Let z₁ and z₂ be complex numbers such that ${\mathrm{z}}_1\ne {\mathrm{z}}_2$ 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

Let ${\mathrm{z}}_1=\mathrm{a}+\mathrm{ib}, \mathrm{a}>0\text{ and }\mathrm{b}\in \mathrm{R}:{\mathrm{z}}_2=\mathrm{c}+\mathrm{id}, \mathrm{d}<0, \mathrm{c}\in \mathrm{R}$

Given |z₁| = |z₂|

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{c}}^2+{\mathrm{d}}^2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}^2-{\mathrm{c}}^2={\mathrm{d}}^2-{\mathrm{b}}^2 \left(1\right)\end{array}$

Now, $\frac{{\mathrm{z}}_1+{\mathrm{z}}_2}{{\mathrm{z}}_1-{\mathrm{z}}_2}=\frac{(\mathrm{a}+\mathrm{c})+\mathrm{i}(\mathrm{b}+\mathrm{d})}{(\mathrm{a}-\mathrm{c})+\mathrm{i}(\mathrm{b}-\mathrm{d})}$

                           $\left[\left({\mathrm{a}}^2-{\mathrm{c}}^2\right)+\left({\mathrm{b}}^2-{\mathrm{d}}^2\right)\right]+$

                      $=\frac{\mathrm{i}\left[\right(\mathrm{a}-\mathrm{c}\left)\right(\mathrm{b}+\mathrm{d})-(\mathrm{a}+\mathrm{c}\left)\right(\mathrm{b}-\mathrm{d}\left)\right]}{(\mathrm{a}-\mathrm{c}{)}^2+(\mathrm{b}-\mathrm{d}{)}^2}$

This values will be purely imaginary as ${\mathrm{a}}^2-{\mathrm{c}}^2={\mathrm{d}}^2-{\mathrm{b}}^2$ or the value will be zero in case a + c = b + d = 0.