Let w(Imw≠ 0) be a complex number. Then, the set of all complex numbers z satisfying the equation w- wz=k(1-z), for some real number k, is

$w\in C$w∈C, $(Imw\neq 0)\Rightarrow w=a+ib:b\neq 0$(Imw=0)⇒w=a+ib:b=0 (suppose)

$$w-\bar{w}z=k(1-z)$$w−wˉz=k(1−z)

If $z=x+iy$z=x+iy

$$\Rightarrow (a+ib)-(a-ib)(x+iy)=k{1-(x+iy)}$$⇒(a+ib)−(a−ib)(x+iy)=k1−(x+iy)

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

Equating real and imaginary parts

$$a-ax-by=k-kx\ldots (1)$$a−ax−by=k−kx…(1)

and $b-ay+bx=-ky\ldots (2)$b−ay+bx=−ky…(2)

From $(1)(k-a)(x-1)=by$(1)(k−a)(x−1)=by

$$(2)(k-a)y=b(x+1)$$(2)(k−a)y=b(x+1)

Dividing the two equations, we get,

$$\frac{x-1}{y}=\frac{y}{x+1}$$yx−1​=x+1y​

$$\Rightarrow {x}^{2}-1={y}^{2}$$⇒x2−1=y2

$$\Rightarrow {x}^{2}+{y}^{2}=1$$⇒x2+y2=1

$$\Rightarrow |z|=1$$⇒∣z∣=1

$z\neq 1+i0$z=1+i0 ($\therefore \mathrm{if}z=1$∴ifz=1 then from original equation $w=w\Rightarrow a+ib=a-ib\Rightarrow b=0$w=w⇒a+ib=a−ib⇒b=0)