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  4. If |z - 6| < |z - 2|, its solution is given by

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Q. If $|z - 6| < |z - 2|$, its solution is given by

Complex Numbers and Quadratic Equations

Solution:

Let $z = a + ib$
Now, $z - 6 = (a - 6) + ib$
and $z - 2 = (a - 2) + ib$
$\therefore | z- 6 | < | z -2|$
$\Rightarrow (a - 6)^2 + b^2 < (a - 2)^2 + b^2$
$\Rightarrow (a -6)^2 < (a - 2)^2$
$\Rightarrow - 8a < - 36 + 4$
$\Rightarrow - 8a < - 32$
$ \Rightarrow a > 4 $
$\therefore Re(z) > 4$.