If z 1- z 2- z 3 be three non-zero complex number- such that z 2- z 1- a - z 1- b - z 2- and c - z 3- suppose that N beginvmatrixa -b - cb -c -a c -a -b lend vmatrix -0 - then z 3-z2 is equal toA- z3- z 1-z2- z 1- z2-z1-z3-z12C- z2-z1-z3-z1D- z3- z 1-z2- z 12

The correct option is D arg(z_3 z_1/z_2 z_1)^2First deduce that a = b = c, then it will be equal to arg (z_3 z_1/z_2 z_1)^2. ...

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Section Formula Using Complex Numbers

If z 1, z 2, ...

Question

If $z_{1}, z_{2}, z_{3}$ be three non-zero complex number, such that $z_{2} \neq z_{1}, a = | z_{1} |, b = | z_{2} | a n d c = | z_{3} |$ suppose that $∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = 0$ , then arg $(\frac{z_{3}}{z_{2}})$ is equal to

a r g (\frac{z_{2} - z_{1}}{z_{3} - z_{1}})^{2}

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a r g (\frac{z_{2} - z_{1}}{z_{3} - z_{1}})

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a r g (\frac{z_{3} - z_{1}}{z_{2} - z_{1}})^{2}

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a r g (\frac{z_{3} - z_{1}}{z_{2} - z_{1}})

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Solution

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z_{1}, z_{2}, z_{3}

z_{2} \neq z_{1}, a = | z_{1} |, b = | z_{2} | a n d c = | z_{3} |

∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = 0, t h e n a r g (\frac{z_{3}}{z_{2}})

z_{1}, z_{2}, z_{3} \in

| z_{1} | = | z_{2} | = | z_{3} | = 1

z_{1} + z_{2} + z_{3} \neq 0

z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 0

| z_{1} + z_{2} + z_{3} |

z_{1}, z_{2}, z_{3}

z_{2} \neq 1, a = | z_{1} |, b = | z_{2} |

c = | z_{3} |

∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = 0

z_{1}, z_{2}, z_{3}

a, b, c

∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣ = 0

z_{1} \neq z_{2}

arg {(\frac{z_{3} - z_{1}}{z_{2} - z_{1}})}^{2} = arg (\frac{z_{3}}{z_{2}})

\frac{a}{| z_{1} - z_{2} |} = \frac{b}{| z_{2} - z_{3} |} = \frac{c}{| z_{3} - z_{1} |} (a, b, c \in R),

then \frac{a^{2}}{z_{1} - z_{2}} + \frac{b^{2}}{z_{2} - z_{3}} + \frac{c^{2}}{z_{3} - z_{1}}

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