Consider the following two statements : Statement I: For any two non-zero complex numbers …

Question

Consider the following two statements :

Statement I: For any two non-zero complex numbers $$z_1, z_2,(|z_1|+|z_2|)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right) \text {, and }$$

Statement II : If $x, y, z$ are three distinct complex numbers and $$\mathrm{a}, \mathrm{b}, \mathrm{c}$$ are three positive real numbers such that $$\frac{\mathrm{a}}{|y-z|}=\frac{\mathrm{b}}{|z-x|}=\frac{\mathrm{c}}{|x-y|}$$, then $$\frac{\mathrm{a}^2}{y-z}+\frac{\mathrm{b}^2}{z-x}+\frac{\mathrm{c}^2}{x-y}=1$$.

Between the above two statements,

Accepted Answer

$$\begin{aligned}
& \frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}=\lambda \
& \Rightarrow a^2=\lambda^2|(y-z)|^2 \
& b^2=\lambda^2|(z-x)|^2 \
& c^2=\lambda^2|(x-y)|^2 \
& \frac{a^2(\overline{y-z})}{(y-z)(y-z)}=\frac{a^2(\bar{y}-\bar{z})}{|y-z|^2}=\frac{a^2(\bar{y}-\bar{z})}{\frac{a^2}{\lambda^2}}=\lambda^2(\bar{y}-\bar{z}) \
& \Rightarrow \sum\left(\frac{a^2}{y-z}\right)=\lambda^2(\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0 \neq 1 \
\end{aligned}$$

Statement I

$$\begin{aligned}
& \left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right) \
& \Rightarrow \quad z_1=\left|z_1\right| e^{i \theta_1} \
& \quad z_2=\left|z_2\right| e^{i \theta_2} \
& \Rightarrow \frac{z_1}{\left|z_1\right|}=e^{i \theta_1} \
& \Rightarrow \frac{z_2}{\left|z_2\right|}=e^{i \theta_2} \
& \Rightarrow\left|e^{i \theta_1}+e^{i \theta_2}\right| \
& \quad=\left|\sqrt{2+2 \cos \left(\theta_1-\theta_2\right)}\right| \
& \quad=\left|2 \cos \left(\frac{\theta_1-\theta_2}{2}\right)\right| \leq 2
\end{aligned}$$

Solutions · 1