If left|frac{z-25}{z-1}right|=5,then |z|= | vedclass

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(A) Given $\left|\frac{z-25}{z-1}\right|=5$. Squaring both sides,we get $\left|\frac{z-25}{z-1}\right|^2 = 25$. This implies $\frac{(z-25)(\bar{z}-25)

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$5$

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(A) Given $\left|\frac{z-25}{z-1}\right|=5$. Squaring both sides,we get $\left|\frac{z-25}{z-1}\right|^2 = 25$. This implies $\frac{(z-25)(\bar{z}-25)}{(z-1)(\bar{z}-1)} = 25$. Expanding the terms: $(z-25)(\bar{z}-25) = 25(z-1)(\bar{z}-1)$. $z\bar{z} - 25z - 25\bar{z} + 625 = 25(z\bar{z} - z - \bar{z} + 1)$. $|z|^2 - 25(z+\bar{z}) + 625 = 25|z|^2 - 25(z+\bar{z}) + 25$. $|z|^2 + 625 = 25|z|^2 + 25$. $24|z|^2 = 600$. $|z|^2 = 25$. Therefore,$|z| = 5$.

If $\left|\frac{z-25}{z-1}\right|=5$,then $|z|=$

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