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

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.

• "Considering r to be a fixed real number" $\;-\;$ Why fixed? The problem says nothing about $r$.
– dxiv
Mar 25 at 5:38
• @dxiv Yeah, that makes sense! Thanks! Mar 25 at 5:47
• Another way in case you are looking for one is to note that $\left(z-\frac{1}{2}\right)^2 \in \mathbb{R}$ from which we conclude that $z \in \mathbb{R}$ or $\Re(z)=\frac{1}{2}$ May 31 at 5:33

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 $$x=\frac 1 2$$.

• Thanks a lot! I do get it now... Mar 25 at 5:41

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 $$\mathrm{Re}(z)=\frac12$$.