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Modulus of a Complex Number

Let z1 and ...

Question

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $\frac{z_{1} - 3 z_{2}}{3 - z_{1} {¯ ¯ ¯ z}_{2}}$ is unimodular. If $z_{2}$ is not unimodular then $| z_{1} |$ is

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Solution

$\begin{matrix} A c c o r d i n g t o q u e s t i o n . . . . . . . . . . . . . \frac{z_{1} - 3 z_{2}}{3 - z_{1} {¯ z}_{2}} = 1 a n d | z_{1} | \neq 1 z_{1} - 3 z_{2} = 3 - z_{1} {¯ z}_{2} s q u a r i n g b o t h s i d e : \Rightarrow {| z_{1} - 3 z_{2} |}^{2} = {| 3 - z_{1} {¯ z}_{2} |}^{2} \Rightarrow {| z_{1} |}^{2} + 9 {| z_{2} |}^{2} - 6 | z_{1} | | z_{2} | = 9 + {| z_{1} |}^{2} {| z_{2} |}^{2} - 6 | z_{1} | | z_{2} | \Rightarrow {| z_{1} |}^{2} + 9 {| z_{2} |}^{2} - 6 | z_{1} | | z_{2} | - 9 - {| z_{1} |}^{2} {| z_{2} |}^{2} + 6 | z_{1} | | z_{2} | = 0 \Rightarrow {| z_{1} |}^{2} - {| z_{1} |}^{2} {| z_{2} |}^{2} + 9 {| z_{2} |}^{2} - 9 = 0 \Rightarrow | z_{1} |^{2} (1 - {| z_{2} |}^{2}) - 9 (1 - {| z_{2} |}^{2}) \Rightarrow (| z_{1} |^{2} - 9) (1 - {| z_{2} |}^{2}) = 0 w e k n o w t h a t {| z_{2} |}^{2} \neq 1 s o, \Rightarrow | z_{1} |^{2} - 9 = 0 \Rightarrow | z_{1} |^{2} = 9 ∴ | z_{1} | = 3 \end{matrix}$

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