Aditya Gupta
Last Activity: 4 Years ago
let z= x+iy
so x+iy + i + (root 2) | z +1|= 0
x + (root 2) | z +1| + i(y+1)= 0
equating real and imaginary parts to 0,
y+1 = 0 and x + (root 2) | z +1| = 0
so y= – 1
and – x= (root 2) | z +1|= (root 2) | x+1 + iy|
x^2= 2[(x+1)^2 + y^2]= 2[(x+1)^2 + 1]= 2(x^2+2x+2)
or x^2= 2x^2+4x+4
(x+2)^2= 0
x= – 2
so, z= – 2 – i
kindly approve :))