Consider the following two statements:Stateme | vedclass

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(C) Statement $I$:We know that by the triangle inequality,for any complex numbers $w_1, w_2$,$|w_1 + w_2| \leq |w_1| + |w_2|$.Let $w_1 = \frac{z_1}{|z

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Statement $I$:We know that by the triangle inequality,for any complex numbers $w_1, w_2$,$|w_1 + w_2| \leq |w_1| + |w_2|$.Let $w_1 = \frac{z_1}{|z_1|}$ and $w_2 = \frac{z_2}{|z_2|}$.Then $|w_1| = |\frac{z_1}{|z_1|}| = 1$ and $|w_2| = |\frac{z_2}{|z_2|}| = 1$.Thus,$|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}| \leq |\frac{z_1}{|z_1|}| + |\frac{z_2}{|z_2|}| = 1 + 1 = 2$.Multiplying both sides by $(|z_1| + |z_2|)$,we get $(|z_1| + |z_2|)|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}| \leq 2(|z_1| + |z_2|)$.Therefore,Statement $I$ is correct.Statement $II$:Given $\frac{a}{|y-z|} = \frac{b}{|z-x|} = \frac{c}{|x-y|} = k$ (where $k > 0$).Then $a = k|y-z|$,$b = k|z-x|$,$c = k|x-y|$.Consider the expression $S = \frac{a^2}{y-z} + \frac{b^2}{z-x} + \frac{c^2}{x-y}$.Since $|w|^2 = w \bar{w}$,we have $a^2 = k^2|y-z|^2 = k^2(y-z)(\bar{y}-\bar{z})$.Substituting this,$S = \frac{k^2(y-z)(\bar{y}-\bar{z})}{y-z} + \frac{k^2(z-x)(\bar{z}-\bar{x})}{z-x} + \frac{k^2(x-y)(\bar{x}-\bar{y})}{x-y}$.$S = k^2(\bar{y}-\bar{z} + \bar{z}-\bar{x} + \bar{x}-\bar{y}) = k^2(0) = 0$.Since $0 
eq 1$,Statement $II$ is incorrect.

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