Part of the solution to a question in my book says $e^{5ix} = -1 $ has five solutions for x. There is no further explanation. How do I arrive at this result?
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1Can you find one solution? If so, you can add any multiple of $\frac{2\pi}{5}$ to it and it will still be a solution. There are actually infinitely many solutions for $x$, but five possible values of $e^{ix}$. – Minus One-Twelfth Apr 28 '19 at 04:25
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The equation has infinitely many solutions: $x=\pi/5 +2k\pi/5$ for $k\in\mathbb{Z}$. – user647486 Apr 28 '19 at 04:30
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$$e^{i5x} = -1$$ is equivalent to $$\cos(5x) + i\sin(5x) = -1$$ because of Euler's formula. This obviously has infinite solutions for $n\geq1$ along $x=\pi\frac{2n-1}{5}$. I think it says $5$ because it has $5$ solutions for $0\leq x \lt 2\pi$.
Ryan Shesler
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You mean $0\le x\lt2\pi$. Of course $0$ and $2\pi$ have the same reference angle. – gen-ℤ ready to perish Apr 28 '19 at 09:25
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@ChaseRyanTaylor $2\pi$ is not a solution so it makes no difference, but I will add it anyways – Ryan Shesler Apr 28 '19 at 13:29
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$e^{i\pi} = e^{3i\pi} = ... = e^{(2n-1)i\pi} = -1$
This has many solutions for,$$ 5x = 2n-1$$ $$x = \frac{2n - 1}{5}, n\epsilon Z$$ e.g.,
$ x = \frac{1}{5}$ or $ x = \frac{3}{5}$ or $ x = \frac{5}{5}$ or $ x = \frac{7}{5}$ or $ x = \frac{9}{5}$
19aksh
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One solution is $$5x=\pi \implies x=\pi /5$$
Other solution come from $$ 5x= 3\pi, 5\pi , 7\pi , 9\pi\implies x=3\pi/5, 5\pi/5, 7\pi/5, 9\pi/5$$
Mohammad Riazi-Kermani
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If $e^{5ix}=e^{iy}$
$5ix=iy+2n\pi i$ where $n$ is any integer
$x=\dfrac{y+2n\pi}5,n=0,1,2,3,4$
Here one of the possible values of $y$ is $\pi$
lab bhattacharjee
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1@Hema, actually $n\equiv0,1,2,3,4\pmod5$ What's the period of $e^{iy}$ – lab bhattacharjee Apr 28 '19 at 04:33
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The period of $e^{ix}$ doesn't limit the number of solutions for $x$ in any way, which are real numbers. – user647486 Apr 28 '19 at 04:34
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If $y_m=e^{i(y+2m\pi)/5},$ $$y_u=y_v\implies$$ $2\pi$ divides $$\dfrac{y+2u\pi}5-\dfrac{y+2v\pi}5$$ $5$ divides $u-v$ – lab bhattacharjee Apr 28 '19 at 04:36
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From https://math.stackexchange.com/questions/192742/how-to-solve-x3-1/192743#192743, $$x^5+1=0$$ doesn't have a repeated root – lab bhattacharjee Apr 28 '19 at 04:38
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The equations $e^{5ix}+1=0$ and $x^5+1=0$ don't have the neither the same set of solutions not the same cardinality. – user647486 Apr 28 '19 at 04:40
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Do you understand that they don't? That $\pi/5+200000\pi/5$ is a very different real number than $\pi/5$? That change of variable gives the solutions $x$ of the original equation as a logarithm of $y$, which has infinitely many values. – user647486 Apr 28 '19 at 04:42
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@user647486, How can $e^{5ix}+1=0$ and $x^5+1=0$ share common solutions! – lab bhattacharjee Apr 28 '19 at 04:47