I'm stuck with this equation:

Find all the $z\in\mathbb{C}$ such that $z-1=z^{-1}$, with $z\neq{0}$

So far, I've tried writing $z=a+bi$ and $z^{-1}=\frac{\bar z}{|z|^{2}}$ but got nowhere.

Then, I wrote:

${re^{i\theta}+e^{\pi i}}=r^{-1}e^{-\theta i}$, where $r=|z|$

Also, I thought of considering the argument on both the LHS and the RHS of the equation, but that's not a good idea when you have a sum.

For some reason I am not being able to continue and/or realize what am I missing.

Thanks for the help!

  • 3
    $\begingroup$ As $z\ne0,$ multiply both sides by $z$ $\endgroup$ Jun 18 '17 at 6:03
  • 1
    $\begingroup$ Would you have approached it differently if the question were phrased "solve $x^2-x-1=0$", instead? $\endgroup$
    – dxiv
    Jun 18 '17 at 6:12
  • 2
    $\begingroup$ Hi, @dxiv . You are right. Both yours and lab bhattacharjee's comment helped me see through it. Maybe it's because it's 3 am, haha. Anyway, thanks a lot :) $\endgroup$ Jun 18 '17 at 6:16

Just multiply throughout by $z$ and solve the resulting quadratic ($z^2 - z - 1 = 0$). There are two real solutions, but real numbers are a proper subset of the complex numbers, so that's fine.


Since $z \ne 0$, we have

$z^2 - z = 1, \tag{1}$


$z^2 - z - 1 = 0; \tag{2}$

now use the quadratic formula:

$z = \dfrac{1}{2}(1 \pm \sqrt 5). \tag{3}$

We check: if

$z = \dfrac{1 + \sqrt 5}{2}, \tag{4}$


$z - 1 = \dfrac{-1 + \sqrt 5}{2}, \tag{5}$

so that

$z(z - 1) = \dfrac{5 - 1}{4} = 1; \tag{6}$

the check for

$z = \dfrac {1 - \sqrt 5}{2} \tag{7}$

is very similar.


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