Let integers $$a,b \in [-3,3]$$ be such that $$a+b \neq 0$$. Then the number of all possible ordered pairs (a, b), for which $$\left|\frac{z-a}{z+b}\right| = 1$$ and $$\begin{vmatrix} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{vmatrix} = 1,\quad z \in \mathbb{C}$$, where $$\omega$$ and $$\omega^{2}$$ are the roots of $$x^{2} + x + 1 = 0$$, is equal to_________.

The integers $$a,b$$ are chosen from $$[-3,3]$$ with the restriction $$a+b \neq 0$$.
For such a pair we have to find a complex number $$z$$ that satisfies both

$$\left|\frac{z-a}{z+b}\right|=1 \qquad -(1)$$
and

$$\begin{vmatrix} z+1 & \omega & \omega^{2}\\ \omega & z+\omega^{2} & 1\\ \omega^{2} & 1 & z+\omega \end{vmatrix}=1, \qquad -(2)$$
where $$\omega,\omega^{2}$$ are the non-real cube roots of unity, i.e. $$\omega^{2}+\omega+1=0$$ and $$\omega^{3}=1$$.

Step 1: Evaluate the determinant in (2).

Let $$D(z)$$ denote the determinant. Expanding along the first row,

$$D(z)=(z+1)\!\begin{vmatrix}z+\omega^{2}&1\\1&z+\omega\end{vmatrix}-\omega\!\begin{vmatrix}\omega&1\\\omega^{2}&z+\omega\end{vmatrix}+\omega^{2}\!\begin{vmatrix}\omega&z+\omega^{2}\\\omega^{2}&1\end{vmatrix}.$$

Compute the three $$2\times 2$$ minors:

$$\begin{aligned} A&=(z+\omega^{2})(z+\omega)-1\\ &=z^{2}+z(\omega+\omega^{2})+\omega\omega^{2}-1\\ &=z^{2}-z+1-1 \quad(\text{because } \omega+\omega^{2}=-1,\;\omega\omega^{2}=1)\\ &=z^{2}-z,\\[6pt] B&=\omega(z+\omega)-\omega^{2}=\omega z+\omega^{2}-\omega^{2}=\omega z,\\[6pt] C&=\omega-\omega^{2}(z+\omega^{2})=\omega-\omega^{2}z-\omega^{4} =\omega-\omega^{2}z-\omega=\!-\,\omega^{2}z. \end{aligned}$$

Hence

$$\begin{aligned} D(z)&=(z+1)(z^{2}-z)-\omega(\omega z)+\omega^{2}(-\omega^{2}z)\\ &=z^{3}-z-\omega^{2}z-\omega z\\ &=z^{3}-z(1+\omega+\omega^{2})\\ &=z^{3}\quad\bigl(1+\omega+\omega^{2}=0\bigr). \end{aligned}$$

Equation (2) therefore gives

$$z^{3}=1 \quad\Longrightarrow\quad z\in\{1,\;\omega,\;\omega^{2}\}.$$(3)

Step 2: Interpret the modulus condition (1).

Because $$a,b$$ are real, points $$a$$ and $$-b$$ lie on the real axis. Writing $$z=x+iy,$$

$$\left|\frac{z-a}{z+b}\right|=1 \;\Longrightarrow\;|z-a|=|z+b|,$$

which is the set of points equidistant from $$a$$ and $$-b$$. The locus is the perpendicular bisector of the segment joining $$a$$ and $$-b$$, i.e. the vertical line

$$x=\frac{a-b}{2}. \qquad -(4)$$

Step 3: Make the locus pass through at least one root of (3).

The three cube roots of unity and their real parts are

$$1 \;(x=1),\qquad \omega=-\tfrac12+\tfrac{\sqrt3}2 i \;(x=-\tfrac12),\qquad \omega^{2}=-\tfrac12-\tfrac{\sqrt3}2 i \;(x=-\tfrac12).$$

Thus a root from (3) lies on the line (4) iff

$$\frac{a-b}{2}=1 \quad\text{or}\quad \frac{a-b}{2}=-\frac12.$$

Hence we need

$$a-b=2 \qquad\text{or}\qquad a-b=-1.\quad -(5)$$

Step 4: List all integer pairs in $$[-3,3]$$ satisfying (5) and $$a+b\neq0$$.

Case 1: $$a-b=2.$$ Put $$b=a-2$$ and keep $$a,b\in[-3,3].$$

Possible $$a$$ values: $$-1,0,1,2,3.$$ Corresponding pairs:

$$(a,b)=(-1,-3),(0,-2),(1,-1),(2,0),(3,1).$$

Reject the pair with $$a+b=0$$, i.e. $$(1,-1).$$ Valid pairs: 4.

Case 2: $$a-b=-1.$$ Put $$b=a+1$$ and keep $$a,b\in[-3,3].$$

Possible $$a$$ values: $$-3,-2,-1,0,1,2.$$ Pairs:

$$(a,b)=(-3,-2),(-2,-1),(-1,0),(0,1),(1,2),(2,3).$$

All have $$a+b\neq0$$, so all 6 are valid.

Step 5: Total number of ordered pairs.

Total valid pairs $$=4+6=10.$$

Answer : 10