My Math Forum 4th root of -1 and 5th root of -1

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 January 18th, 2017, 07:20 AM #1 Newbie   Joined: Jan 2017 From: Germany Posts: 1 Thanks: 0 4th root of -1 and 5th root of -1 can you help me please to solve 4th root of -1?
January 18th, 2017, 07:30 AM   #2
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Quote:
 Originally Posted by robax25 can you help me please to solve 4th root of -1?
$-1 = e^{i \pi}$

$\sqrt[4]{-1} = \left(e^{i \pi}\right)^{1/4} = e^{i \frac{\pi+2k\pi}{4}},~k=0,1,2,3$

$\sqrt[4]{-1} = 1+i,~ -1+i,~ -1-i, ~1-i$

see if you can find $\sqrt[5]{-1}$ based on above

 January 18th, 2017, 09:10 AM #3 Senior Member   Joined: May 2016 From: USA Posts: 673 Thanks: 283 Umm $(1 + i)^4 = \{(1 + i)^2\}^2 = (1^2 + 2i + i^2)^2 = (\cancel 1 + 2i - \cancel 1)^2 = (2i)^2 = - 4.$ What I believe romsek meant was: $\sqrt[4]{-\ 1} = \dfrac{\sqrt{2}}{2} * (1 + i)\ and\ so\ on\ for\ the\ other\ three\ roots.$ $\left \{ \dfrac{\sqrt{2}}{2} * (1 + i) \right \}^4 = \left ( \dfrac{2}{4} * 2i \right )^2 = \left( \dfrac{\cancel 4 i}{\cancel 4} \right )^2 = i^2 = -\ 1.$ Thanks from topsquark and romsek
 January 18th, 2017, 03:45 PM #4 Senior Member   Joined: Aug 2012 Posts: 1,414 Thanks: 342 OP, the key to these problems is to know that when you multiply complex numbers, you multiply their lengths and add their angles. Since in this case $-1$ has length $1$, we are just looking for some number whose angle added together four times has the same angle as $-1$, which has angle $\pi$ radians. I'm skipping a discussion of the polar form of complex numbers, but that's the essential thing to know. You can read about it here. https://en.wikipedia.org/wiki/Complex_number#Polar_form So, what angle times $4$ is $\pi$? It's $\frac{\pi}{4}$ obviously, so $e^{i \frac{\pi}{4}}$ is one of the fourth roots. The others are the integer powers of that. If you are not familiar with this geometric interpretation of complex multiplication, it's very helpful to know. Then when you see a problem like this you just go, "Well, $-1$ makes an angle of $\pi$ with the positive x-axis, and one fourth of that is $\frac{\pi}{4}$ so the principle fourth root is $e^{i \frac{\pi}{4}}$. And the other fourth roots are the integer powers of that." You just see it right away, you don't even need to think about it. Thanks from topsquark and JeffM1
 January 18th, 2017, 06:32 PM #5 Member   Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 In its most general form, we can write \displaystyle \begin{align*} -1 = \mathrm{e}^{\left( 2\,n + 1 \right) \,\pi\,\mathrm{i}} \end{align*}, where \displaystyle \begin{align*} n \in \mathbf{Z} \end{align*} - in other words, all the complex numbers of magnitude 1 which have an argument that is an odd multiple of \displaystyle \begin{align*} \pi \end{align*}, thus \displaystyle \begin{align*} \sqrt[4]{-1} &= \left( -1 \right) ^{\frac{1}{4}} \\ &= \left[ \mathrm{e}^{\left( 2\,n + 1 \right) \,\pi\,\mathrm{i}} \right] ^{\frac{1}{4}} \\ &= \mathrm{e}^{ \frac{\left( 2\,n + 1 \right) \,\pi}{4}\,\mathrm{i} } \end{align*} in other words, all the complex numbers that have a magnitude of 1 and an argument that is an odd multiple of \displaystyle \begin{align*} \frac{\pi}{4} \end{align*}. Thus the four principal solutions are \displaystyle \begin{align*} \mathrm{e}^{-\frac{3\,\pi}{4}\,\mathrm{i}} &= \cos{ \left( -\frac{3\,\pi}{4} \right) } + \mathrm{i}\,\sin{ \left( -\frac{3\,\pi}{4} \right) } \\ &= -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\,\mathrm{i} \\ \\ \mathrm{e}^{-\frac{\pi}{4}\,\mathrm{i}} &= \cos{ \left( -\frac{\pi}{4} \right) } + \mathrm{i}\,\sin{ \left( -\frac{\pi}{4} \right) } \\ &= \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\,\mathrm{i} \\ \\ \mathrm{e}^{\frac{\pi}{4}\,\mathrm{i}} &= \cos{ \left( \frac{\pi}{4} \right) } + \mathrm{i}\,\sin{ \left( \frac{\pi}{4} \right) } \\ &= \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\,\mathrm{i} \\ \\ \mathrm{e}^{\frac{3\,\pi}{4}\,\mathrm{i}} &= \cos{ \left( \frac{3\,\pi}{4} \right) } + \mathrm{i}\,\sin{ \left( \frac{3\,\pi}{4} \right) } \\ &= -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\,\mathrm{i} \end{align*} As for the fifth roots, we do the same thing again... \displaystyle \begin{align*} \sqrt[5]{-1} &= \left( -1 \right) ^{\frac{1}{5}} \\ &= \left[ \mathrm{e}^{\left( 2\,n + 1 \right) \,\pi } \right] ^{\frac{1}{5}} \\ &= \mathrm{e}^{ \frac{\left( 2\,n + 1 \right) \,\pi}{5}\,\mathrm{i} } \end{align*} in other words, all the numbers which have a magnitude of 1 and an argument that is an odd multiple of \displaystyle \begin{align*} \frac{\pi}{5} \end{align*}. It is handy to know that \displaystyle \begin{align*} \cos{ \left( \frac{\pi}{5} \right) } = \frac{1 + \sqrt{5}}{4} \end{align*} and \displaystyle \begin{align*} \sin{ \left( \frac{\pi}{5} \right) } = \frac{\sqrt{ 10 - 2\,\sqrt{5} }}{4} \end{align*} Do you think you can find the five principal solutions? Thanks from greg1313 and topsquark Last edited by Prove It; January 18th, 2017 at 06:46 PM.

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