Question

If $△=\left|\begin{array}{ccc}arg{z}_1& arg{z}_2& arg{z}_3\\ arg{z}_2& arg{z}_3& arg{z}_1\\ arg{z}_3& arg{z}_1& arg{z}_2\end{array}\right|$, the, $△$ is divided by:

• A. $arg(z_1+z_2+z_3)$
• B. $arg(z_1.z_2.z_3)$
• C. $(arg{z}_1+arg{z}_2+arg{z}_3)$
• D. N.O.T

Answer 1

The correct option is B $arg(z_1.z_2.z_3)$

$△=\left|\begin{array}{ccc}arg{z}_1& arg{z}_2& arg{z}_3\\ arg{z}_2& arg{z}_3& arg{z}_1\\ arg{z}_3& arg{z}_1& arg{z}_2\end{array}\right|$

$\Rightarrow △=(arg{z}_1+arg{z}_2+arg{z}_3)\left|\begin{array}{ccc}1& arg{z}_2& arg{z}_3\\ 1& arg{z}_3& arg{z}_1\\ 1& arg{z}_1& arg{z}_2\end{array}\right|$

Using $C_1\to C_1+C_2+C_3$

$\Rightarrow △=arg\left(z_1{z}_2{z}_3\right)\left|\begin{array}{ccc}1& arg{z}_2& arg{z}_3\\ 1& arg{z}_3& arg{z}_1\\ 1& arg{z}_1& arg{z}_2\end{array}\right|$

Hence, $△$ is divisible by $arg\left(z_1{z}_2{z}_3\right)$