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સમીકરણ $\left| {\frac{{z - 12}}{{z - 8i}}} \right| = \frac{5}{3},\left| {\frac{{z - 4}}{{z - 8}}} \right| = 1$ નું સમાધાન કરે તેવી સંકર સંખ્યા $Z$ મેળવો.
$6$
$6 \pm 8i$
$6 + 8i,\,6 + 17i$
એકપણ નહીં.
Solution
(c)We have $\left| {\frac{{z – 12}}{{z – 8i}}} \right| = \frac{5}{3}$and $\left| {\frac{{z – 4}}{{z – 8}}} \right| = 1$
Let $z = x + iy$, then
$\left| {\frac{{z – 12}}{{z – 8i}}} \right| = \frac{5}{3} \Rightarrow 3|z – 12| = 5|z – 8i|$
==> $3|(x – 12) + iy| = 5|x + (y – 8)i|$
==> $9{(x – 12)^2} + 9{y^2} = 25{x^2} + 25{(y – 8)^2}$ ….$(i)$
and $\left| {\frac{{z – 4}}{{z – 8}}} \right| = 1 \Rightarrow |z – 4| = |z – 8|$
==> $|x – 4 + iy| = |x – 8 + iy|$
==> ${(x – 4)^2} + {y^2} = {(x – 8)^2} + {y^2} \Rightarrow x = 6$
Putting $x = 6$in $(i)$, we get ${y^2} – 25y + 136 = 0$
$y = 17,8$
Hence $z = 6 + 17i$or $z = 6 + 8i$
Trick : Check it with options.
