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

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(c) यहाँ $|{z_k}| = 1,k = 1,\,\,2,....n$

$&rArr; |{z_k}{|^2} = 1\,\, \Rightarrow {z_k}{\overline z _k} = 1\, \Rightarrow {\overline z _k} = \frac{1

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) यहाँ $|{z_k}| = 1,k = 1,\,\,2,....n$

$&rArr; |{z_k}{|^2} = 1\,\, \Rightarrow {z_k}{\overline z _k} = 1\, \Rightarrow {\overline z _k} = \frac{1}{{{z_k}}}$

इसलिए $|{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|$

वैकल्पिक : माना ${z_k} = \cos {	heta _k} + i\sin {	heta _k},\,\,\,k = 1,\,\,2,....n$

ताकि $|{z_k}| = \sqrt {{{\cos }^2}{	heta _k} + {{\sin }^2}{	heta _k}}  = 1$

 तब $\frac{1}{{{z_k}}} = {(\cos {	heta _k} + i\sin {	heta _k})^{ - 1}} = (\cos {	heta _k} - i\sin {	heta _k})$

अब ${z_1} + {z_2} + ..... + {z_n}$

$ = (\cos {	heta _1} + ..... + \cos {	heta _n}) - i(\sin {	heta _1} + ..... + \sin {	heta _n})$ और $\left( {\frac{1}{{{z_1}}}} ight) + \left( {\frac{1}{{{z_2}}}} ight) + ..... + \left( {\frac{1}{{{z_n}}}} ight)$

$ = (\cos {	heta _1} + ..... + \cos {	heta _n}) - i(\sin {	heta _1} + ..... + \sin {	heta _n})$

अत: $|{z_1} + {z_2} + ..... + {z_n}| = \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + ..... + \frac{1}{{{z_n}}}} ight|$ 

प्रत्येक पक्ष $\sqrt {{{(\cos {	heta _1} + ..... + \cos {	heta _n})}^2} + {{(\sin {	heta _1} + .... + \sin {	heta _n})}^2}} $ के बराबर है।

यदि $|{z_1}| = |{z_2}| = .......... = |{z_n}| = 1,$ तो $|{z_1} + {z_2} + {z_3} + ............. + {z_n}|$=

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