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4.1 complex nubers — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHS4.1 complex nubers1 Mark
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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