Nth root of a complex number:

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June 8th 2009, 10:18 AM
Neversh

Nth root of a complex number:

I need to find the fourth root of -81. Thank you in advance.
June 8th 2009, 10:28 AM
Moo

Hello,

Quote:

Originally Posted by Neversh

I need to find the fourth root of -81. Thank you in advance.

-81=(-1)*(81)
The 4th root of 81 is 3 (because 81=9x9=3x3x3x3)

Now, what is the 4th root of -1 ? Write $e^{i\pi+2ik\pi}=-1$ , where k is an integer
And then take the fourth root of [tex]e^{i\pi+2ik\pi}... : $e^{i\pi/4},e^{i\pi/2},e^{3i\pi/4},e^{i\pi}$ (k=0,1,2,3) then as k grows up, you'll get back to these solutions...
June 8th 2009, 10:33 AM
HallsofIvy

[QUOTE=Moo;327307]Hello,

-81=(-1)*(81)
The 4th root of 81 is 3 (because 81=9x9=3x3x3x3)

Now, what is the 4th root of -1 ? Write $e^{i\pi+2k\pi}=-1$ , where k is an integer
And then take the fourth root of $e^{i\pi+2k\pi}...$
Typo: This should be either $e^{i(\pi+ 2k\pi)}= -1$ or $e^{i\pi+ 2ki\pi}$

Quote:

: $e^{i\pi/4},e^{i\pi/2},e^{3i\pi/4},e^{i\pi}$ (k=0,1,2,3) then as k grows up, you'll get back to these solutions...
June 8th 2009, 10:38 AM
pankaj

Let x be the fourth root of -81.

$x^4=-81=81(\cos \pi +i\sin \pi)=3^4(\cos(2k\pi+\pi)+i\sin(2k\pi+\pi);k=0,1,2, 3$

$x=3(\cos(2k\pi+\pi)+i\sin(2k\pi+\pi))^{\frac{1}{4} }$

By De-Moivre's theorem

$ x=3(\cos(\frac{(2k+1)\pi}{4})+i\sin(\frac{(2k+1)\p i}{4})) $ ;k=0,1,2,3

$ k=0;x=3(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}) $

$ k=1;x=3(-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}) $

$ k=2;x=3(-\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}) $

$ k=3;x=3(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}}) $