# Factoring z^4+1 over the complex numbers

• Feb 26th 2011, 11:48 AM
ChrisYee08
Factoring z^4+1 over the complex numbers
to do this problem you just set z= x+yi then just factor away right? i just need to know because homework is worth 40% of my grade in this class....ty very much!
• Feb 26th 2011, 12:14 PM
FernandoRevilla
Find the four roots $z_0,z_1,z_2,z_3$ of $z^4=-1$ i.e. $z_j=\sqrt [4] {-1}$ then,

$z^4+1=(z-z_0)(z-z_1)(z-z_2)(z-z_3)$
• Feb 26th 2011, 08:14 PM
lanierms
It might be hard to find the four roots. Here is what it would look like if I do what FernandoRevilla said.

Solution:

$z^4+1$

$=z^4-i^2$

$=(z^2+i)(z^2-i)$

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

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

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

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

$=(z+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)(z-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)(z+\frac{\s qrt{2}}{2}+\frac{\sqrt{2}}{2}i)(z-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)$

Why is Latex so confusing.....
• Feb 26th 2011, 10:20 PM
TheCoffeeMachine
Quote:

Originally Posted by lanierms
Solution:

$z^4+1$

$=z^4-i^2$

$=(z^2+i)(z^2-i)$

Why not at this step write $z^2+i = 0 \Rightarrow z^2 = -i \Rightarrow z = \pm i\sqrt{i}$ or
$z^2-i \Rightarrow z^2 = i \Rightarrow z = \pm \sqrt{i}$ (or it's just the difference of two squares).
(So the four roots of the given equation are $\sqrt{i}$, $-\sqrt{i}$, $i\sqrt{i}$, and $-i\sqrt{i} )$ Thus:
$z^4+1 = (z^2+i)(z^2-i) = (z-i\sqrt{i})(z+i\sqrt{i})(z-\sqrt{i})(z+\sqrt{i})$.
• Feb 27th 2011, 02:22 AM
FernandoRevilla
An alternative:

$\sqrt[4]{-1}=\sqrt[4]{\cos \pi/4+i\sin \pi/4}=$

$\left\{{z_k=\cos (\pi/4+k\pi/2)+i\sin (\pi/4+k\pi/2):k=0,1,2,3}\right\}=$

$\left\{{\sqrt{2}/2(1+i),\sqrt{2}/2(-1+i),\sqrt{2}/2(-1-i),\sqrt{2}/2(1-i)}\right\}$

So,

$z^4+1=(z-z_0)(z-z_1)(z-z_2)(z-z_3)$

in the form lanierms provided i.e. separated real and imaginary parts of the roots.
• Feb 27th 2011, 04:47 AM
HallsofIvy
Note that ChrisYee08 said "homework is worth 40% of my grade in this class". It would have been better to give some hints rather than give the entire answer.
• Feb 27th 2011, 07:42 AM
FernandoRevilla
Quote:

Originally Posted by HallsofIvy
It would have been better to give some hints rather than give the entire answer.

I agree, so I did im my first post. After the complete solution of lanierms it was irrelevant to add or not to add different alternatives.
• Feb 28th 2011, 02:05 PM
ChrisYee08
wow cool thanks. its ok if its just the solution. when i have the solution im always able to go back and figure out what i did wrong and how to do it correctly. i was jw how did you some up with roots?
• Mar 1st 2011, 03:09 AM
HallsofIvy
No, it's not okay! It is very easy to look at a solution given by someone else and tell yourself that you understand it now, but even if you do understand every step, you have not had the experience of finding the steps for yourself- and that is where learning occurs.