Consider the following two statements : Statement I : For any two non-zero complex numbers z_1, z_2 ( |z_1 |+ |z_2 | ) |z_1 |z_1 | +z_2 |z_2 | | 2 ( |z_1 |+ |z_2 | ) and Statement II : If x, y, z are three distinct complex numbers and a, b, c are three positive real numbers such that a |y-z| =b |z-x| =c |x-y| , then a^2 y-z + b^2 z-x + c^2 x-y =1 Between the above two statements,

Consider the following two statements : Statement I : For any two…

MCQ

Consider the following two statements :

Statement $I$ : For any two non-zero complex numbers $\mathrm{z}_1, \mathrm{z}_2$

$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$ and

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

$\frac{\mathrm{a}^2}{\mathrm{y}-\mathrm{z}}+\frac{\mathrm{b}^2}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{c}^2}{\mathrm{x}-\mathrm{y}}=1$

Between the above two statements,

A
both Statement $I$ and Statement $II$ are incorrect.
B
Statement $I$ is incorrect but Statement $II$ is correct.
✓
Statement $I$ is correct but Statement $II$ is incorrect.
D
both Statement $I$ and Statement $II$ are correct.

✓

Answer

Correct option: C.

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

c
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 $\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq\left|\frac{z_1}{\left|z_1\right|}\right|+\left|\frac{z_2}{\left|z_2\right|}\right|$

$ \left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq \frac{\left|z_1\right|}{\left|z_1\right|}+\frac{\left|z_2\right|}{\left|z_2\right|} $

$ \left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2$

$\left(\left|z_1\right|+\left|z_2\right|\right)\left(\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right|\right) \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$

$\therefore$ 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\right)=\lambda(y-z)(\bar{y}-\bar{z}) $

$ b^2=\lambda(z-x)(\bar{z}-\bar{x}) \text { 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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