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. $2n\pi +\frac{\pi }{3},n\in I$
• 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 , ${\tan }^{-1}\left(y/x\right)+{\cot }^{-1}\left(u/v\right)$
$={\tan }^{-1}\left(\frac{u-v}{u+v}\right)+{\cot }^{-1}\left(u/v\right)$
$={\tan }^{-1}\left(\frac{1-\frac{u}{v}}{1+\frac{v}{u}}\right)+{\cot }^{-1}\left(u/v\right)=$
${\tan }^{-1}={\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(\frac{v}{u}\right)+{\tan }^{-1}\left(\frac{v}{u}\right)$
$=n\pi +\pi 4,n\in I$