Let integers a, b ∈ [-3,3] be such that a+b ≠ 0. Then the number of all possible ordered pairs (a, b

Q: Why does the determinant simplify so easily? Is there a trick?

It is not a trick, but a beautiful manifestation of the symmetry of the cube roots of unity! When you add the rows, you get $z + (1 + \omega + \omega^2)$ in every entry of the first row. Since $1 + \omega + \omega^2 = 0$, the row becomes $[z, z, z]$. It is a perfect collapse that turns a terrifying matrix into a simple $z^3$ expression. Always look for row or column sums in determinant problems involving $\omega$!

Q: Why do we square the modulus equation $|z-a| = |z+b|$?

Great question! The modulus represents a distance in the complex plane. Working with square roots is messy and prone to error. By squaring both sides, we use the identity $|w|^2 = w\bar{w}$. This allows us to expand the terms algebraically and isolate the real part of $z$, which is much easier to handle than dealing with complex magnitudes directly.

Q: Why do we reject the pair $(1, -1)$ in the first case?

The problem explicitly states the constraint $a+b \neq 0$. In the case $z=1$, we derived the condition $a-b=2$. If we test the pair $(1, -1)$, we see that $a+b = 1 + (-1) = 0$. This violates the problem's condition, so we must exclude it. Always check your final set of solutions against the initial constraints provided in the question!

Question

Let integers $a, b\in[-3,3]$ be such that $a+b
e0$. Then the number of all possible ordered pairs $(a, b)$, for which $|\frac{z-a}{z+b}|=1$ and $|\begin{matrix}z+1&\omega&\omega^{2}\ \omega&z+\omega^{2}&1\ \omega^{2}&1&z+\omega\end{matrix}|=1, z\in \mathbb{C}$, where $\omega$ and $\omega^{2}$ are the roots of $x^{2}+x+1=0$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are presented with a complex determinant $D$ and a geometric condition involving integers. The determinant is defined as:

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

At first glance, the expression appears algebraically dense. However, the cyclic nature of the elements suggests a simplification through row operations.

### The Determinant's Secret

We perform the row operation $R_1 	o R_1 + R_2 + R_3$. The first row becomes $[z+1+\omega+\omega^2, z+\omega^2+1+\omega, z+\omega+1+\omega^2]$.

Recalling the identity $1 + \omega + \omega^2 = 0$, the first row simplifies to $[z, z, z]$. The determinant now takes the form:

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

Factoring out $z$ from the first row and performing subsequent column operations, the determinant collapses into $D = z^3$. Given $|D|=1$, we conclude $|z^3|=1$, which implies $|z|=1$. Thus, $z$ is constrained to the unit circle.

### The Geometry of Distance

We now address the condition $|z-a| = |z+b|$. Geometrically, this represents the locus of points equidistant from $a$ and $-b$.

Squaring both sides, we obtain $(z-a)(\bar{z}-a) = (z+b)(\bar{z}+b)$. Expanding this yields:

$$|z|^2 - a(z+\bar{z}) + a^2 = |z|^2 + b(z+\bar{z}) + b^2$$

The $|z|^2$ terms cancel out, leaving $a^2 - b^2 = (a+b)(z+\bar{z})$. Assuming $a+b 
eq 0$, we divide by $(a+b)$ to find:

$$a-b = z+\bar{z} = 2Re(z)$$

### The Systematic Hunt

Since $|z|=1$ and the determinant structure implies $z^3=1$, the possible values for $z$ are $1, \omega, \omega^2$.

If $z=1$, then $Re(z)=1$, leading to $a-b=2$. Testing integer pairs $(a, b)$ from the set $\{-3, -2, -1, 0, 1, 2, 3\}$, we identify the pairs: $(3, 1), (2, 0), (1, -1), (0, -2), (-1, -3)$.

We must exclude $(1, -1)$ because $a+b=0$ is forbidden. This leaves **4 valid pairs** for $z=1$.

If $z=\omega$ or $z=\omega^2$, then $Re(z)=-1/2$, leading to $a-b=2(-1/2)=-1$. Testing the pairs again, we find: $(2, 3), (1, 2), (0, 1), (-1, 0), (-2, -1), (-3, -2)$.

None of these pairs result in $a+b=0$, so all **6 pairs** are valid. Summing these possibilities, we arrive at a total of $4 + 6 = 10$.

The final number of valid integer pairs is **10**.

Enter Numerical Value:

Visualized Solution

Problem Setup & Determinant

Simplifying the Determinant

Applying Cube Root Properties

Evaluating the Determinant

Possible Values of $z$

Geometric Interpretation

Simplifying the Modulus Equation

Case 1: $z = 1$

Case 2: $z = ω, ω^{2}$

Final Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Determinant's Secret

The Geometry of Distance

The Systematic Hunt

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