I've got an excercise:
In complex plane visualise the following equation:
$\left | z - 1 \right | = \left | z + 1 \right | + 3$
I know it has no solution. But I have no clue how to get to this result.
Thank You for Your help,
Dominik
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4$\begingroup$ I'd visualize them by drawing the "height lines" (that is, the lines with constant value of the left and right side). Then you'll see that the lines of equal height don't intersect (intersections would be solutions of the equation). To formally proof that the equation cannot hold, write $|z-1|$ as $|(z+1)+(-2)|$ and use the triangle inequality. $\endgroup$– celtschkAug 2, 2016 at 15:06
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$\begingroup$ @DonAntonio Copied to answer $\endgroup$– z100Aug 2, 2016 at 15:17
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$\begingroup$ @celtschk You should write that as an answer. $\endgroup$– DonAntonioAug 2, 2016 at 15:17
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$\begingroup$ @DonAntonio: Done. $\endgroup$– celtschkAug 2, 2016 at 15:29
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2 Answers
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I'd visualize them by drawing the "height lines" (that is, the lines with constant value of the left and right side). Then you'll see that the lines of equal height don't intersect (intersections would be solutions of the equation). To formally proof that the equation cannot hold, write $|z−1|$ as $|(z+1)+(−2)|$ and use the triangle inequality.
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Take points $z,-1,1$ then distances are $d(z,1)=\left | z-1 \right |$, $d(z,-1)=\left |z+1\right |$, $d(-1,1)=2$, use triangle inequality and add $1$ on the right side.
