mathrm{iR} \\ \begin{array}{l}{\mathrm{Z}}_2=\mathrm{c}+\mathrm{id},\mathrm{d}<\mathrm{o},\mathrm{c}\in \mathrm{iR}\\ \left|{\mathrm{Z}}_1\right|=\left|{\mathrm{Z}}_2\right|\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{c}}^2+{\mathrm{d}}^2\Rightarrow {\mathrm{a}}^2-{\mathrm{c}}^2={\mathrm{b}}^2-{\mathrm{d}}^2\Rightarrow (\mathrm{a}+\mathrm{c})(\mathrm{a}-\mathrm{c})=(\mathrm{b}+\mathrm{d})(\mathrm{b}-\mathrm{d})\end{array} \\ \frac{{\mathrm{Z}}_1+{\mathrm{Z}}_2}{{\mathrm{Z}}_1-{\mathrm{Z}}_2}=\frac{(\mathrm{a}+\mathrm{c})+\mathrm{i}(\mathrm{b}+\mathrm{d})}{(\mathrm{a}-\mathrm{c})+\mathrm{i}(\mathrm{b}-\mathrm{d})}\times \frac{(\mathrm{a}-\mathrm{c})-\mathrm{i}(\mathrm{b}-\mathrm{d})}{(\mathrm{a}-\mathrm{c})-\mathrm{i}(\mathrm{b}-\mathrm{d})}=\frac{\left({\mathrm{a}}^2-{\mathrm{c}}^2\right)+\left({\mathrm{b}}^2-{\mathrm{d}}^2\right)+\mathrm{i}\left[\right(\mathrm{a}-\mathrm{c}\left)\right(\mathrm{b}+\mathrm{d})-(\mathrm{a}+\mathrm{c}\left)\right(\mathrm{b}-\mathrm{d}\left)\right]}{(\mathrm{a}-\mathrm{c}{)}^2+(\mathrm{b}-\mathrm{d}{)}^2} \end{gathered}$$
$=\frac{\mathrm{i}(2\mathrm{ad}-2\mathrm{bc})}{(\mathrm{a}-\mathrm{c}{)}^2+(\mathrm{b}-\mathrm{d})}$ is zero or purley imaginary