STD 11 - 4.1 complex nubers — JEE कक्षा 11 साइन्स — Question

माना $z = x + iy =  - 1 + i$

 $\therefore r\cos \theta  =  - 1$एवं $r\sin \theta =1$ $\therefore \theta  = \frac{{3\pi }}{4}$और $r = \sqrt 2 $

तब $\frac{{1 + 7i}}{{{{(2 - i)}^2}}} = \sqrt 2 \left[ {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right]$

वैकल्पिक :  $\left| {\frac{{1 + 7i}}{{{{(2 - i)}^2}}}} \right| = \left| {\frac{{1 + 7i}}{{3 - 4i}}} \right| = \sqrt 2 $

एवं $arg\left( {\frac{{1 + 7i}}{{3 - 4i}}} \right) = {\tan ^{ - 1}}7 - {\tan ^{ - 1}}\left( { - \frac{4}{3}} \right)$

$ = {\tan ^{ - 1}}7 + {\tan ^{ - 1}}\frac{4}{3} = \frac{{3\pi }}{4}$

$\therefore \frac{{1 + 7i}}{{{{(2 - i)}^2}}}$$ = \sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)$