Let w I m   w ≠0 be a complex number. Then, the set of all complex numbers z satisfying the…

$w\in C$,  $\left(Im w\ne 0\right) \Rightarrow w=a+ib:b\ne 0$ (suppose)

$w-\overline{w} z=k\left(1-z\right)$

If $z=x+iy$

$\Rightarrow \left(a+ib\right)-\left(a-ib\right)\left(x+iy\right)=k\{1-\left(x+iy\right)\}$

$\Rightarrow \left(a+ib\right)-\left(ax+iay-ibx+by\right)=k-kx-iky$

Equating real and imaginary parts

$a-ax-by=k-kx \dots \left(1\right)$

and $b-ay+bx=-ky \dots \left(2\right)$

From $\left(1\right) \left(k-a\right)\left(x-1\right)=by$

 $\left(2\right) \left(k-a\right)y=b(x+1)$
Dividing the two equations, we get, 

$\frac{x-1}{y}=\frac{y}{x+1}$

$\Rightarrow x^2-1=y^2$

$\Rightarrow x^2+y^2=1$

$\mathrm{\Rightarrow }\left|z\right|=1$

$z\ne 1+i0$ ($\therefore \mathrm{if} z=1$ then from original equation   $w=\overline{w} \Rightarrow a+ib=a-ib\Rightarrow b=0$)