If z1 = 2 + 3i, z2 = 3 - 2i, z3 = - 1 - 2 i √3 then which of the following hold good?

Question

If z1 = 2 + 3i, z2 = 3 - 2i, z3 = - 1 - 2 i √3 then which of the following hold good? (A) arg((z3/z2)) = arg((z2/z1)) (B) arg((z3/z2)) = 2arg((z3- z1/z2-z1)) (C) arg((z3/z2)) = arg((z3- z1/z2-z1)) (D) arg((z3/z2)) = (1/2)arg((z3- z1/z2-z1))

Tardigrade Admin · Accepted Answer

z1 = 2 + 3i ∴ |z1| = √13 z2 = 3 - 2i ∴ |z2| = √13 and z3 = -1 - 2i √3 ∴ | z3| = √13 So | z1 | = | z2 | = | z3| = √13 ⇒ z1, z2, z3 lies on a circle of radius √13 centre (0, 0) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d1e40f6d14e9d96cd3ba17e0d804ea11-.png" /> Now, arg((z3/z2)) = 2 arg((z3- z1/z3-z1)) ∴ arg((z3/z2)) = 2 arg ((z3-z1/z2-z1))

Q. If $z_{1} = 2 + 3 i, z_{2} = 3 - 2 i, z_{3} = - 1 - 2 i 3$ then which of the following hold good?

$a r g (\frac{z _{3}}{z _{2}}) = a r g (\frac{z _{2}}{z _{1}})$

$a r g (\frac{z _{3}}{z _{2}}) = 2 a r g (\frac{z _{3} - z _{1}}{z _{2} - z _{1}})$

$a r g (\frac{z _{3}}{z _{2}}) = a r g (\frac{z _{3} - z _{1}}{z _{2} - z _{1}})$

$a r g (\frac{z _{3}}{z _{2}}) = \frac{1}{2} a r g (\frac{z _{3} - z _{1}}{z _{2} - z _{1}})$

Q. If z1​=2+3i,z2​=3−2i,z3​=−1−2i3​ then which of the following hold good?

arg(z2​z3​​)=arg(z1​z2​​)

arg(z2​z3​​)=2arg(z2​−z1​z3​−z1​​)

arg(z2​z3​​)=arg(z2​−z1​z3​−z1​​)

arg(z2​z3​​)=21​arg(z2​−z1​z3​−z1​​)

Q. If $z_{1} = 2 + 3 i, z_{2} = 3 - 2 i, z_{3} = - 1 - 2 i 3$ then which of the following hold good?

$a r g (\frac{z _{3}}{z _{2}}) = a r g (\frac{z _{2}}{z _{1}})$

$a r g (\frac{z _{3}}{z _{2}}) = 2 a r g (\frac{z _{3} - z _{1}}{z _{2} - z _{1}})$

$a r g (\frac{z _{3}}{z _{2}}) = a r g (\frac{z _{3} - z _{1}}{z _{2} - z _{1}})$

$a r g (\frac{z _{3}}{z _{2}}) = \frac{1}{2} a r g (\frac{z _{3} - z _{1}}{z _{2} - z _{1}})$