What is $\frac{z_1+z_2}{z_1−z_2}$ if $z_1,z_2\in\mathbb{C}$, such that $z_1\neq z_2,|z_1|\neq|z_2|$

Let $$z_1$$ and $$z_2$$ be two complex numbers such that $$z_1≠z_2$$ and $$|z_1|\neq|z_2|$$. If $$z_1$$ has positive real part and $$z_2$$ has negative imaginary part, then $$\frac{z_1+z_2}{z_1−z_2}$$ may be ___________

a) zero or purely imaginary

b) real and +ve

c) real and -ve

My reference gives the solution "zero or purely imaginary". If it was $$|z_1|=|z_2|$$ I can easily find the solution from geometry as the complex numbers $$z_1$$ and $$z_2$$ form a rhombus and $$z_1+z_2$$ and $$z_1-z_2$$ be its diagonals and they intersect orthoganally, thus $$\arg\Big(\frac{z_1+z_2}{z_1-z_2}\Big)=\arg(z_1+z_2)-\arg(z_1-z_2)=\pm\frac{\pi}{2}$$, thus purely imaginary. But, in this case how do I find the solution ?

• For $z_1 = 2$ and $z_2 = -i$ I get $(z_1+z_2)/(z_1 - z_2) = (3+4 i)/5$, therefore I cannot see how "zero or purely imaginary" can be the correct solution. – Martin R Jan 31 at 10:31
• @MartinR My reference gives that solution, i'm confused. – ss1729 Jan 31 at 10:34
• Are you sure that it is not $z_1\neq z_2$ and $|z_1|=|z_2|$? – Reinhard Meier Jan 31 at 11:20
• I agree with Reinhard Meier, there is apparently a typo. If $|z_1|\neq|z_2|,\;$ all three possibilities can occur. – user376343 Jan 31 at 22:38