Let $n \in \mathbb N$. Determine all complex numbers $z \in \mathbb C $ such that $ |z| ^{n-2} = 1.$
How would i begin this question, thanks!
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user136088
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If a positive number $r$ is such that $r^{n-2} = 1$, what can you say about $r$? – Najib Idrissi Apr 02 '14 at 07:20
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Let $r=|z|$ then notice that $r$ is a nonnegative real number.
$r^{n-2}=1$ if $n=2$ then it is true for all $r>0\implies $ it is true for all $z\neq 0$
if $n\ne2$ then $r=1\implies$ it is true for all $z=xi+y$ such that $x^2+y^2=1$
mesel
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@user136088: it is the final answer, it has infinitly many roots in both case. what do yo know about $n$? in your question, it can be anynumber. – mesel Apr 02 '14 at 07:28
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Yeah the question doesn't say anything about n. Thus for this equation, there are infinitely many roots right? – user136088 Apr 02 '14 at 07:29
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Oh wait, i missed the part where it said n is a natural number. How would this affect the answer? – user136088 Apr 02 '14 at 07:30
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@user136088 it doesn't matter, it's the absolute value that causes the entire unit circle to be a solution. – orion Apr 02 '14 at 07:33
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If $n = 2$, then $\vert z \vert = 1$ for all $z \ne 0$, and perhaps also for $z = 0$, depending on how one looks at $0^0$. In any event, the case $n =2$ stands by itself, especially in light of the fact that $\vert z \vert^{n -2} = 1$ implies $\vert z \vert = 1$ for $n \ne 2$; then any unimodular $z$ is a solution, and only such $z$ are solutions. Thus $z = e^{i \theta}$ for any real $\theta$; of course, since $e^{i(\theta + 2\pi)} = e^{i \theta}$, we only need consider $\theta \in [0, 2\pi)$; such $\theta$ yield all solutions to $\vert z \vert^{n - 2} = 1$ when $n \ne 2$.
Hope this helps. Cheerio,
and as always, Fiat Lux!!!**
Robert Lewis
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