Question

# If $$\triangle =\left| \begin{matrix} arg{ z }_{ 1 } & arg{ z }_{ 2 } & arg{ z }_{ 3 } \\ arg{ z }_{ 2 } & arg{ z }_{ 3 } & arg{ z }_{ 1 } \\ arg{ z }_{ 3 } & arg{ z }_{ 1 } & arg{ z }_{ 2 } \end{matrix} \right|$$, the, $$\triangle$$ is divided by:

A
arg(z1+z2+z3)
B
arg(z1.z2.z3)
C
(argz1+argz2+argz3)
D
N.O.T

Solution

## The correct option is B $$arg(z_{1}.z_{2}.z_{3})$$$$\triangle =\left |\begin{matrix} arg{z}_{1}&arg{z}_{2}&arg{z}_{3}\\arg{z}_{2}&arg{z}_{3}&arg{z}_{1} \\ arg{z}_{3}&arg{z}_{1}&arg{z}_{2}\end{matrix}\right |$$$$\Rightarrow \triangle =(arg{z}_{1}+arg{z}_{2}+arg{z}_{3}) \left |\begin{matrix} 1&arg{z}_{2}&arg{z}_{3}\\1&arg{z}_{3}&arg{z}_{1} \\ 1&arg{z}_{1}&arg{z}_{2}\end{matrix}\right |$$Using $$C_1\rightarrow C_1+C_2+C_3$$$$\Rightarrow \triangle =arg({z}_{1}{z}_{2}{z}_{3}) \left |\begin{matrix} 1&arg{z}_{2}&arg{z}_{3}\\1&arg{z}_{3}&arg{z}_{1} \\ 1&arg{z}_{1}&arg{z}_{2}\end{matrix}\right |$$Hence, $$\triangle$$ is divisible by $$arg(z_1z_2z_3)$$ Mathematics

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