If $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then $|z|=$?

Question

If z is a complex number and $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then find the value of $|z|$ .

I tried to put $ \frac {z^2 + z+ 1} {z^2 -z +1} =k $ then solve for $z$ and tried to find |z|, but it gets messy and I am stuck. The answer given is |z|=1

If $z$ is any real number, then $\dfrac{z^2+z+1}{z^2-z+1}$ is also real. Hence $|z|$ could be any non-negative number. — JimmyK4542, Aug 08 '14 at 05:15
@JimmyK4542 But note that for real $z$, the function $\frac{z^2+z+1}{z^2-z+1}$ can only take on values in $[1/3,3]$. So that argument doesn't really hold. — Ryan, Aug 08 '14 at 05:25
@Ryan but his point still holds. OP, you should amend the claim so that it's actually correct. Either stipulate that $z$ is nonreal, or write the proposition I wrote in my answer. — anon, Aug 08 '14 at 06:12

score 4 · Answer 1 · answered Aug 08 '14 at 06:13

First note:

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

So the given fraction is real if and only if the fraction $\frac{z}{z^2 - z + 1}$ is real. But a fraction is real if and only if its reciprocal is, so we need:

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

To be a real number. So we get:

$$\boxed{\text{The fraction is real if and only if } z + z^{-1} \text{ is real (or } z= 0).}$$

Now if $z = a + bi$, with $z \neq 0$, the imaginary part of $z + z^{-1}$ is $b - \frac{b}{a^2 + b^2} = \frac{b(a^2 + b^2 - 1)}{a^2 + b^2} = \frac{b(|z| - 1)}{|z|}$. Hence we get:

$$\boxed{\text{Either } |z| = 1 \text{ or } z \text{ is real.}}$$

It's much better to determine the imaginary part with polar form. — anon, Aug 08 '14 at 06:18

score 2 · Accepted Answer · answered Aug 08 '14 at 06:08

2

Proposition. $\frac{z^2+z+1}{z^2-z+1}$ is real if and only if $z\in\Bbb R$ or $|z|=1$.

I'll let you handle the $\Leftarrow$ direction. For $\Rightarrow$, write $k=\frac{z^2+z+1}{z^2-z+1}=\frac{(z+z^{-1})+1}{(z+z^{-1})-1}\implies z+z^{-1}=\frac{k+1}{k-1}$.

When is $z+z^{-1}$ real for nonreal $z$? (What's its imaginary part?)

answered Aug 08 '14 at 06:08

anon

151,657

great job , thanks – raj Aug 08 '14 at 06:14

score 1 · Answer 3 · answered Aug 08 '14 at 05:43

Let's come up with a more interesting question (which might be what you intended to ask).

Find an $\alpha \in \mathbb{R}$ such that $\forall z \in \mathbb{C}: |z| = \alpha \implies \dfrac{z^2+z+1}{z^2-z+1} \in \mathbb{R} \cup \{\infty\}$.

I'll show that $\alpha = 1$ works. If $|z| = 1$, then $z = e^{i\theta}$ for some $\theta \in [0,2\pi)$.

Thus, $\dfrac{z^2+z+1}{z^2-z+1} = \dfrac{z+1+z^{-1}}{z-1+z^{-1}} = \dfrac{e^{i\theta}+e^{-i\theta}+1}{e^{i\theta}+e^{-i\theta}-1} = \dfrac{2\cos\theta+1}{2\cos\theta-1} \in \mathbb{R} \cup \{\infty\}$, as desired.

Therefore, if $|z| = 1$, then $\dfrac{z^2+z+1}{z^2-z+1}$ is either purely real or undefined.

i need to do this other way round i.e. if $\frac{z^2+z+1}{z^2-z+1} $ is real then |z|=1 — raj, Aug 08 '14 at 05:51
@raj but that's not true, e.g. the quotient is real for $z=0$ but $|0|\ne1$ — anon, Aug 08 '14 at 06:01

score 0 · Answer 4 · answered Aug 08 '14 at 06:01

Suppose $|z|=r$, so that $z\bar{z}=r^2$ or $\bar{z}=r^2/z$.

We have that $$\overline{\frac{z^2+z+1}{z^2-z+1}}=\frac{\bar{z}^2+\bar{z}+1}{\bar{z}^2-\bar{z}+1}=\frac{r^4/z^2+r^2/z+1}{r^4/z^2-r^2/z+1}=\frac{z^2+r^2 z+r^4}{z^2-r^2z+r^4}. $$

But if the expression $$ \frac{z^2+z+1}{z^2-z+1}$$ is real, we must have

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

This should help you determining $r=|z|$.

score 0 · Answer 5 · answered Aug 08 '14 at 07:02

raj

Since the fraction is real, so it is equal to its conjugate. That is, $\frac{z^2+z+1}{z^2-z+1}={\bar{(\frac{z^2+z+1}{z^2-z+1})}}$. Then by simple calculations one can obtain

$(z\bar{z}-1)(z-\bar{z})=0$. Since $z$ is complex so $z\bar{z}=\mid{{z}\mid^2}=1$

If $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then $|z|=$?

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