COMEDK Exam 2014 Solved Paper

Given, equation is
1+i=(x+iy)(u+iv)
⇒1+i=(xu−yv)+i(xv+yv)
Comparing real and imaginary parts, we get
xu−yv=1 ..........(i)
xv+yu=1...........(ii)
Multiply Eq. (i) by u and Eq. (ii) by v and then adding,
we get
x(u2+v2)=u+v⇒x=

u+v

u2+v2

From Eq. (i)
y=xu=

1

v

=

u−v

u2+v2

(Substituting the value of x)
Now, tan−1(

y

x

)+cot−1(

u

v

)
tan−1(

u−v

u+v

)+cot−1(

u

v

)
=tan−1(

1− v u

1+ v u

)+cot−1(

u

v

)
=tan−1(1)−tan−1(

v

u

)+tan−1(

v

u

)
=nπ+

π

4

,n∊I