If - z-2 - -2- z-1 - where z is a complex number- prove - z - 2 - 4 3 Rez

|x + iy 2| = 2|x+iy 1| #160; |x 2+iy|=2|x 1+iy| (x 2)^2+y^2=4((x 1)^2+y^2 x^2+4 4x+y^2 = 4x^2+4 8x+4y^2 #160; 4x=3x^2+3y^2 #16 ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Complex Numbers

If | z-2 | ...

Question

If $| z - 2 | = 2 | z - 1 |$ , where $z$ is a complex number, prove ${| z |}^{2} = \frac{4}{3}$ $R e (z)$

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Solution

$| x + i y - 2 | = 2 | x + i y - 1 |$
$| x - 2 + i y | = 2 | x - 1 + i y |$
$(x - 2)^{2} + y^{2} = 4 ((x - 1)^{2} + y^{2}$
$x^{2} + 4 - 4 x + y^{2} = 4 x^{2} + 4 - 8 x + 4 y^{2}$
$4 x = 3 x^{2} + 3 y^{2}$
$x^{2} + y^{2} = \frac{4}{3} x$
${| z |}^{2} = \frac{4}{3}$ $R e (z)$

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If |z−2|=2|z−1|, where z is a complex number, prove |z|2=43 Re(z)

|x+iy−2|=2|x+iy−1| |x−2+iy|=2|x−1+iy|(x−2)2+y2=4((x−1)2+y2x2+4−4x+y2=4x2+4−8x+4y2 4x=3x2+3y2 x2+y2=43x|z|2=43 Re(z)

If $| z - 2 | = 2 | z - 1 |$ , where $z$ is a complex number, prove ${| z |}^{2} = \frac{4}{3}$ $R e (z)$

$| x + i y - 2 | = 2 | x + i y - 1 |$
$| x - 2 + i y | = 2 | x - 1 + i y |$
$(x - 2)^{2} + y^{2} = 4 ((x - 1)^{2} + y^{2}$
$x^{2} + 4 - 4 x + y^{2} = 4 x^{2} + 4 - 8 x + 4 y^{2}$
$4 x = 3 x^{2} + 3 y^{2}$
$x^{2} + y^{2} = \frac{4}{3} x$
${| z |}^{2} = \frac{4}{3}$ $R e (z)$