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. (c)

Solution:

  1. 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.



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