Question

solve this problem
hurry

good evening

Answer 1

$\huge\mathfrak {Answer:-}$

If $z= x+iy$ $- - - (1)$

|z|² $= x^2+ y^2 = 1$ $- - - (Given)$

$z-1=(x-1)+iy=a+iy$ $- - - (2)$

$z+1=(x+1)+iy=c+iy$

Let ω= ${(z-1)/(z+1)}$ $- - - (3)$

Re(ω)

= Re { $(a+iy)$ $(c-iy)$ / $(c^2+y^2)$} = $ac + y^2$

$Now,$

$x-1=c$ $- - - (4)$

$x+1=a$

$x^2-1$ $=ac$

Re(ω) $- - - (5)$

$=ac+y^2$

$=$ $x^2-1+y^2$

$=$ $(x^2+y^2)^{-1}$

$=(1)-1$

$=0$

$Therefore,$

ω is imaginary.

$\huge {Be\:Brainly}$❤️

Answer 2

Heya ...

===== Solution ======

z = x+iy

|z|^2 = x^2+y^2 ........1

z-1 =(x-1)+iy = a+iy........2

z+1 =(x+1)+iy= ciy

Let... w

= (z-1)/(z+1)........3

Re(w)

= Re{(a+iy)(c-iy)/c^2+y^2 } = ac+y^2

x-1=c.... 4
x+1 =a
x^2-1 = ac

Re(w) = 5

ac^2 + y^2

x^2-1+y^2

(x^2+y^2)

= 1-1 = 0

So, it's imaginary ( kalpanik )

Thank you