w∈C, (Imw=0)⇒w=a+ib:b=0 (suppose)
w−wˉz=k(1−z)
If z=x+iy
⇒(a+ib)−(a−ib)(x+iy)=k1−(x+iy)
⇒(a+ib)−(ax+iay−ibx+by)=k−kx−iky
Equating real and imaginary parts
a−ax−by=k−kx…(1)
and b−ay+bx=−ky…(2)
From (1)(k−a)(x−1)=by
(2)(k−a)y=b(x+1)
Dividing the two equations, we get,
yx−1=x+1y
⇒x2−1=y2
⇒x2+y2=1
⇒∣z∣=1
z=1+i0 (∴ifz=1 then from original equation w=w⇒a+ib=a−ib⇒b=0)