Mathematics - Complex Number Question with Solution | TestHub

Statement I : $\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}\right|$
Since $\left|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}\right|\le \left|\frac{z_1}{|z_1|}\right|+\left|\frac{z_2}{|z_2|}\right|$ $\begin{array}{rl}& \left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right|\le \frac{\left|z_1\right|}{\left|z_1\right|}+\frac{\left|z_2\right|}{\left|z_2\right|}\\ & \left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right|\le 2\end{array}$ $\left(\left|z_1\right|+\left|z_2\right|\right)\left(\left|\frac{z_1}{|z_1|}+\frac{z_2}{|z_2|}\right|\right)\le 2\left(\left|z_1\right|+\left|z_2\right|\right)$ $\therefore$ statement $\mathrm{I}$ is correct For Statement II : $\begin{array}{rl}& \frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}\\ & \frac{a^2}{|y-z{|}^2}=\frac{b^2}{|z-x{|}^2}=\frac{c^2}{|x-y{|}^2}=\lambda \\ & a^2=\lambda \left(|y-z{|}^2\right)=\lambda (y-z)(\bar{y}-\bar{z})\\ & b^2=\lambda (z-x)(\bar{z}-\bar{x})\text{ and }{c}^2=\lambda (x-y)(\bar{x}-\bar{y})\\ & \frac{a^2}{y-z}+\frac{b^2}{z-x}+\frac{c^2}{x-y}=\lambda (\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0\end{array}$ Statement II is false