1
$\begingroup$

If $n$ is an integer and $n>2$, then $$\bar{z}=z^{n-1}$$ has $n+1$ solutions. Does this have something to do with the rational root theorem?

$\endgroup$
  • $\begingroup$ $|z|=0$ or $|z|=1$. If $|z|=1$, we have $\overline{z}=\frac{1}{z}$, so that $z^{n}=1$, it gives $n$ solutions and $|z|=0$ gives the $(n+1)$-th solution. $\endgroup$ – MoebiusCorzer Nov 26 '15 at 16:20
5
$\begingroup$

Since $|z|=|\overline{z}|$, taking absolute values gives $$ |z|=|z|^{n-1} $$ which, as $|z|$ is a non-negative real number, means $|z|=0$ or $|z|=1$.

The equation is obviously satisfied when $z=0$. If $|z|=1$, then $\overline{z}=\frac{1}{z}$, which means the equation reduces to $$ \frac{1}{z}=z^{n-1} $$ whose solutions are the $n$th roots of unity.

$\endgroup$
2
$\begingroup$

Let $z=re^{it}$. We then have $$re^{-it} = r^{n-1}e^{(n-1)it}$$ This means $$r=0 \text{ or }r^{n-2} e^{nit} = 1$$ Now $r^{n-2} e^{nit} = 1$ gives us $r=1$ and $t = \dfrac{2 \pi k}n$, where $k \in \{0,1,\ldots,n-1\}$, which gives us a total of $n+1$ solutions (including the solution $z=0$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.