Complex Numbers 2 Question 23
24.
Let $z _1$ and $z _2$ be complex numbers such that $z _1 \neq z _2$ and $\left|z _1\right|=\left|z _2\right|$. 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
(1986, 2M)
(a) zero
(b) real and positive
(c) real and negative
(d) purely imaginary
Show Answer
Answer:
Correct Answer: 24. (a,d)
Solution:
- Given, $\left|z _1\right|=\left|z _2\right|$
Now, $\frac{z _1+z _2}{z _1-z _2} \times \frac{\bar{z} _1-\bar{z} _2}{\bar{z} _1-\overline{z _2}}=\frac{z _1 \bar{z} _1-z _1 \bar{z} _2+z _2 \bar{z} _1-z _2 \bar{z} _2}{\left|z _1-z _2\right|^{2}}$
$ \begin{aligned} & =\frac{\left|z _1\right|^{2}+\left(z _2 \bar{z} _1-z _1 \bar{z} _2\right)-\left|z _2\right|^{2}}{\left|z _1-z _2\right|^{2}} \\ & =\frac{z _2 \bar{z} _1-z _1 \bar{z} _2}{\left|z _1-z _2\right|^{2}} \quad\left[\because\left|z _1\right|^{2}=\left|z _2\right|^{2}\right] \end{aligned} $
As, we know $z-\bar{z}=2 i \operatorname{Im}(z)$
$ \begin{aligned} & \therefore \quad z _2 \bar{z} _1-z _1 \bar{z} _2=2 i \operatorname{Im}\left(z _2 \bar{z} _1\right) \\ & \therefore \quad \frac{z _1+z _2}{z _1-z _2}=\frac{2 i \operatorname{Im}\left(z _2 \bar{z} _1\right)}{\left|z _1-z _2\right|^{2}} \end{aligned} $
which is purely imaginary or zero.
