Question

If \( z_{1} \) and \( z_{2} \) are two distinct non-zero complex numbers such that \( \left|z_{1}\right|=\left|z_{2}\right| \), then \( \frac{z_{1}+z_{2}}{z_{1}-z_{2}} \) is always 

(A) purely real 

(B) purely imaginary 

(C) equal to zero 

(D) none of these

Answer 1

z₁ & z₂ are two non zero distinct complex numbers and |z₁| = |z₂|

Now, \(\frac{z_1+z_2}{z_1-z_2}=\frac{(z_1+z_2)}{(z_1-z_2)}\times\frac{\overline{z_1-z_2}}{\overline{z_1-z_2}}\)

\(=\frac{(z_1+z_2)(\bar{z_1}-\bar{z_2})}{|z_1-z_2|}\) (\(\because z\bar z = |z|^2\))

\(=\frac{z_1\bar{z_1}-z_1\bar{z_2}+z_2\bar{z_1}-z_2\bar z_2}{|z_1-\bar z_2|^2}\) 

\(=\frac{|z_1|^2-|z_2|^2-z_1\bar z_2+z_2\bar z_1}{|z_1-z_2|^2}\)

\(=\frac{z_2\bar z_1-z_1\bar z_2}{|z_1-z_2|^2}\)....(1)

Let z₁ = a + ib & z₂ = c + id

Then \(\bar z_1\) = a - ib & \(\bar z_2\) = c - id

\(\therefore\) \(z_1\bar z_2\) = (a + ib)(c - id) = ac + bd + i(bc - ad)

& \(z_2\bar z_1 = \) (c + id) (c - id) = ac + bd + i(ad - bc)

\(\therefore\) \(z_2\bar z_1-z_1\bar z_2 = \) (ac + bd + i(ad - bc)) - (ac + bd) - i(ad - bc)

 = 2i(ad - bc)

\(\therefore\) From (1)

\(\frac{z_1+z_2}{z_1-z_2}=\frac{2i(ad-bc)}{|z_1-z_2|^2}\) which is a purely imaginary number.