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(B) Let $w = a + ib$. The condition states that $|f(z) - z| = |f(z) - 0|$.Substituting $f(z) = wz$,we get $|wz - z| = |wz|$.$|z||w - 1| = |z||w|$.Assu

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$403$

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(B) Let $w = a + ib$. The condition states that $|f(z) - z| = |f(z) - 0|$.Substituting $f(z) = wz$,we get $|wz - z| = |wz|$.$|z||w - 1| = |z||w|$.Assuming $z 
eq 0$,we have $|w - 1| = |w|$.$|(a - 1) + ib|^2 = |a + ib|^2$.$(a - 1)^2 + b^2 = a^2 + b^2$.$a^2 - 2a + 1 + b^2 = a^2 + b^2$.$-2a + 1 = 0 \Rightarrow a = \frac{1}{2}$.Given $|a + bi| = 10$,we have $a^2 + b^2 = 10^2 = 100$.Substituting $a = \frac{1}{2}$,we get $(\frac{1}{2})^2 + b^2 = 100$.$\frac{1}{4} + b^2 = 100 \Rightarrow b^2 = 100 - \frac{1}{4} = \frac{399}{4}$.Since $b^2 = \frac{p}{q}$,we have $p = 399$ and $q = 4$. $	ext{gcd}(399, 4) = 1$.Therefore,$p + q = 399 + 4 = 403$.

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