Find the complex number z satisfying the equa | vedclass

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(C) Given equations are $\left| \frac{z - 12}{z - 8i} \right| = \frac{5}{3}$ and $\left| \frac{z - 4}{z - 8} \right| = 1$. Let $z = x + iy$. From the

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$6 + 8i, 6 + 17i$

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(C) Given equations are $\left| \frac{z - 12}{z - 8i} \right| = \frac{5}{3}$ and $\left| \frac{z - 4}{z - 8} \right| = 1$. Let $z = x + iy$. From the second equation,$|z - 4| = |z - 8|$,which implies $|(x - 4) + iy| = |(x - 8) + iy|$. Squaring both sides,$(x - 4)^2 + y^2 = (x - 8)^2 + y^2$. $x^2 - 8x + 16 = x^2 - 16x + 64$,which gives $8x = 48$,so $x = 6$. Now substitute $x = 6$ into the first equation: $3|z - 12| = 5|z - 8i|$. $3|(6 - 12) + iy| = 5|6 + (y - 8)i|$. $3|-6 + iy| = 5|6 + (y - 8)i|$. Squaring both sides,$9(36 + y^2) = 25(36 + (y - 8)^2)$. $324 + 9y^2 = 25(36 + y^2 - 16y + 64)$. $324 + 9y^2 = 25(y^2 - 16y + 100)$. $324 + 9y^2 = 25y^2 - 400y + 2500$. $16y^2 - 400y + 2176 = 0$. Dividing by $16$,$y^2 - 25y + 136 = 0$. $(y - 8)(y - 17) = 0$. Thus,$y = 8$ or $y = 17$. Therefore,$z = 6 + 8i$ or $z = 6 + 17i$.

Find the complex number $z$ satisfying the equations $\left| \frac{z - 12}{z - 8i} \right| = \frac{5}{3}$ and $\left| \frac{z - 4}{z - 8} \right| = 1$.

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