1995 HSC Complex Number Problem (1 Viewer)

frog1944 · Dec 3, 2016

Hi,

I was looking through past papers and found the 1995 MX2 paper question 4 a, and I wasn't sure how to tackle it. The question is;
$\text{Find the least positive integer k, such that } \cos(\frac{4\pi}{7}) + i\sin(\frac{4\pi}{7}) \text{is a solution to} z^k=1.$

How do you do it?

Thanks

InteGrand · Dec 3, 2016

Re: 1995 Complex Number Problem

frog1944 said:
Hi,

I was looking through past papers and found the 1995 MX2 paper question 4 a, and I wasn't sure how to tackle it. The question is;
$\text{Find the least positive integer k, such that } \cos(\frac{4\pi}{7}) + \sin(\frac{4\pi}{7}) \text{is a solution to} z^k=1.$

How do you do it?

Thanks

$\noindent You meant to put an $i$ before the sin, right? In that case, let $\omega = \mathrm{cis}\left(\frac{4\pi}{7}\right)$, then $\omega$ is a seventh root of unity and $7$ is prime, so $\omega$ is automatically a \emph{primitive} seventh root of unity (all non-real $p$-th roots of unity are primitive, where $p$ is an odd prime, as you can show as an exercise). So the answer is $k = 7$.$

frog1944 · Dec 3, 2016

Re: 1995 Complex Number Problem

Yeah I did, I've fixed it now. How can you be certain that it doesn't "overlap" over the previous roots, such as it is a root but not a "principal" root?

InteGrand · Dec 3, 2016

Re: 1995 Complex Number Problem

frog1944 said:
Yeah I did, I've fixed it now. How can you be certain that it doesn't "overlap" over the previous roots, such as it is a root but not a "principal" root?

Because 7 is prime.

$\noindent Also in general, for positive integers $j$ and integers $n \geq 2$, we have that $\mathrm{cis}\left(\frac{j}{n}\cdot (2\pi)\right)$ is a primitive $n$-th root of unity if and only if $j$ and $n$ are coprime.$

frog1944 · Dec 5, 2016

Re: 1995 Complex Number Problem

Wow! Seems much more involved then I thought it would have been. Thanks InteGrand!

1995 HSC Complex Number Problem (1 Viewer)

frog1944

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