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!

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- Feb 26th 2011, 10:48 AMChrisYee08Factoring 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, 11:14 AMFernandoRevilla
Find the four roots $\displaystyle z_0,z_1,z_2,z_3$ of $\displaystyle z^4=-1$ i.e. $\displaystyle z_j=\sqrt [4] {-1}$ then,

$\displaystyle z^4+1=(z-z_0)(z-z_1)(z-z_2)(z-z_3)$ - Feb 26th 2011, 07:14 PMlanierms
It might be hard to find the four roots. Here is what it would look like if I do what FernandoRevilla said.

Solution:

$\displaystyle z^4+1$

$\displaystyle =z^4-i^2$

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

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

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

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

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

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

Why is Latex so confusing..... - Feb 26th 2011, 09:20 PMTheCoffeeMachine
Why not at this step write $\displaystyle z^2+i = 0 \Rightarrow z^2 = -i \Rightarrow z = \pm i\sqrt{i} $ or

$\displaystyle z^2-i \Rightarrow z^2 = i \Rightarrow z = \pm \sqrt{i}$ (or it's just the difference of two squares).

(So the four roots of the given equation are $\displaystyle \sqrt{i}$, $\displaystyle -\sqrt{i}$, $\displaystyle i\sqrt{i}$, and $\displaystyle -i\sqrt{i} )$ Thus:

$\displaystyle z^4+1 = (z^2+i)(z^2-i) = (z-i\sqrt{i})(z+i\sqrt{i})(z-\sqrt{i})(z+\sqrt{i})$. - Feb 27th 2011, 01:22 AMFernandoRevilla
An alternative:

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

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

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

So,

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

in the form**lanierms**provided i.e. separated real and imaginary parts of the roots. - Feb 27th 2011, 03:47 AMHallsofIvy
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, 06:42 AMFernandoRevilla
- Feb 28th 2011, 01:05 PMChrisYee08
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, 02:09 AMHallsofIvy
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.