1. Interpret the first condition

We are given $\left|\frac{z-a}{z+b}\right|=1.$ This implies $|z-a|=|z+b|.$

Let $z=x+iy$. Then $|z-a|^2=(x-a)^2+y^2,$ $|z+b|^2=(x+b)^2+y^2.$ So, $(x-a)^2=(x+b)^2.$ Expanding, $x^2-2ax+a^2=x^2+2bx+b^2$ $-2(a+b)x=(b^2-a^2)=(b-a)(a+b).$ Thus, $(a+b)\big(-2x-(b-a)\big)=0.$ Since $a+b\ne 0$, we get $-2x-(b-a)=0 \implies x=\frac{a-b}{2}.$

Hence the first condition means that $z$ lies on the vertical line $\Re(z)=\frac{a-b}{2}.$

2. Evaluate the determinant in the second condition

We need $\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1,$ where $\omega^3=1$ and $1+\omega+\omega^2=0.$

Let $D=\begin{vmatrix}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{vmatrix}.$

Observe that this matrix can be written as $D=\det(zI+M),$ where $M=\begin{pmatrix}1&\omega&\omega^2\\ \omega&\omega^2&1\\ \omega^2&1&\omega\end{pmatrix}.$

Now each row of $M$ is a cyclic shift of $(1,\omega,\omega^2)$, and also $1+\omega+\omega^2=0.$ So one eigenvalue of $M$ is $0$ (since row sum is $0$).

Let us find the others using circulant matrix theory. For first row $(1,\omega,\omega^2)$, the eigenvalues are $\lambda_k=1+\omega\,\xi_k+\omega^2\,\xi_k^2,$ where $\xi_k\in\{1,\omega,\omega^2\}$.

• For $\xi=1$: $\lambda_1=1+\omega+\omega^2=0.$
• For $\xi=\omega$: $\lambda_2=1+\omega^2+\omega^4=1+\omega^2+\omega=0.$
• For $\xi=\omega^2$: $\lambda_3=1+\omega^3+\omega^6=1+1+1=3.$

So eigenvalues of $M$ are $0,0,3$. Therefore eigenvalues of $zI+M$ are $z,z,z+3.$ Hence $D=z^2(z+3).$

The condition is $|D|=1 \implies |z^2(z+3)|=1,$ i.e. $|z|^2|z+3|=1.$

3. Find when such a complex number $z$ exists with prescribed real part

From step 1, $z=x+iy$ must satisfy $x=\frac{a-b}{2}.$ We need to know for which real numbers $x$ there exists some $y\in\mathbb R$ such that $|z|^2|z+3|=1.$

Write $z=x+iy.$ Then $|z|^2=x^2+y^2,$ $|z+3|=\sqrt{(x+3)^2+y^2}.$ So define $f(y)=(x^2+y^2)\sqrt{(x+3)^2+y^2}.$ We need some $y$ such that $f(y)=1$.

Since $f(y)$ is continuous in $y$ and $f(y)\to\infty$ as $|y|\to\infty$, such a $y$ exists iff the minimum value of $f(y)$ is $\le 1$.

Now for fixed $x$, both factors increase with $y^2$, so minimum occurs at $y=0$. Thus $f_{\min}=x^2|x+3|.$ Therefore a solution exists iff $x^2|x+3|\le 1.$

4. Possible values of $x=\dfrac{a-b}{2}$

Since $a,b\in[-3,3]\cap\mathbb Z$, the difference $a-b$ ranges from $-6$ to $6$, so $x\in\left\{-3,-\frac52,-2,-\frac32,-1,-\frac12,0,\frac12,1,\frac32,2,\frac52,3\right\}.$

Now test $x^2|x+3|\le 1.$

• $x=-3$: $9\cdot 0=0\le 1$ ✓
• $x=-\frac52$: $\frac{25}{4}\cdot \frac12=\frac{25}{8}>1$ ✗
• $x=-2$: $4\cdot 1=4>1$ ✗
• $x=-\frac32$: $\frac94\cdot \frac32=\frac{27}{8}>1$ ✗
• $x=-1$: $1\cdot 2=2>1$ ✗
• $x=-\frac12$: $\frac14\cdot \frac52=\frac58\le 1$ ✓
• $x=0$: $0\cdot 3=0\le 1$ ✓
• $x=\frac12$: $\frac14\cdot \frac72=\frac78\le 1$ ✓
• $x=1$: $1\cdot 4=4>1$ ✗
• Larger $x$ clearly fail.

So the allowed real parts are $x\in\left\{-3,-\frac12,0,\frac12\right\}.$

Thus $\frac{a-b}{2}\in\left\{-3,-\frac12,0,\frac12\right\},$ so $a-b\in\{-6,-1,0,1\}.$ But $a+b\ne 0$.

5. Count ordered pairs $(a,b)$

We count integer pairs $a,b\in[-3,3]$ satisfying each difference.

Case 1: $a-b=-6$

Then $a=b-6$. Within $[-3,3]$, only $(a,b)=(-3,3).$ But here $a+b=0$, forbidden. So count $=0$.

Case 2: $a-b=-1$

Then $a=b-1$. Possible pairs in range: $(-3,-2),(-2,-1),(-1,0),(0,1),(1,2),(2,3).$ Exclude those with $a+b=0$: $(-1,0)\to -1 \neq 0,$ $(0,1)\to 1 \neq 0,$ Actually none of these have sum $0$ except if $a=-b$, which does not occur here. So count $=6$.

Case 3: $a-b=0$

Then $a=b$. Possible pairs: $(-3,-3),(-2,-2),(-1,-1),(0,0),(1,1),(2,2),(3,3).$ Condition $a+b\ne 0$ excludes only $(0,0)$. So count $=6$.

Case 4: $a-b=1$

Then $a=b+1$. Possible pairs: $(-2,-3),(-1,-2),(0,-1),(1,0),(2,1),(3,2).$ Again none have sum $0$. So count $=6$.

Total count would be $0+6+6+6=18.$

6. But check the determinant interpretation carefully

Let us directly compute the determinant to avoid any issue with circulant ordering.

Expand: \begin{align*} D&=(z+1)\big((z+\omega^2)(z+\omega)-1\big)-\omega\big(\omega(z+\omega)-\omega^2\big)+\omega^2\big(\omega\cdot 1-(z+\omega^2)\omega^2\big). \end{align*}

Now, $(z+\omega^2)(z+\omega)=z^2+z(\omega+\omega^2)+\omega^3=z^2-z+1.$ So $(z+\omega^2)(z+\omega)-1=z^2-z.$ Hence first term is $(z+1)(z^2-z)=z^3-z.$

Second term: $-\omega\big(\omega z+\omega^2-\omega^2\big)=-\omega(\omega z)=-\omega^2 z.$

Third term: $\omega^2\big(\omega-(z+\omega^2)\omega^2\big)=\omega^2\big(\omega-z\omega^2-\omega^4\big).$ Since $\omega^4=\omega$, $\omega-z\omega^2-\omega= -z\omega^2,$ so third term is $\omega^2(-z\omega^2)=-z\omega^4=-z\omega.$

Thus $D=z^3-z-\omega^2 z-\omega z=z^3-z(1+\omega+\omega^2)=z^3.$ Since $1+\omega+\omega^2=0$, $D=z^3.$

Therefore the second condition is actually $|z^3|=1 \implies |z|=1.$

7. Combine both conditions correctly

Now we need $z$ on the line $\Re(z)=\frac{a-b}{2}$ and also on the unit circle $|z|=1.$

Such a point exists iff $\left|\frac{a-b}{2}\right|\le 1,$ i.e. $|a-b|\le 2.$ Also given $a+b\ne 0$.

So count ordered integer pairs $(a,b)$ with $a,b\in[-3,3]$, $|a-b|\le 2$, and $a+b\ne 0$.

8. Count all pairs with $|a-b|\le 2$

For each $a\in\{-3,-2,-1,0,1,2,3\}$, count valid $b$ in $[-3,3]$ with $|a-b|\le 2$:

• $a=-3$: $b\in\{-3,-2,-1\}$ → $3$
• $a=-2$: $b\in\{-3,-2,-1,0\}$ → $4$
• $a=-1$: $b\in\{-3,-2,-1,0,1\}$ → $5$
• $a=0$: $b\in\{-2,-1,0,1,2\}$ → $5$
• $a=1$: $b\in\{-1,0,1,2,3\}$ → $5$
• $a=2$: $b\in\{0,1,2,3\}$ → $4$
• $a=3$: $b\in\{1,2,3\}$ → $3$

Total $3+4+5+5+5+4+3=29.$

Now exclude pairs with $a+b=0$, i.e. $(a,-a)$. Among these, also need $|a-(-a)|=|2a|\le 2$, so $|a|\le 1$. Thus excluded pairs are $(-1,1),(0,0),(1,-1),$ which are $3$ pairs.

Therefore required count is $29-3=26.$

9. Final answer

The number of ordered pairs is $\boxed{26}.$

This does not match the stored answer $10$. The likely issue is that the determinant simplifies to $z^3$, not to any more complicated expression; after that the counting is straightforward and gives $26$.