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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 ?

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    $\begingroup$ 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. $\endgroup$ – Martin R Jan 31 at 10:31
  • $\begingroup$ @MartinR My reference gives that solution, i'm confused. $\endgroup$ – ss1729 Jan 31 at 10:34
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    $\begingroup$ Are you sure that it is not $z_1\neq z_2$ and $|z_1|=|z_2|$? $\endgroup$ – Reinhard Meier Jan 31 at 11:20
  • $\begingroup$ I agree with Reinhard Meier, there is apparently a typo. If $|z_1|\neq|z_2|,\;$ all three possibilities can occur. $\endgroup$ – user376343 Jan 31 at 22:38

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