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(B) Given $a, b \in \{-3, -2, -1, 0, 1, 2, 3\}$ and $a+b 
eq 0$.The condition $|\frac{z-a}{z+b}|=1$ implies $|z-a|=|z+b|$,which means $z$ lies on the

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$11$

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(B) Given $a, b \in \{-3, -2, -1, 0, 1, 2, 3\}$ and $a+b 
eq 0$.The condition $|\frac{z-a}{z+b}|=1$ implies $|z-a|=|z+b|$,which means $z$ lies on the perpendicular bisector of the segment joining $a$ and $-b$ on the complex plane. This is the line $	ext{Re}(z) = \frac{a-b}{2}$.Now,consider the determinant $D = \left|\begin{array}{ccc}z+1 & \omega & \omega^2 \ \omega & z+\omega^2 & 1 \ \omega^2 & 1 & z+\omega\end{array}ight|$.Adding $C_2$ and $C_3$ to $C_1$,we get $C_1 	o C_1+C_2+C_3$:$D = \left|\begin{array}{ccc}z+1+\omega+\omega^2 & \omega & \omega^2 \ z+\omega+\omega^2 & z+\omega^2 & 1 \ z+\omega^2+\omega & 1 & z+\omega\end{array}ight| = \left|\begin{array}{ccc}z & \omega & \omega^2 \ z & z+\omega^2 & 1 \ z & 1 & z+\omega\end{array}ight| = z \left|\begin{array}{ccc}1 & \omega & \omega^2 \ 1 & z+\omega^2 & 1 \ 1 & 1 & z+\omega\end{array}ight|$.Performing $R_2 	o R_2-R_1$ and $R_3 	o R_3-R_1$:$D = z \left|\begin{array}{ccc}1 & \omega & \omega^2 \ 0 & z+\omega^2-\omega & 1-\omega^2 \ 0 & 1-\omega & z+\omega-\omega^2\end{array}ight| = z((z+\omega^2-\omega)(z+\omega-\omega^2) - (1-\omega^2)(1-\omega))$.Simplifying the expression,we get $D = z(z^2 - (\omega^2-\omega)^2 - (1-\omega-\omega^2+\omega^3)) = z(z^2 - (\omega^4+\omega^2-2\omega^3) - (1-\omega-\omega^2+1)) = z(z^2 - (\omega+\omega^2-2) - (2-\omega-\omega^2)) = z(z^2 - (-1-2) - (2-(-1))) = z(z^2+3-3) = z^3$.Given $z^3=1$,so $z \in \{1, \omega, \omega^2\}$.If $z=1$,$|1-a|=|1+b| \implies (1-a)^2 = (1+b)^2 \implies 1-2a+a^2 = 1+2b+b^2 \implies a^2-b^2 = 2(a+b) \implies (a-b)(a+b) = 2(a+b)$.Since $a+b 
eq 0$,$a-b=2$,i.e.,$a=b+2$.Possible pairs $(a, b)$ with $a, b \in \{-3, \dots, 3\}$ and $a+b 
eq 0$:$(-1, -3), (0, -2), (1, -1), (2, 0), (3, 1)$. ($5$ pairs).If $z=\omega$,$|\omega-a|=|\omega+b| \implies (\omega-a)(\bar{\omega}-a) = (\omega+b)(\bar{\omega}+b) \implies |\omega|^2 - a(\omega+\bar{\omega}) + a^2 = |\omega|^2 + b(\omega+\bar{\omega}) + b^2$.$1 - a(-1) + a^2 = 1 + b(-1) + b^2 \implies a^2+a = b^2-b \implies a^2-b^2+a+b=0 \implies (a+b)(a-b+1)=0$.Since $a+b 
eq 0$,$a-b = -1$,i.e.,$b=a+1$.Pairs: $(-3, -2), (-2, -1), (-1, 0), (0, 1), (1, 2), (2, 3)$. ($6$ pairs).If $z=\omega^2$,$|\omega^2-a|=|\omega^2+b| \implies a^2+a = b^2-b$ (same as $z=\omega$),so $b=a+1$. ($6$ pairs).Total unique pairs: $5 + 6 = 11$.

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