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

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(c) $R.H.S =$ $\frac{1}{2}|{(\sqrt {{z_1}} + \sqrt {{z_2}} )^2}| + \frac{1}{2}|{(\sqrt {{z_1}} - \sqrt {{z_2}} )^2}|$
$ = \frac{1}{2}|\sqrt {{z_1}} +

Vedclass · Accepted Answer

$|{z_1}| + |{z_2}|$

Vedclass · Answer

(c) $R.H.S =$ $\frac{1}{2}|{(\sqrt {{z_1}} + \sqrt {{z_2}} )^2}| + \frac{1}{2}|{(\sqrt {{z_1}} - \sqrt {{z_2}} )^2}|$
$ = \frac{1}{2}|\sqrt {{z_1}} + \sqrt {{z_2}} {|^2} + \frac{1}{2}|\sqrt {{z_1}} - \sqrt {{z_2}} {|^2}$$\{ \because |{z^2}| = |z{|^2}\} $
$ = \frac{1}{2}2\,[|\sqrt {{z_1}} {|^2} + |\sqrt {{z_2}} {|^2}] = |{z_1}| + |{z_2}|$

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

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