Question

If $f \left(z\right)=\frac{1-z^{3}}{1-z} ,$ where $z=x+iy$ with $z\ne1,$ then $Re\left\{\overline{f \left(z\right)}\right\}=0$ reduces to

• A. $x^{2}+y^{2}+x+1=0$
• B. $x^{2}-y^{2}+x+1=0$
• C. $x^{2}-y^{2}-x+1=0$
• D. $x^{2}-y^{2}+x-1=0$

Answer 1

The Correct Option is D

$f(z) =\frac{1-z^{3}}{1-z}=\frac{(1-z)\left(1+z+z^{2}\right)}{(1-z)} $
$=1+z+z^{2} $
Put $z=x+i y$, we get
$f(z)=1+x+i y+(x+i y)^{2}$
$=1+x+i y+x^{2}+2 x y i+i^{2} y^{2} $
$=\left(1+x+x^{2}-y^{2}\right)+i(y+2 x y) $
$\Rightarrow f \overline{(z)} =\left(1+x+x^{2}-y^{2}\right)-i(y+2 x y) $
Now, $\operatorname{Re}\{f \overline{(z)}\}=0$
$ \Rightarrow 1+x+x^{2}-y^{2}=0 $
$ \Rightarrow x^{2}-y^{2}+x+1=0 $