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Prove that if $z+\frac1z$ is real, then either $|z|=1$ or $z$ is real.

Original image

I am not sure whether my proof is sufficient. So far, I have shown that

$$z+\frac1z = \frac{z^2+1}{z}=\frac{|z|+1}{z}$$

However, I don't think the proof enough, and also I seemed to have proven that they both have to be real and not either... Please advise.

Note: I realised that my proof is wrong, as it was kindly mentioned that $z^2$ doesn't equal $|z|$. I remembered wrongly, it should be $zz*=|z|$ with $z*$ being a conjugate of $z$.

CCC
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    It is $z^2\neq|z|$ – Vincenzo Tibullo Dec 28 '15 at 08:38
  • $|z|$ is the norm of z? – ÝTAN Dec 28 '15 at 08:39
  • |z| is the modulus of z in an argand diagram. – CCC Dec 28 '15 at 08:39
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    $|z|$ is always real. You mean $|z| = 1$ or $z$ is real, I think. – Robert Israel Dec 28 '15 at 08:40
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    Proof of @RobertIsrael's proposed correction is straightforward from certain points of view, but you may not yet be familiar with them, or the exercise may want you to take a particular approach. Do you know how the reciprocal of a complex number is related to the number's conjugate and its absolute value (aka modulus)? – Blue Dec 28 '15 at 08:56
  • @Blue, no I don't. Would you be able to offer me some advice on it? Many thanks. I just know that $zz*=|z|^2$ – CCC Dec 28 '15 at 09:02
  • @Blue, I understand now. Reciprocal of a complex number is its conjugate number and has the same modulus but opposite arguments – CCC Dec 29 '15 at 14:21

4 Answers4

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As already mentioned in the comments, there are some errors in your calculation ($z^2 \ne |z|$, and $|z|$ is always a real number).

You can get the desired result by exanding $1/z$ with the complex conjugate to make the denominator real:

$$ \text{Im} \left( z + \frac 1z \right) = \text{Im} \left( z + \frac {\overline z}{z \overline z} \right) = \text{Im} \, z - \frac{\text{Im} \, z}{|z|^2} = \text{Im} \, z \cdot \left( 1 - \frac{1}{|z|^2} \right) $$ is zero if and only if $$\text{Im} \, z = 0 \text{ or } |z| = 1 \, .$$

Therefore $$ z + \frac 1z \in \Bbb R \Longleftrightarrow z \in \Bbb R \text{ or } |z| = 1 \, . $$

Martin R
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If $z+\dfrac1z$ is real $\iff z+\dfrac1z=\overline{z+\dfrac1z}=\bar z+\dfrac1{\bar z}$

$$\iff (z-\bar z)\left(z\bar z-1\right)=0$$

Now if $z-\bar z=0,z$ is real

Else use $z\bar z=|z|^2$

lab bhattacharjee
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By direct evaluation,

$$\Im\left(a+bi+\frac{a-bi}{a^2+b^2}\right)=b-\frac b{a^2+b^2}=0.$$

Then $b=0$ or $a^2+b^2=1$.

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Another way: \begin{align} z+\frac{1}{z}&=\rho(\cos\theta+i\sin\theta)+\frac{1}{\rho}(\cos\theta-i\sin\theta)\\ &\left(\rho+\frac{1}{\rho}\right)\cos\theta+i\left(\rho-\frac{1}{\rho}\right)\sin\theta \end{align} And $$ \left(\rho-\frac{1}{\rho}\right)\sin\theta=0\implies \rho=1 \text{ or } \theta=0 \text{ or } \theta=\pi $$

Vincenzo Tibullo
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