If z1=2+3i, z2=3-2i and z3=-1-2√3i , then which of the following is true? (where, i2=-1 )

Question

If z1=2+3i, z2=3-2i and z3=-1-2√3i , then which of the following is true? (where, i2=-1 ) (A) arg((z2/z3))=arg((z2 - z1/z3 - z1)) (B) arg((z2/z3))=arg((z3/z1)) (C) (1/2)arg((z2/z3))=arg((z2 - z1/z3 - z1)) (D) 2arg((z3/z2))=arg((z3 - z1/z2 - z1))

Tardigrade Admin · Accepted Answer

Here, |z1|=|z1|=|z3|=√13 Hence, z1,z2,z3 lie on a circle with center (0 , 0) and r=√13 as shown <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-hfqfdvewqep0.png" /> Now, arg(z2/z3)=2arg(z2 - z1/z3 - z1) ∴ arg(z3/z 2)=2arg(z3 - z1/z2 - z1)

Q. If $z_{1} = 2 + 3 i, z_{2} = 3 - 2 i$ and $z_{3} = - 1 - 23 i$ , then which of the following is true? (where, $i^{2} = - 1$ )

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

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

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

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

Q. If z1​=2+3i,z2​=3−2i and z3​=−1−23​i , then which of the following is true? (where, i2=−1 )

arg(z3​z2​​)=arg(z3​−z1​z2​−z1​​)

arg(z3​z2​​)=arg(z1​z3​​)

21​arg(z3​z2​​)=arg(z3​−z1​z2​−z1​​)

2arg(z2​z3​​)=arg(z2​−z1​z3​−z1​​)

Here, ∣z1​∣=∣z1​∣=∣z3​∣=13​ Hence, z1​,z2​,z3​ lie on a circle with center (0,0) and r=13​ as shown Now, argz3​z2​​=2argz3​−z1​z2​−z1​​ ∴ argz2z3​​=2argz2​−z1​z3​−z1​​

Q. If $z_{1} = 2 + 3 i, z_{2} = 3 - 2 i$ and $z_{3} = - 1 - 23 i$ , then which of the following is true? (where, $i^{2} = - 1$ )

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

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

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

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