Describe the set of complex numbers (locus) $z$ such that $z(1-z)$ is a real number.

Question

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.

Answer 1

Your answer is wrong becasue $z(1-z)$ is a real number whenever $z$ is a real number, which shows there are infinitely many solutions.

$z(1-z)$ is a real if and only if its imaginary part is $0$ which gives you $b=0$ or $a=\frac 1 2$. The locus is, therefore, a union of two straight lines: the $x-$ axis an the vertical line $x=\frac 1 2$.

Answer 2

The equation $$ z^2-z+r=0 $$ has roots $$ z=\frac{1\pm\sqrt{1-4r}}2 $$ If $r\le\frac14$, then $z\in\mathbb{R}$.

If $r\gt\frac14$, then $\mathrm{Re}(z)=\frac12$.

That is, the roots are either real or have real part equal to $\frac12$.

Let's check each of these.

If $z\in\mathbb{R}$, then $z(1-z)\in\mathbb{R}$.

If $z=\frac12+iy$, then $z(1-z)=y^2+\frac14\in\mathbb{R}$.

Thus, $z(1-z)\in\mathbb{R}$ if and only if either $z\in\mathbb{R}$ or $\mathrm{Re}(z)=\frac12$.

Answer 3

$4z(1-z) \in \ \mathbb{R} \Rightarrow [z+1-z]^2-[z-(1-z)]^2 \in \ \mathbb{R} $

$\Rightarrow (2z-1)^2 \in \ \mathbb{R} $ so that $z \in \ \mathbb{R} $ or $2z-1 = ki, k \in \ \mathbb{R} \Rightarrow z = \dfrac{1}{2}+\dfrac{ki}{2} $

We can see that the real axis and the line $x=\dfrac{1}{2}$ are the loci