Question

Let ${z_1},{z_2},{z_3} \in C$ and $f\left( {{z_1},{z_2},{z_3}} \right) = \operatorname{sgn} \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  {{z_1}}&{{z_2}}&{{z_3}} \\
  {{{\bar z}_1}}&{{{\bar z}_2}}&{{{\bar z}_3}}
\end{array}} \right|$ where $\operatorname{sgn} x = \dfrac{x}{{\left| x \right|}}$ if, $x \ne 0{\text{ and }}\operatorname{sgn} \left( 0 \right) = 0$, then for $a \in C$
$\left( a \right)f\left( {{z_1},{z_2},{z_3}} \right) = f\left( {{z_1} + a,{z_2} + a,{z_3} + a} \right)$
$\left( b \right)f\left( {{z_1},{z_2},{z_3}} \right) = \left| a \right|f\left( {{z_1},{z_2},{z_3}} \right)$
$\left( c \right)f\left( {{z_1},{z_2},{z_3}} \right) = \left| a \right|f\left( {{a_1},{a_2},{a_3}} \right)$ where ${a_1},{a_2},{a_3}$ is a permutation of ${z_1},{z_2},{z_3}$
$\left( d \right)$ None of these.

Answer 1

Hint: In this particular question use the concept of determinant simplification using some basic determinant rules such as \[{C_2} \to {C_2} - {C_1},{C_3} \to {C_3} - {C_1}\], then check options one by one so use these concepts to reach the solution of the question.

Complete step by step answer:
Given data:
$f\left( {{z_1},{z_2},{z_3}} \right) = \operatorname{sgn} \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  {{z_1}}&{{z_2}}&{{z_3}} \\
  {{{\bar z}_1}}&{{{\bar z}_2}}&{{{\bar z}_3}}
\end{array}} \right|$
Now apply determinant rule i.e. \[{C_2} \to {C_2} - {C_1},{C_3} \to {C_3} - {C_1}\] so we have,
$f\left( {{z_1},{z_2},{z_3}} \right) = \operatorname{sgn} \left| {\begin{array}{*{20}{c}}
  1&0&0 \\
  {{z_1}}&{{z_2} - {z_1}}&{{z_3} - {z_1}} \\
  {{{\bar z}_1}}&{{{\bar z}_2} - {{\bar z}_1}}&{{{\bar z}_3} - {{\bar z}_1}}
\end{array}} \right|$
Now expand the determinant we have,
$f\left( {{z_1},{z_2},{z_3}} \right) = \operatorname{sgn} \left| {\begin{array}{*{20}{c}}
  1&0&0 \\
  {{z_1}}&{{z_2} - {z_1}}&{{z_3} - {z_1}} \\
  {{{\bar z}_1}}&{{{\bar z}_2} - {{\bar z}_1}}&{{{\bar z}_3} - {{\bar z}_1}}
\end{array}} \right| = \operatorname{sgn} \left( {1\left| {\begin{array}{*{20}{c}}
  {{z_2} - {z_1}}&{{z_3} - {z_1}} \\
  {{{\bar z}_2} - {{\bar z}_1}}&{{{\bar z}_3} - {{\bar z}_1}}
\end{array}} \right| - 0 + 0} \right)$
$f\left( {{z_1},{z_2},{z_3}} \right) = \operatorname{sgn} \left( {1\left| {\begin{array}{*{20}{c}}
  {{z_2} - {z_1}}&{{z_3} - {z_1}} \\
  {{{\bar z}_2} - {{\bar z}_1}}&{{{\bar z}_3} - {{\bar z}_1}}
\end{array}} \right|} \right) = \operatorname{sgn} \left[ {\left( {{z_2} - {z_1}} \right)\left( {{{\bar z}_3} - {{\bar z}_1}} \right) - \left( {{z_3} - {z_1}} \right)\left( {{{\bar z}_2} - {{\bar z}_1}} \right)} \right]$........ (1)
Now consider option (a)
$ \Rightarrow f\left( {{z_1},{z_2},{z_3}} \right) = f\left( {{z_1} + a,{z_2} + a,{z_3} + a} \right)$
So from equation (1) we have,
$ \Rightarrow \operatorname{sgn} \left[ {\left( {{z_2} - {z_1}} \right)\left( {{{\bar z}_3} - {{\bar z}_1}} \right) - \left( {{z_3} - {z_1}} \right)\left( {{{\bar z}_2} - {{\bar z}_1}} \right)} \right] = f\left( {{z_1} + a,{z_2} + a,{z_3} + a} \right)$
Now consider the RHS of the above equation we have,
$ \Rightarrow f\left( {{z_1} + a,{z_2} + a,{z_3} + a} \right)$
Now if, $f\left( {{z_1},{z_2},{z_3}} \right) = \operatorname{sgn} \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  {{z_1}}&{{z_2}}&{{z_3}} \\
  {{{\bar z}_1}}&{{{\bar z}_2}}&{{{\bar z}_3}}
\end{array}} \right|$
$ \Rightarrow f\left( {{z_1} + a,{z_2} + a,{z_3} + a} \right) = \operatorname{sgn} \left| {\begin{array}{*{20}{c}}
  1&1&1 \\
  {{z_1} + a}&{{z_2} + a}&{{z_3} + a} \\
  {{{\bar z}_1} + a}&{{{\bar z}_2} + a}&{{{\bar z}_3} + a}
\end{array}} \right|$
Now apply determinant rule i.e. \[{C_2} \to {C_2} - {C_1},{C_3} \to {C_3} - {C_1}\] so we have,
$ \Rightarrow f\left( {{z_1} + a,{z_2} + a,{z_3} + a} \right) = \operatorname{sgn} \left| {\begin{array}{*{20}{c}}
  1&0&0 \\
  {{z_1} + a}&{{z_2} - {z_1}}&{{z_3} - {z_1}} \\
  {{{\bar z}_1} + a}&{{{\bar z}_2} - {{\bar z}_1}}&{{{\bar z}_3} - {{\bar z}_1}}
\end{array}} \right|$
Now expand the determinant we have,
$ \Rightarrow f\left( {{z_1} + a,{z_2} + a,{z_3} + a} \right) = \operatorname{sgn} \left| {\begin{array}{*{20}{c}}
  1&0&0 \\
  {{z_1} + a}&{{z_2} - {z_1}}&{{z_3} - {z_1}} \\
  {{{\bar z}_1} + a}&{{{\bar z}_2} - {{\bar z}_1}}&{{{\bar z}_3} - {{\bar z}_1}}
\end{array}} \right| = \operatorname{sgn} \left( {1\left| {\begin{array}{*{20}{c}}
  {{z_2} - {z_1}}&{{z_3} - {z_1}} \\
  {{{\bar z}_2} - {{\bar z}_1}}&{{{\bar z}_3} - {{\bar z}_1}}
\end{array}} \right| - 0 + 0} \right)$
$ \Rightarrow f\left( {{z_1} + a,{z_2} + a,{z_3} + a} \right) = \operatorname{sgn} \left( {1\left| {\begin{array}{*{20}{c}}
  {{z_2} - {z_1}}&{{z_3} - {z_1}} \\
  {{{\bar z}_2} - {{\bar z}_1}}&{{{\bar z}_3} - {{\bar z}_1}}
\end{array}} \right|} \right) = \operatorname{sgn} \left[ {\left( {{z_2} - {z_1}} \right)\left( {{{\bar z}_3} - {{\bar z}_1}} \right) - \left( {{z_3} - {z_1}} \right)\left( {{{\bar z}_2} - {{\bar z}_1}} \right)} \right]$
= LHS
Hence, $f\left( {{z_1},{z_2},{z_3}} \right) = f\left( {{z_1} + a,{z_2} + a,{z_3} + a} \right)$
So this is the required answer.

So, the correct answer is “Option A”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic determinant rule i.e. how to simplify the determinant, how to expand the determinant which is all stated above, so simply apply them and simplify the given problem then also apply these rules as well as in the options we will get the required answer.