Question

Let $z_1$ and $z_2$ be complex numbers such that $z_1\ne z_2$ and $|z_1|=|z_2|$. 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
• B. Real and positive
• C. Real and negative
• D. Purely imaginary

Answer 1

Answer: (A), (D)

Purely imaginary

Given, $|z_1|=|z_2|$
Now, $\frac{z_1+z_2}{z_1-z_2}\times \frac{\overline{z_1}-\overline{z_2}}{\overline{z_1}-\overline{z_2}}=\frac{z_1\overline{z_1}-z_1\overline{z_2}+z_2\overline{z_1}-z_2\overline{z_2}}{|z_1-z_2{|}^2}=\frac{|z_1{|}^2+(z_2\overline{z_1}-z_1\overline{z_2})-|z_2{|}^2}{|z_1-z_2{|}^2}=\frac{z_2\overline{z_1}-z_1\overline{z_2}}{|z_1-z_2{|}^2}$
As we know $z-\overline{z}=2iIm\left(z\right)\therefore z_2\overline{z_1}-z_1\overline{z_2}=2iIm\left(z_2\overline{z_1}\right)\therefore \frac{z_1+z_2}{z_1-z_2}=\frac{2iIm\left(z_2\overline{z_1}\right)}{|z_1-z_2{|}^2}$

Which is purely imaginary or zero.
Therefore, (a) and (d) are correct answers.