If z_1 and z_2 are complex numbers such that ∣z_2 ∣≠1 and ∣(z_1 −2z_2 )/(2−z_1 z_2 ^� )∣=1, then ∣z_1 ∣ is equal to ____


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Question Number 140442 by EnterUsername last updated on 07/May/21

$If z_{1} and z_{2} are complex numbers such that ∣ z_{2} ∣\neq 1 and$ $∣ (z_{1} - 2 z_{2}) / (2 - z_{1} {\bar{z}}_{2}) ∣= 1, then ∣ z_{1} ∣ is equal to_____.$
	Answered by mr W last updated on 08/May/21

	$∣ \frac{r_{1} e^{θ_{1} i} - 2 r_{2} e^{θ_{2} i}}{2 - r_{1} e^{θ_{1} i} r_{2} e^{- θ_{2} i}} ∣= 1$ $∣ \frac{(r_{1} e^{(θ_{1} - θ_{2}) i} - 2 r_{2}) e^{θ_{2} i}}{2 - r_{1} r_{2} e^{(θ_{1} - θ_{2}) i}} ∣= 1$ $∣ \frac{(r_{1} e^{θ_{3} i} - 2 r_{2}) e^{θ_{2} i}}{2 - r_{1} r_{2} e^{θ_{3} i}} ∣= 1$ $∣ \frac{\sqrt{{(r_{1} \cos θ_{3} - 2 r_{2})}^{2} + {(r_{1} \sin θ_{3})}^{2}} e^{θ_{4} i} e^{θ_{2} i}}{\sqrt{{(2 - r_{1} r_{2} \cos θ_{3})}^{2} + {(- r_{1} r_{2} \sin θ_{3})}^{2}} e^{θ_{5} i}} ∣= 1$ $∣ \frac{\sqrt{r_{1}^{2} + 4 r_{2}^{2} - 4 r_{1} r_{2} \cos θ_{3}}}{\sqrt{4 + r_{1}^{2} r_{2}^{2} - 4 r_{1} r_{2} \cos θ_{3}}} e^{θ_{6} i} ∣= 1$ $\frac{\sqrt{r_{1}^{2} + 4 r_{2}^{2} - 4 r_{1} r_{2} \cos θ_{3}}}{\sqrt{4 + r_{1}^{2} r_{2}^{2} - 4 r_{1} r_{2} \cos θ_{3}}} = 1$ $r_{1}^{2} + 4 r_{2}^{2} = 4 + r_{1}^{2} r_{2}^{2}$ $r_{1}^{2} (1 - r_{2}^{2}) = 4 (1 - r_{2}^{2})$ $r_{1}^{2} = 4 s i n c e r_{2} \neq 1$ $\Rightarrow r_{1} = 2 =∣ z_{1} ∣$
	Commented by EnterUsername last updated on 12/May/21

	$Thank you Sir$
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