If |{z_1}| = |{z_2}| = .......... = |{z_n}| = | vedclass

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(c) We have $|{z_k}| = 1,k = 1,\,\,2,....n$
==> $|{z_k}{|^2} = 1\,\, \Rightarrow {z_k}{\overline z _k} = 1\, \Rightarrow {\overline z _k} = \frac{

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$\left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + ......... + \frac{1}{{{z_n}}}} \right|$

Vedclass · Answer

(c) We have $|{z_k}| = 1,k = 1,\,\,2,....n$
==> $|{z_k}{|^2} = 1\,\, \Rightarrow {z_k}{\overline z _k} = 1\, \Rightarrow {\overline z _k} = \frac{1}{{{z_k}}}$
Therefore $|{z_1} + {z_2} + .... + {z_n}| = |\overline {{z_1} + {z_2} + .... + {z_n}} |$
$(\because \,\,\,|z|\, = \,|\,\overline z |)$
$ = |{\overline z _1} + \overline {{z_2}} + ..... + {\overline z _n}| = \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + .... + \frac{1}{{{z_n}}}} ight|$
Aliter : Let ${z_k} = \cos {	heta _k} + i\sin {	heta _k},\,\,\,k = 1,\,\,2,....n$
So that $|{z_k}| = \sqrt {{{\cos }^2}{	heta _k} + {{\sin }^2}{	heta _k}} = 1$
Then $\frac{1}{{{z_k}}} = {(\cos {	heta _k} + i\sin {	heta _k})^{ - 1}} = (\cos {	heta _k} - i\sin {	heta _k})$
Now, ${z_1} + {z_2} + ..... + {z_n}$
$ = (\cos {	heta _1} + ..... + \cos {	heta _n}) - i(\sin {	heta _1} + ..... + \sin {	heta _n})$
and $\left( {\frac{1}{{{z_1}}}} ight) + \left( {\frac{1}{{{z_2}}}} ight) + ..... + \left( {\frac{1}{{{z_n}}}} ight)$
$ = (\cos {	heta _1} + ..... + \cos {	heta _n}) - i(\sin {	heta _1} + ..... + \sin {	heta _n})$
Hence $|{z_1} + {z_2} + ..... + {z_n}| = \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + ..... + \frac{1}{{{z_n}}}} ight|$
Since each side is equal to
$\sqrt {{{(\cos {	heta _1} + ..... + \cos {	heta _n})}^2} + {{(\sin {	heta _1} + .... + \sin {	heta _n})}^2}} $

List-$I$	List-$II$
($P$) $\|z\|^2$ is equal to	($1$) $12$
($Q$) $\|z-\bar{z}\|^2$ is equal to	($2$) $4$
($R$) $\|z\|^2+\|z+\bar{z}\|^2$ is equal to	($3$) $8$
($S$) $\|z+1\|^2$ is equal to	($4$) $10$
	($5$) $7$

If $|{z_1}| = |{z_2}| = .......... = |{z_n}| = 1,$ then the value of $|{z_1} + {z_2} + {z_3} + ............. + {z_n}|$=

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