Verified answer

l1+i*zl^ν =1

so l1+i*zl =1

let z = x + iy

so l1+i*(x+iy)l =1

so l1+i*(x+iy)l =1

so | (1-y) + ix| = 1

so (1-y)^2 + x^2 = 1

or x^2 + (y-1)^2 = 1

math kp

|1+iz| = v th root of 1 = 1. Because LHS is a real number so RHS is also a real and +ve number being modulus of a complex number.

So 1 + iz = 1

so iz = 0

so z = 0.