જો √ 3 + i = (a + ib)(c + id) , તો {tan ^{ - 1}}left( {frac{b}{a}} right) + {tan ^{

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4-1.Complex numbers

hard

જો $\sqrt 3 + i = (a + ib)(c + id)$, તો ${\tan ^{ - 1}}\left( {\frac{b}{a}} \right) + $ ${\tan ^{ - 1}}\left( {\frac{d}{c}} \right)$ = . . .

$\frac{\pi }{3} + 2n\pi ,n \in I$

$n\pi + \frac{\pi }{6},n \in I$

$n\pi - \frac{\pi }{3},n \in I$

$2n\pi - \frac{\pi }{3},n \in I$

Solution

(b)We have $\sqrt 3 + i = (a + ib)(c + id)$
$\therefore ac – bd = \sqrt 3 $and $ad + bc = 1$
Now $tan^{-1}$ $\left( {\frac{b}{a}} \right) + {\tan ^{ – 1}}\left( {\frac{d}{c}} \right)$
$\begin{array}{l} = {\tan ^{ – 1}}\left( {\frac{{\frac{b}{a} + \frac{d}{c}}}{{1 – \frac{b}{a}.\frac{d}{c}}}} \right) = {\tan ^{ – 1}}\left( {\frac{{bc + ad}}{{ac – bd}}} \right) = {\tan ^{ – 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)\\\end{array}$
$ = n\pi + \frac{\pi }{6},n \in I$

Standard 11

Mathematics