Evaluate: left| frac{1}{2}(z_1 + z_2) + sqrt{ | vedclass

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(C) Let $z_1 = a^2$ and $z_2 = b^2$,where $a = \sqrt{z_1}$ and $b = \sqrt{z_2}$.Then the expression becomes $\left| \frac{1}{2}(a^2 + b^2) + ab \right

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$|z_1| + |z_2|$

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(C) Let $z_1 = a^2$ and $z_2 = b^2$,where $a = \sqrt{z_1}$ and $b = \sqrt{z_2}$.Then the expression becomes $\left| \frac{1}{2}(a^2 + b^2) + ab \right| + \left| \frac{1}{2}(a^2 + b^2) - ab \right|$.This simplifies to $\frac{1}{2} |a^2 + b^2 + 2ab| + \frac{1}{2} |a^2 + b^2 - 2ab|$.$= \frac{1}{2} |(a+b)^2| + \frac{1}{2} |(a-b)^2|$.Since $|z^2| = |z|^2$,this is $\frac{1}{2} |a+b|^2 + \frac{1}{2} |a-b|^2$.Using the parallelogram law $|u+v|^2 + |u-v|^2 = 2(|u|^2 + |v|^2)$,we get:$= \frac{1}{2} \cdot 2(|a|^2 + |b|^2) = |a|^2 + |b|^2$.Substituting back $a^2 = z_1$ and $b^2 = z_2$,we get $|z_1| + |z_2|$.

Evaluate: $\left| \frac{1}{2}(z_1 + z_2) + \sqrt{z_1 z_2} \right| + \left| \frac{1}{2}(z_1 + z_2) - \sqrt{z_1 z_2} \right|$

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