If z1 - x1 - iy1- z2 - x2 - iy2- then 2ix2 y2 x1 y1 equals

The correct option is C z̅_̅1̅z_2 z_1z̅_̅2̅ Given that #160; z_ 1 =x_ 1 +iy_ 1 , z_ 2 =x_ 2 +iy_ 2 Let z=2i x_ 1 amp; y_ 1 x_ 2 ...

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Cramer's Rule

If z1 = x1 ...

Question

If $z_{1} = x_{1} + i y_{1}, z_{2} = x_{2} + i y_{2}$ , then

$2 i ∣ ∣ ∣ \begin{matrix} x_{2} & y_{2} x_{1} & y_{1} \end{matrix} ∣ ∣ ∣$ equals

_{1} z_{2} - z_{1}_{2}

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z_{1}_{2} - z_{2}_{1}

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| z_{1} |^{2} - | z_{2} |^{2}

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| z_{1} |^{2} - | z_{1} - z_{2} |^{2}

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Solution

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Similar questions

| 1 + z_{1}_{2} |^{2} + | z_{1} - z_{2} |^{2} = (1 + | z_{1} |^{2}) (1 + | z_{2} |^{2})

x_{1}, x_{2}, y_{1}, y_{2} ϵ R

x_{1}

x_{2}, y_{1} = y_{2}

z_{1} = x_{1} + i y_{1}, z_{2} = x_{2} + i y_{2}

z_{3} = \frac{1}{2} (z_{1} + z_{2})

z_{1}, z_{2}, z_{3}

z_{1} = 2 - i, z_{2} = - 2 + i

I m (\frac{1}{z_{1}_{2}})

z_{1}, z_{2}

\frac{z_{1}_{2} +_{1} z_{2} + z_{1} z_{2} +_{1}_{2}}{| z_{1} z_{2} |}

{| z_{1} + z_{2} |}_{1}^{2} = {| z_{1} |}_{1}^{2} + {| z_{2} |}_{2}^{2}

| z_{1} |, | z_{2} | \neq 0

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