Complex Numbers 2 Question 39
40. If $z _1$ and $z _2$ are two complex numbers such that $\left|z _1\right|<1<\mid z _2$, then prove that $\left|\frac{1-z _1 \bar{z} _2}{z _1-z _2}\right|<1$.
(2003, 2M)
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Solution:
- Given, $\left|z _1\right|<1$ and $\left|z _2\right|>1$
Then, to prove
$$ \begin{array}{lll} & \frac{1-z _1 \bar{z} _2}{z _1-z _2}<1 \quad \text { using } \frac{z _1}{z _2}=\frac{\left|z _1\right|}{\left|z _2\right|} \\ \Rightarrow \quad & \left|1-z _1 \bar{z} _2\right|<\left|z _1-z _2\right| & \ldots \text { (ii) } \end{array} $$
On squaring both sides, we get,
$$ \begin{array}{rc} & \left(1-z _1 \bar{z} _2\right)\left(1-\bar{z} _1 z _2\right)<\left(z _1-z _2\right)\left(\bar{z} _1-\bar{z} _2\right) \quad\left[\text { using }|z|^{2}=z \bar{z}\right] \\ \Rightarrow & 1-z _1 \bar{z} _2-\bar{z} _1 z _2+z _1 \bar{z} _1 z _2 \bar{z} _2<z _1 \bar{z} _1-z _1 \bar{z} _2-z _2 \bar{z} _1+z _2 \bar{z} _2 \\ \Rightarrow & 1+\left|z _1\right|^{2}\left|z _2\right|^{2}<\left|z _1\right|^{2}+\left|z _2\right|^{2} \\ \Rightarrow & 1-\left|z _1\right|^{2}-\left|z _2\right|^{2}+\left|z _1\right|^{2}\left|z _2\right|^{2}<0 \\ \Rightarrow & \left(1-\left|z _1\right|^{2}\right)\left(1-\left|z _2\right|^{2}\right)<0 \end{array} $$
which is true by Eq. (i) as $\left|z _1\right|<1$ and $\left|z _2\right|>1$
$$ \therefore \quad\left(1-\left|z _1\right|^{2}\right)>0 \text { and }\left(1-\left|z _2\right|^{2}\right)<0 $$
$\therefore$ Eq. (iii) is true whenever Eq. (ii) is true.
$\Rightarrow \quad \frac{1-z _1 \bar{z} _2}{z _1-z _2}<1$
Hence proved.
