Let u=2z+iz-ki,z=x+iy and k>0 If the curve represented by Re(u)+Im(u)=1 intersects the y-axis at the points P and Q wherePQ=5, then the value of k is:
4
12
2
32
Explanation for the correct option:
Given that, u=2z+iz-ki
u=2x+1(2y+1)x+i(y-k)×x-i(y-k)x-i(y-k)u=2x2+(2y+1)(y-k)+i{2xy+x-2xy+2xk}x2+(y-k)2
Re(u)+Img(u)=1⇒2x2+(2y+1)(y-k)+x+2xk=x2+(y-k)2
at Y axis,x=0
(2y+1)(y-k)=(y-k)22y2+y-2yk-k=y2+k2-2yk
Roots of y2+y-(k+k2)=0 are y1 and y2
Difference of roots =5
(1+4k+4k2)=54k2+4k=24k2+k-6=0(k+3)(k-2)=0⇒k=2
Hence, option (C) is the correct answer.