Describe the set of complex numbers (locus) $z$ such that $z(1-z)$ is a real number.
My solution goes like this:
Let $z=a+ib$ then $z(1-z)=z-z^2=(a+ib)-(a^2-b^2+2abi)=a+ib-a^2+b^2-2abi=r,$ where $r\in \Bbb R.$ Thus, $a-a^2+b^2+i(b-2ab)=r$ and comparing the real and imaginary parts in LHS and RHS of the equation, we get, $a-a^2+b^2=r$ and $b(1-2a)=0.$ Now if $b(1-2a)=0,$ then $b=0$ or $a=\frac 12.$ If $b=0$ then from $a-a^2+b^2=r$ we have, $a+a^2-r=0.$ This is a quadratic equation in $a$ and $a\in \Bbb R$ so, the discriminant must be $\geq 0$ and hence, $1+4r\geq 0.$ Considering $r$ to be a fixed real number in the question snd assuming $r$ satisfies $1+4r\geq 0,$ we have atmost two values of $a.$ So, if $b=0$ we obtain atmost $2$ points on the $x$ axis (or real axis in Argand plane). Now, if $a=\frac 12, $ then from $a-a^2+b^2=r$ and assuming $r$ is so chosen such that $b\in\Bbb R,$ then we have atmost $2$ values of $b.$ Thus, we again get atmost two points on the line $x=\frac 12.$ So, we have atmost $4$ complex numbers satisfying the condition in the question.
Is the above solution valid? If not, where is it going wrong? I feel something is off with the solution but I can't seem to understand why ? I was essentially asked to find the locus but I just found out, that atmost $4$ complex numbers satisfies the condition for a fixed given real number.