Question

Let $w(Imw\ne 0)$ be a complex number. Then the set of all complex number z satisfying the equation $w-\overline{w}z=k(1-z)$, for some real number k, is :

• A. $\{z:|z|=1\}$
• B. $\{z:z=\overline{z}\}$
• C. $\{z:z\ne 1\}$
• D. $\{z:|z|=1,z\ne 1\}$

Answer 1

Answer: (D)

$\{z:|z|=1,z\ne 1\}$

let 'w' be $a+ib$ and z be ${}^{\prime }x+i{y}^{\prime }$

$a+ib-(a-ib)(x+iy)=k(1-x-iy)$

$a+ib-[ax+ayi-bxi+by]=k(1-x)-iky$

$a+ib-ax-ayi+bxi-by=k(1-x)-kyi$

$(a-ax-by)+i(b-ay+bx)=k(1-x)-kyi$

.
comparing real and imaginary coefficients from both sides.

$k=\frac{(a-ax-by)}{(1-x)},k=\left(\frac{ay-bx-b}{y}\right)$

$\frac{a-ax-by}{1-x}=\frac{ay-bx-b}{y}$

$ay-axy-b{y}^2=ay-bx-b-ayx+b{x}^2+bx$

$-b{y}^2=b{x}^2-b$

$b=b{y}^2+b{x}^2$

$x^2+y^2=1$

$|z|=1$