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Mathematics Question on Complex Numbers and Quadratic Equations

If f(z)=1−z31−z,f \left(z\right)=\frac{1-z^{3}}{1-z} , where z=x+iyz=x+iy with z≠1,z\ne1, then Re\left\\{\overline{f \left(z\right)}\right\\}=0 reduces to

A

x2+y2+x+1=0x^{2}+y^{2}+x+1=0

B

x2−y2+x+1=0x^{2}-y^{2}+x+1=0

C

x2−y2−x+1=0x^{2}-y^{2}-x+1=0

D

x2−y2+x−1=0x^{2}-y^{2}+x-1=0

Answer

x2−y2+x−1=0x^{2}-y^{2}+x-1=0

Explanation

Solution

f(z)=1−z31−z=(1−z)(1+z+z2)(1−z)f(z) =\frac{1-z^{3}}{1-z}=\frac{(1-z)\left(1+z+z^{2}\right)}{(1-z)}
=1+z+z2=1+z+z^{2}
Put z=x+iyz=x+i y, we get
f(z)=1+x+iy+(x+iy)2f(z)=1+x+i y+(x+i y)^{2}
=1+x+iy+x2+2xyi+i2y2=1+x+i y+x^{2}+2 x y i+i^{2} y^{2}
=(1+x+x2−y2)+i(y+2xy)=\left(1+x+x^{2}-y^{2}\right)+i(y+2 x y)
⇒f(z)‾=(1+x+x2−y2)−i(y+2xy)\Rightarrow f \overline{(z)} =\left(1+x+x^{2}-y^{2}\right)-i(y+2 x y)
Now, Re⁡f(z)‾=0\operatorname{Re}\\{f \overline{(z)}\\}=0
⇒1+x+x2−y2=0\Rightarrow 1+x+x^{2}-y^{2}=0
⇒x2−y2+x+1=0\Rightarrow x^{2}-y^{2}+x+1=0