Question

Given that $z_{1},z_{2}$ and $z_{3}$ are complex numbers with $\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1,z_{1}+z_{2}+z_{3}=1$ and $z_{1}z_{2}z_{3}=1$ , then $\left|\left(z_{1} + 2\right) \left(z_{2} + 2\right) \left(z_{3} + 2\right)\right|$ is equal to

• A. $14$
• B. $15$
• C. $20$
• D. $9$

Answer 1

Answer: (B)

$\left(z_{1} + 2\right)\left(z_{2} + 2\right)\left(z_{3} + 2\right)=z_{1}z_{2}z_{3}+2\left(z_{1} z_{2} + z_{2} z_{3} = z_{3} z_{1}\right)+4\left(z_{1} + z_{2} + z_{3}\right)+8$
$=1+2\left(z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}\right)+4+8=13+2\left(z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}\right)$
Now $z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=z_{1}z_{2}z_{3}\left(\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}}\right)=\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}$
$=\frac{\bar{z}_{1}}{\left|z_{1}\right|^{2}}+\frac{\bar{z}_{2}}{\left|z_{2}\right|^{2}}+\frac{\bar{z}_{3}}{\left|z_{3}\right|^{2}}=\bar{z}_{1}+\bar{z}_{2}+\bar{z}_{3}=z_{1}+z_{2}+z_{3}=1$