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Range of Quadratic Expression

If | z1| 1,...

Question

If $| z_{1} | \neq 1, ∣ ∣ \frac{z_{1} - z_{2}}{1 - {¯ z}_{1} z_{2}} ∣ ∣ = 1$ , then find $z_{2}$

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Solution

$∣ ∣ ∣ \frac{z_{1} - z_{2}}{1 - {¯ ¯ ¯ z}_{1} z_{2}} ∣ ∣ ∣ = 1 \Rightarrow ∣ ∣ ∣ \frac{a + i b - c - i d}{1 - (a - i) (c + i d)} ∣ ∣ ∣ = 1$
$\Rightarrow ∣ ∣ ∣ \frac{(a - c) + i (b - d)}{1 - (a c + b d + i (a d - b c)} ∣ ∣ ∣ = 1 \Rightarrow ∣ ∣ ∣ \frac{(a - c) + i (b - d)}{(1 - a c - b d) + i (b c - a d)} ∣ ∣ ∣ = 1$
$| (a - c) + i (b - d) | = | (1 - a c - b d) + i (b c - a d) |$
${(a - c)}^{2} + {(b - d)}^{2} = {(1 - a c - b d)}^{2} + {(b c - a d)}^{2} \Rightarrow {(a)}^{2} + {(b)}^{2} + {(c)}^{2} + {(d)}^{2} - 2 (a c + b d) = 1 + {(a)}^{2} {(c)}^{2} + {(b)}^{2} {(d)}^{2} - 2 a c - 2 b d + 2 a b c d + {(b)}^{2} {(c)}^{2} + {(a)}^{2} {(d)}^{2} - 2 a b c d$
$\Rightarrow {(a)}^{2} + {(b)}^{2} + {(c)}^{2} + {(d)}^{2} = 1 + a^{2} (c^{2} + d^{2}) + b^{2} (c^{2} + d^{2})$
$\Rightarrow {| z_{1} |}^{2} + {| z_{2} |}^{2} = 1 + (c^{2} + d^{2}) (a^{2} + b^{2})$
$\Rightarrow {| z_{1} |}^{2} + {| z_{2} |}^{2} = 1 + {| z_{1} |}^{2} {| z_{2} |}^{2} \Rightarrow {| z_{1} |}^{2} [1 - {| z_{2} |}^{2}]$
$= [1 - {| z_{2} |}^{2}] \Rightarrow [1 - {| z_{2} |}^{2}] [1 - {| z_{2} |}^{2}] = 0$
$| z_{1} | \neq 0 \Rightarrow {| z_{2} |}^{2} - 1 = 0 \Rightarrow | z_{2} | = 1$

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