Question

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

• A. $arg\left(\frac{z_{2}}{z_{3}}\right)=arg\left(\frac{z_{2} - z_{1}}{z_{3} - z_{1}}\right)$
• B. $arg\left(\frac{z_{2}}{z_{3}}\right)=arg\left(\frac{z_{3}}{z_{1}}\right)$
• C. $\frac{1}{2}arg\left(\frac{z_{2}}{z_{3}}\right)=arg\left(\frac{z_{2} - z_{1}}{z_{3} - z_{1}}\right)$
• D. $2arg\left(\frac{z_{3}}{z_{2}}\right)=arg\left(\frac{z_{3} - z_{1}}{z_{2} - z_{1}}\right)$

Answer 1

Answer: (C)

Here, $\left|z_{1}\right|=\left|z_{1}\right|=\left|z_{3}\right|=\sqrt{13}$
Hence, $z_{1},z_{2},z_{3}$ lie on a circle with center $\left(0 , 0\right)$
and $r=\sqrt{13}$ as shown

Now, $arg\frac{z_{2}}{z_{3}}=2arg\frac{z_{2} - z_{1}}{z_{3} - z_{1}}$
$\therefore $ $arg\frac{z_{3}}{z 2}=2arg\frac{z_{3} - z_{1}}{z_{2} - z_{1}}$