# Finding the rootsof complex number

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• Nov 14th 2010, 03:11 AM
Paymemoney
Finding the rootsof complex number
Hi

How would i do the following question:

Find the roots of the equation: $z^4-1$

This is what i have done:
$z^4 = -1$

$z^4 = cis(\frac{\pi}{4})$, $cis(\frac{5\pi}{4})$, $cis(\frac{9\pi}{4})$, $cis(\frac{13\pi}{4}$

$z^4 = cis(\frac{\pi}{16})$, $cis(\frac{5\pi}{16})$, $cis(\frac{9\pi}{16})$, $cis(\frac{13\pi}{16}$

P.S
• Nov 14th 2010, 03:44 AM
Unknown008
I would have used the difference of two squares twice:

$z^4 - 1 = (z^2+1)(z^2 -1) = (z^2+1)(z+1)(z-1)$

Then, z = 1 and -1.

For the other two complex roots, we get:

$z^2 = -1$

Which you should be able to solve.
• Nov 14th 2010, 04:31 AM
tonio
Quote:

Originally Posted by Unknown008
I would have used the difference of two squares twice:

$z^4 - 1 = (z^2+1)(z^2 -1) = (z^2+1)(z+1)(z-1)$

Then, z = 1 and -1.

For the other two complex roots, we get:

$z^2 = -1$

Which you should be able to solve.

I think from what she/he wrote, though, that the equation she's trying to solve is

$z^4=-1=cis(\pi i+2k\pi i)=cis(\pi i(2k+1))\,,\,\,k=0,1,2,3$ ,

with solutions $\displaystyle{z_k=cis\left(\frac{\pi}{4}i(2k+1)\ri ght)}\,,\,\,k=01,2,3$

Tonio
• Nov 14th 2010, 06:21 AM
HallsofIvy
Quote:

Originally Posted by Paymemoney
Hi

How would i do the following question:

Find the roots of the equation: $z^4-1$

This isn't an equation. I suppose you mean $z^4- 1= 0$

Quote:

This is what i have done:
$z^4 = -1$

$z^4 = cis(\frac{\pi}{4})$, $cis(\frac{5\pi}{4})$, $cis(\frac{9\pi}{4})$, $cis(\frac{13\pi}{4}$
I wrote what I did about what you had not being an equation because you seem to have a problem with writing exactly what you mean- and it hurt you here. Those four numbers are the fourth roots- you have already solved the problem at this point:
$z^4= cis(\pi)$ and
$z = cis(\frac{\pi}{4})$, $cis(\frac{5\pi}{4})$, $cis(\frac{9\pi}{4})$, $cis(\frac{13\pi}{4})$

Quote:

$z^4 = cis(\frac{\pi}{16})$, $cis(\frac{5\pi}{16})$, $cis(\frac{9\pi}{16})$, $cis(\frac{13\pi}{16}$

P.S
• Nov 14th 2010, 06:24 AM
HallsofIvy
Quote:

Originally Posted by Unknown008
I would have used the difference of two squares twice:

$z^4 - 1 = (z^2+1)(z^2 -1) = (z^2+1)(z+1)(z-1)$

almost:

$z^4+ 1= (z^2+ i)(z^2- i)= (z-(\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2})(z+(\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}))(z-(\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2})(z+(\frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2})$

Quote:

Then, z = 1 and -1.

For the other two complex roots, we get:

$z^2 = -1$

Which you should be able to solve.
• Nov 14th 2010, 06:30 AM
Unknown008
Quote:

Originally Posted by HallsofIvy
almost
$z^4+ 1= (z^2+ i)(z^2- i)= (z-(\sqrt{2}{2}+ i\sqrt{2}{2})(z+(\sqrt{2}{2}+ i\sqrt{2}{2})(z-(\sqrt{2}{2}- i\sqrt{2}{2})(z+(\sqrt{2}{2}- i\sqrt{2}{2})$

Um... I don't understand why you look for $z^4+1$ while it's $z^4-1$... (Itwasntme)

EDIT: Ah, okay. I understand now.
• Nov 14th 2010, 11:48 AM
Paymemoney
yes my bad, its meant to be $z^4+1=0$ srry for the confusion