SATHEE: Complex Numbers 2 Question 39

Complex Numbers 2 Question 39

40. If $z_{1}$ and $z_{2}$ are two complex numbers such that $| z_{1} | < 1 <∣ z_{2}$ , then prove that $| \frac{1 - z_{1} {\bar{z}}_{2}}{z_{1} - z_{2}} | < 1$ .

(2003, 2M)

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Solution:

Given, $| z_{1} | < 1$ and $| z_{2} | > 1$

Then, to prove

$\begin{array}{lll} \frac{1 - z_{1} {\bar{z}}_{2}}{z_{1} - z_{2}} < 1 using \frac{z_{1}}{z_{2}} = \frac{| z_{1} |}{| z_{2} |} \\ \Rightarrow & | 1 - z_{1} {\bar{z}}_{2} | < | z_{1} - z_{2} | & \dots (ii) \end{array}$

On squaring both sides, we get,

$\begin{array}{rc} (1 - z_{1} {\bar{z}}_{2}) (1 - {\bar{z}}_{1} z_{2}) < (z_{1} - z_{2}) ({\bar{z}}_{1} - {\bar{z}}_{2}) [using | z |^{2} = z \bar{z}] \\ \Rightarrow & 1 - z_{1} {\bar{z}}_{2} - {\bar{z}}_{1} z_{2} + z_{1} {\bar{z}}_{1} z_{2} {\bar{z}}_{2} < z_{1} {\bar{z}}_{1} - z_{1} {\bar{z}}_{2} - z_{2} {\bar{z}}_{1} + z_{2} {\bar{z}}_{2} \\ \Rightarrow & 1 + {| z_{1} |}^{2} {| z_{2} |}^{2} < {| z_{1} |}^{2} + {| z_{2} |}^{2} \\ \Rightarrow & 1 - {| z_{1} |}^{2} - {| z_{2} |}^{2} + {| z_{1} |}^{2} {| z_{2} |}^{2} < 0 \\ \Rightarrow & (1 - {| z_{1} |}^{2}) (1 - {| z_{2} |}^{2}) < 0 \end{array}$

which is true by Eq. (i) as $| z_{1} | < 1$ and $| z_{2} | > 1$

$∴ (1 - {| z_{1} |}^{2}) > 0 and (1 - {| z_{2} |}^{2}) < 0$

$∴$ Eq. (iii) is true whenever Eq. (ii) is true.

$\Rightarrow \frac{1 - z_{1} {\bar{z}}_{2}}{z_{1} - z_{2}} < 1$

Hence proved.

Complex Numbers 2 Question 39

40. If $z_{1}$ and $z_{2}$ are two complex numbers such that $| z_{1} | < 1 <∣ z_{2}$ , then prove that $| \frac{1 - z_{1} {\bar{z}}_{2}}{z_{1} - z_{2}} | < 1$ .

Solution:

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Complex Numbers 2 Question 39

40. If z1 and z2 are two complex numbers such that |z1|<1<∣z2, then prove that |1−z1z¯2z1−z2|<1.

Solution:

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40. If $z_{1}$ and $z_{2}$ are two complex numbers such that $| z_{1} | < 1 <∣ z_{2}$ , then prove that $| \frac{1 - z_{1} {\bar{z}}_{2}}{z_{1} - z_{2}} | < 1$ .