z_1 =3+i z_2 =1−2i determinant ((((2z_2 +z_1 −5−i)/(2z_1 −z


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Question Number 47391 by gunawan last updated on 09/Nov/18

$z_{1} = 3 + i$ $z_{2} = 1 - 2 i$ ${\| \begin{matrix} \frac{2 z_{2} + z_{1} - 5 - i}{2 z_{1} - z_{2} + 3 - i} \end{matrix} \|}^{2} = . .$
Commented by maxmathsup by imad last updated on 10/Nov/18

$w e h a v e 2 z_{2} + z_{1} - 5 - i = 2 - 4 i + 3 + i - 5 - i = - 4 i a n d$ $2 z_{1} - z_{2} + 3 - i = 6 + 2 i - 1 + 2 i + 3 - i = 8 + 3 i \Rightarrow$ $∣ \frac{2 z_{2} + z_{1} - 5 - i}{2 z_{1} - z_{2} + 3 - i} ∣ = \frac{4}{∣ 8 + 3 i ∣} = \frac{4}{\sqrt{64 + 9}} = \frac{4}{\sqrt{73}} \Rightarrow∣ \frac{2 z_{2} + z_{1} - 5 - i}{2 z_{1} - z_{2} + 3 - i} ∣^{2} = \frac{8}{73} .$
	Answered by mr W last updated on 11/Nov/18

	$\frac{2 z_{2} + z_{1} - 5 - i}{2 z_{1} - z_{2} + 3 - i} = \frac{2 - 4 i + 3 + i - 5 - i}{6 + 2 i - 1 + 2 i + 3 - i} = \frac{- 4 i}{8 + 3 i} = \frac{- 4 i (8 - 3 i)}{73} = \frac{4 (- 3 - 8 i)}{73}$ $∣ \frac{2 z_{2} + z_{1} - 5 - i}{2 z_{1} - z_{2} + 3 - i} ∣= \frac{4 ∣ - 3 - 8 i ∣}{73} = \frac{4 \sqrt{9 + 64}}{73} = \frac{4}{\sqrt{73}}$ $∣ \frac{2 z_{2} + z_{1} - 5 - i}{2 z_{1} - z_{2} + 3 - i} ∣^{2} = \frac{16}{73}$
	Answered by ajfour last updated on 10/Nov/18

	$=∣ \frac{2 - 4 i + 3 + i - 5 - i}{6 + 2 i - 1 + 2 i + 3 - i} ∣$ $=∣ \frac{- 4 i}{8 + 3 i} ∣ = ∣ \frac{- 12 - 32 i}{64 + 9} ∣$ $= \frac{4 \sqrt{9 + 64}}{64 + 9} = \frac{4}{\sqrt{73}}$ $S o {\| \begin{matrix} \frac{2 z_{2} + z_{1} - 5 - i}{2 z_{1} - z_{2} + 3 - i} \end{matrix} \|}^{2} = \frac{16}{73} .$
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