Question

If $1 + i = (x + iy)(u + iv) $ then $\tan^{-1} \left( \frac{y}{x} \right) + \cot^{-1} \left(\frac{u}{v} \right) $ has the value

• A. $n \pi + \frac{\pi}{6} , n \in I$
• B. $2 n \pi + \frac{\pi}{3} , n \in I$
• C. $n \pi + \frac{\pi}{4} , n \in I$
• D. $n \pi - \frac{\pi}{3} , n \in I$

Answer 1

Answer: (C)

Given complex equation is
$1 + i = (x + iy)(u + iv) $ or $1 + i = (xu + yv) + i(xv + yu) $
Compare real and imaginary parts, we get
$xu - yv= 1 \, \, \, \, \, \, \, \, \, \, \, \, \, $ ...(i)
$xv - yu = 1 \, \, \, \, \, \, \, \, \, \, \, \, \, $ ...(ii)
Multiply $(i)$ by $u$ and $(ii)$ by $v$ and then add, we get $x(u^2 + v^2) = u + v$
$\Rightarrow \: x = \frac{u + v}{u^2 + v^2}$
from (i), $ y = \frac{xu - 1}{v} = \frac{u -v}{u^2 + v^2} $
(substituting the value of x)
Now , $\tan^{-1} \left(y /x\right) + \cot^{-1}\left(u/ v \right)$
$=\tan^{-1}\left(\frac{u-v}{u+v}\right) + \cot^{-1} \left(u/ v\right) $
$= \tan^{-1} \left(\frac{1- \frac{u}{v}}{1+ \frac{v}{u}}\right) + \cot^{-1} \left(u/ v\right) = $
$\tan^{-1}= \tan ^{-1}\left(1\right) - \tan ^{-1} \left(\frac{v}{u}\right) +\tan ^{-1}\left(\frac{v}{u}\right) $
$= n\pi + \pi 4 , n \in I$