If z1-z2-z3 are three complex numbers- prove that z 1 Imz- - -2- z 3 - z 2 Imz- - -3- z 1 - z 3 Imz- - -1- z 2 -0 where Imw- imaginary part of w-w being a complex number-

z _ 1 = x _ 1 + iy _ 1 , z _ 2 z _ 3 =( x _ 2 + iy _ 2 ) ( x _ 3 + iy _ 3 ) ∴ Im( z̅ ̅_̅ ̅2̅ z _ 3 ) =( x _ 2 y _ 3 x _ 3 y _ 2 ) ∴ z _ 1 ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

If z1,z2,z3...

Question

If $z_{1}, z_{2}, z_{3}$ are three complex numbers, prove that
$z_{1} I m (_{2} z_{3}) + z_{2} I m (_{3} z_{1}) + z_{3} I m (_{1} z_{2}) = 0$ where $I m (w) =$ imaginary part of $w, w$ being a complex number.

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Solution

$z_{1} = x_{1} + {i y}_{1}, z_{2} z_{3} = (x_{2} + {i y}_{2}) (x_{3} + {i y}_{3})$
$∴ I m (_{2} z_{3}) = (x_{2} y_{3} - x_{3} y_{2})$
$∴ z_{1} I m (_{2} z_{3}) = (x_{1} + {i y}_{1}) (x_{2} y_{3} - x_{3} y_{2})$
$= x_{1} (x_{2} y_{3} - x_{3} y_{2}) + {i y}_{1} (x_{2} y_{3} - x_{3} y_{2})$
$∴ \sum z_{1} I m (_{2} z_{3}) = 0$
Each sigma will have six terms, three $+ i v e$ and three $- i v e$ which will cancel.

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If z1,z2,z3 are three complex numbers, prove that z1Im(¯z2z3)+z2Im(¯z3z1)+z3Im(¯z1z2)=0 where Im(w)= imaginary part of w,w being a complex number.

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