Question

if |z^2-1|=|z^2|+1 then show that z lies on imaginary​

Answer 1

Step-by-step explanation:

|z^2-1| = |z|^2+1

Square both side,

(z^2-1)^2 = (zz+1)^2

Z^2⋅Z^2−Z^2−Z^2+1=(zZ)^2+1+2ZZ

Z^2+ z^2 + 2zZ = 0

(Z+z)^2=0

z=x−iy

Z=x+iy

=> (x+iy) = -(x-iy)

x +iy = -x + iy =>x=0

Thus, Z is pure imaginary.