Consider the following two statements :

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Question

Statement $I$ :

$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right|$

Since $\l

Vedclass · Accepted Answer

Statement $I$ is correct but Statement $II$ is incorrect.

Vedclass · Answer

Statement $I$ :

$\left(\left|z_1ight|+\left|z_2ight|ight)\left|\frac{z_1}{\left|z_1ight|}+\frac{z_2}{\left|z_2ight|}ight|$

Since $\left|\frac{z_1}{\left|z_1ight|}+\frac{z_2}{\left|z_2ight|}ight| \leq\left|\frac{z_1}{\left|z_1ight|}ight|+\left|\frac{z_2}{\left|z_2ight|}ight|$

$ \left|\frac{z_1}{\left|z_1ight|}+\frac{z_2}{\left|z_2ight|}ight| \leq \frac{\left|z_1ight|}{\left|z_1ight|}+\frac{\left|z_2ight|}{\left|z_2ight|} $

$ \left|\frac{z_1}{\left|z_1ight|}+\frac{z_2}{\left|z_2ight|}ight| \leq 2$

$\left(\left|z_1ight|+\left|z_2ight|ight)\left(\left|\frac{z_1}{\left|z_1ight|}+\frac{z_2}{\left|z_2ight|}ight|ight) \leq 2\left(\left|z_1ight|+\left|z_2ight|ight)$

$	herefore$ statement $I$ is correct

For Statement $II$ :

$ \frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|} $

$ \frac{a^2}{|y-z|^2}=\frac{b^2}{|z-x|^2}=\frac{c^2}{|x-y|^2}=\lambda $

$ a^2=\lambda\left(|y-z|^2ight)=\lambda(y-z)(\bar{y}-\bar{z}) $

$ b^2=\lambda(z-x)(\bar{z}-\bar{x}) 	ext { and } c^2=\lambda(x-y)(\bar{x}-\bar{y}) $

$ \frac{a^2}{y-z}+\frac{b^2}{z-x}+\frac{c^2}{x-y}=\lambda(\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0$

Statement $II$ is false

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