Let integers a,b [-3,3] be such that a+b 0.

Question

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

Answer: 10

Accepted Answer

$\begin{gathered} a,b\in I,-3\le a,b\le 3,a+b
e 0 \ |z-a|=|z+b| \ |\begin{array}{ccc}z+1 \quad  \omega  \quad  {\omega }^{2} \ \omega  \quad  z+{\omega }^{2} \quad  1 \ {\omega }^{2} \quad  1 \quad  z+\omega \end{array}|=1 \ \Rightarrow |\begin{array}{ccc}z \quad  z \quad  z \ \omega  \quad  z+{\omega }^{2} \quad  1 \ {\omega }^{2} \quad  1 \quad  z+\omega \end{array}|=1 \end{gathered}$

$\begin{gathered} \Rightarrow z|\begin{array}{ccc}1 \quad  1 \quad  1 \ \omega  \quad  z+{\omega }^{2} \quad  1 \ {\omega }^{2} \quad  1 \quad  z+\omega \end{array}|=1 \ \Rightarrow z|\begin{array}{ccc}1 \quad  0 \quad  0 \ \omega  \quad  z+{\omega }^{2}-\omega  \quad  1-\omega \ {\omega }^{2} \quad  1-{\omega }^{2} \quad  z+\omega -{\omega }^{2}\end{array}|=1 \end{gathered}$

$\begin{gathered} \Rightarrow {z}^{3}=1 \ \Rightarrow z=\omega ,{\omega }^{2},1 \ \ |1-a|=|1+b| \ \Rightarrow 10 \end{gathered}$

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