Question

If $f\left(z\right)=\frac{1-z^3}{1-z}$, where $z=x+iy$ with $z\ne 1$, then $Re\overline{\left\{f\left(z\right)\right\}}=0$ 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. $x^2-y^2+x+1=0$
• E. $x^2-y^2+x+2=0$

Answer 1

Answer: (D)

$x^2-y^2+x+1=0$

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