यदि z_1 तथा {z_2} दो सम्मिश्र संख्याएँ इस प्र | vedclass

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Question

(c) $\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} \right| = 1$

$&rArr;\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}} = \cos \alpha  + i\sin \alph

Vedclass · Accepted Answer

-$2{	an ^{ - 1}}k$

Vedclass · Answer

(c) $\left| {\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}}} ight| = 1$

$&rArr;\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}} = \cos \alpha  + i\sin \alpha $

$\frac{{2{z_1}}}{{ - 2{z_2}}} = \frac{{\cos \alpha  + i\sin \alpha  + 1}}{{\cos \alpha  - 1 + i\sin \alpha }}$

(योगान्तरानुपात से)

$&rArr;\frac{{{z_1}}}{{{z_2}}} = i\cot \frac{\alpha }{2}$

$&rArr;i{z_1} =  - \left( {\cot \frac{\alpha }{2}} ight)\,\,{z_2}$

लेकिन $i{z_1} = k{z_2}&rArr; k =  - \cot \frac{\alpha }{2}$

अब $k =  - \cot \frac{\alpha }{2} \Rightarrow $$\cot \frac{\alpha }{2} =  - k$&THORN;$	an \alpha  = \frac{{ + 2k}}{{{k^2} - 1}}$

$&rArr;  	an \alpha  = \frac{{ - 2k}}{{1 - {k^2}}}$

$&rArr; \alpha  = {	an ^{ - 1}}\left( {\frac{{ - 2k}}{{1 - {k^2}}}} ight) =  - 2{	an ^{ - 1}}k$

अब $\frac{{{z_1} - {z_2}}}{{{z_1} + {z_2}}} = \cos \alpha  + i\sin \alpha $

$&rArr;{z_1} - {z_2}$ एवं ${z_1} + {z_2}$के बीच का कोण $\alpha $ है

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