If $(z_{1}, z_{2})$ and $(z_{3}, z_{4})$ are two pairs of conjugate complex numbers, then show that
$arg \frac{z_{1}}{z_{3}} + arg \frac{z_{2}}{z_{4}} = 0$ .

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Solution

$a r g \frac{z_{1}}{z_{3}} + a r g \frac{z_{2}}{z_{4}} = a r g \frac{z_{1}}{z_{3}} . \frac{z_{2}}{z_{4}}$
But $1_{1} . z_{2}$ are conjugate ${∴ z}_{1} z_{2} = {| z_{1} |}^{2}$
$L . H . S = a r g \frac{{| z_{1} |}^{2}}{{| z_{3} |}^{2}} = 0$
( $∵ a r g$ of purely real number is zero).

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If (z1,z2) and (z3,z4) are two pairs of conjugate complex numbers, then show that argz1z3+argz2z4=0.

argz1z3+argz2z4=argz1z3.z2z4But 11.z2 are conjugate ∴z1z2=|z1|2 L.H.S=arg|z1|2|z3|2=0(∵ arg of purely real number is zero).

If $(z_{1}, z_{2})$ and $(z_{3}, z_{4})$ are two pairs of conjugate complex numbers, then show that
$arg \frac{z_{1}}{z_{3}} + arg \frac{z_{2}}{z_{4}} = 0$ .

$a r g \frac{z_{1}}{z_{3}} + a r g \frac{z_{2}}{z_{4}} = a r g \frac{z_{1}}{z_{3}} . \frac{z_{2}}{z_{4}}$
But $1_{1} . z_{2}$ are conjugate ${∴ z}_{1} z_{2} = {| z_{1} |}^{2}$
$L . H . S = a r g \frac{{| z_{1} |}^{2}}{{| z_{3} |}^{2}} = 0$
( $∵ a r g$ of purely real number is zero).