3+ix−1​+3−iy−1​=i⇒(3+i)(3−i)(x−1)(3−i)+(y−1)(3+i)​=i⇒9−i23x−ix−3+i+3y+iy−3−i​=i⇒3x+3y−ix+iy−6=i[9−(−1)]⇒3x+3y−ix+iy−6=10i⇒3(x+y)+i(−x+y)=6+10iOn comparing both the sides, we get 3(x+y)=6 and −x+y=10⇒x+y=2.....(i)and −x+y=10.....(ii) On solving Eqs. (i) and (ii); we get2y=12⇒y=6 and x=−4