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I am trying to prove that If $z\in \mathbb{C}-\mathbb{R}$ such that $\frac{z^2+z+1}{z^2-z+1}\in \mathbb{R}$. Show that $|z|=1$.

1 method , through which I approached this problem is to assume $z=a+ib$ and to see that $$\frac{z^2+z+1}{z^2-z+1}=1+\frac{2z}{z^2-z+1}$$.

So problem reduces to show that $|z|=1$ whenever $\frac{2z}{z^2-z+1}\in \mathbb{R}$

I put $z=a+ib$ and then rationalise to get the imaginary part of $\frac{2z}{z^2-z+1}$ be $\frac{b-b^3-a^2b}{something}$. I equated this to zero and got my answer.

Is there any better method?

GraduateStudent
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2 Answers2

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We are given that the imaginary part of $\frac{z^2+z+1}{z^2-z+1}$ is zero. Therefore $\frac{z^2+z+1}{z^2-z+1}$ is equal to its own conjugate: $$\frac{z^2+z+1}{z^2-z+1} = \frac{\bar z^2+\bar z+1}{\bar z^2-\bar z+1}.$$

Cross-multiply (multiply both sides by $(z^2-z+1)(\bar z^2-\bar z+1)$) and cancel all terms on the right. The result is $$ 2z\bar z^2 - 2z^2\bar z - 2\bar z + 2z = 0,$$ which you can simplify to $$(\bar z - z)z\bar z - \bar z + z = 0.$$

Since the imaginary part of $z$ is not zero, it follows that $\bar z - z \neq 0$ and you can divide both sides of the equation by $\bar z - z$ to obtain $$ z\bar z - 1 = 0.$$

David K
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  • I wonder if you noticed that a possible duplicate target was already pointed out ... – Martin R Feb 11 '19 at 17:07
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    @MartinR The “duplicate” has a lot of discussion about the fact that the question neglected to specify that $z$ must not be real. I thought that was distracting. Possibly it would have been better to add a sentence to deal with that detail and post there. I still could do that and delete this answer. (I don’t care much about the rep points.) – David K Feb 11 '19 at 18:14
  • Thank you David – GraduateStudent Feb 12 '19 at 01:22
  • Yes, Martin I wondered why they didn't consider the case that $z$ isn't purely real. – GraduateStudent Feb 12 '19 at 01:23
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Let $w=\frac{z^2+z+1}{z^2-z+1}=1+\frac{2z}{z^2-z+1}$

Since $\frac{z^2+z+1}{z^2-z+1}\in \mathbb{R}$ , so $\operatorname{Im}(w)=0$ $$\iff w-\bar{w}=0$$

Now, let's solve $1+\frac{2z}{z^2-z+1}=\overline{1+\frac{2z}{z^2-z+1}}$

$$\implies \frac{z}{z^2-z+1}=\overline{\frac{z}{z^2-z+1}}$$

Since $z^2-z+1=0 \ when\ z= \frac{1 \pm i \sqrt{3}}{2}$

Assume $z \neq \frac{1 \pm i \sqrt{3}}{2}$ $$\implies \frac{z}{z^2-z+1}= \frac{\overline{z}}{\overline{z^2}-\overline{z}+1}$$

$$\implies z(\overline{z^2}-\overline{z}+1)=\overline{z}(z^2-z+1) $$

After some simplifications, we have $$ (z-\overline{z})(|z|-1)=0$$ $$\implies (z-\overline{z})=0 \ or \ |z|-1=0$$

Since $(z-\overline{z})=0\ $must be true by the above statement, so we just need to solve $|z|-1=0 \implies |z|=1$

Bello Bello
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