Question

For any complex numbers $z_{1}$ and $z_{2}$, the maximum value of $\frac{z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}}{\left|z_{1}\right|\left|z_{2}\right|}$ is

• A. $1 / 2$
• B. $1$
• C. $3 / 2$
• D. $2$

Answer 1

Answer: (D)

$\frac{z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}}{\left|z_{1}\right|\left|z_{2}\right|}$
$=2 \text{Re}\left(\frac{z_{1} \bar{z}_{2}}{\left|z_{1} z_{2}\right|}\right)$
$\leq 2 \cdot \frac{\left|z_{1} \bar{z}_{2}\right|}{\left|z_{1} z_{2}\right|}=2$
Alternatively, $z_{1}=r_{1} e^{i \alpha}$
and $z_{2}=r_{2} e^{i \beta}$
$\therefore \frac{z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}}{\left|z_{1} \| z_{2}\right|}$
$=e^{i(\alpha-\beta)}+e^{-i(\alpha-\beta)}$
$=2 \cos (\alpha-\beta)$
$\leq 2$