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Condition for Coplanarity of Four Points

Let Z1 and ...

Question

Let $Z_{1}$ and $Z_{2}$ be two complex numbers such that $∣ ∣ ∣ \frac{Z_{1} - 2 Z_{2}}{2 - z_{1} {¯ ¯ ¯ Z}_{2}} ∣ ∣ ∣ = 1$ and $| Z_{2} | \neq 1$ , find $| Z_{1} |$

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Solution

Given: $∣ ∣ ∣ \begin{matrix} \frac{Z_{1} - 2 Z_{2}}{2 - Z_{1}_{2}} \end{matrix} ∣ ∣ ∣ = 1$
To find: $| \begin{matrix} Z_{1} \end{matrix} |$
On squaring we get
${| \begin{matrix} Z_{1} - 2 Z_{2} \end{matrix} |}^{2} = {| \begin{matrix} 2 - Z_{1}_{2} \end{matrix} |}^{2}$

Now we know that ${| \begin{matrix} Z \end{matrix} |}^{2} = Z ¯ Z$
$\Rightarrow (Z_{1} - 2 Z_{2}) (_{1} - 2 ¯ Z_{2}) = (2 - Z_{1} ¯ Z_{2}) (2 -_{1} Z_{2})$

${\Rightarrow | \begin{matrix} Z_{1} \end{matrix} |}^{2} - 2 Z_{1}_{2} - 2 Z_{2}_{1} + 4 {| \begin{matrix} Z_{2} \end{matrix} |}^{2} = 4 - 2_{1} Z_{2} - 2 Z_{1}_{2} + {| \begin{matrix} Z_{1} \end{matrix} |}^{2} {| \begin{matrix} Z_{2} \end{matrix} |}^{2}$

${\Rightarrow | \begin{matrix} Z_{1} \end{matrix} |}^{2} - {| \begin{matrix} Z_{1} \end{matrix} |}^{2} {| \begin{matrix} Z_{2} \end{matrix} |}^{2} = 4 - 4 {| \begin{matrix} Z_{2} \end{matrix} |}^{2}$

${\Rightarrow | \begin{matrix} Z_{1} \end{matrix} |}^{2} {1 - {| \begin{matrix} Z_{2} \end{matrix} |}^{2}} = 4 {1 - {| \begin{matrix} Z_{2} \end{matrix} |}^{2}}$

Since $| \begin{matrix} Z_{2} \end{matrix} | \neq 1$

${\Rightarrow | \begin{matrix} Z_{1} \end{matrix} |}^{2} = 4$

$\Rightarrow | \begin{matrix} Z_{1} \end{matrix} | = 2$

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Condition for Coplanarity of Four Points

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