Vi Nguyen Posts: 48, Reputation: 2 Junior Member #1 Jul 3, 2009, 04:43 AM
roots of polynomials
Need to find the roots of :

1) ( z^4)+(4z^2)+16=0

2) (z+i)^5-(z-i)^5=0

thanks :)
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #2 Jul 3, 2009, 09:26 AM

Well, I've not done complex no.s yet, but consulting my book, I think you can get it that way;

1)
$z^4 + 4z^2 + 16 =0$

$z^2 = \frac{-4 \pm \sqrt{4^2 - 4(1)(16)}}{2(1)}$

$z^2 = \frac{-4 \pm \sqrt{16- 64}}{2}$

$z^2 = \frac{-4 \pm \sqrt{-48}}{2}$

$z^2 =-2 \pm 2\sqrt{3}i$

So,

$z = -\sqrt{-2 \pm 2\sqrt{3}i}\ or\ +\sqrt{-2 \pm 2\sqrt{3}i}$

Then you simplify. Note that I'm not sure if that's how it is done, I'm trying.
 Vi Nguyen Posts: 48, Reputation: 2 Junior Member #3 Jul 4, 2009, 10:57 AM
roots of polynomials
Can someone show me how to find the roots of:
(Z^4)+(Z^3)+(Z^2)+Z+1

Please show how you worked this out. Thanks
 Vi Nguyen Posts: 48, Reputation: 2 Junior Member #4 Jul 4, 2009, 11:01 AM
This isn't the answer given in my book. The answer given is
plus or minus one plus or minus i root three. Thanks for trying but, spinning me out too :)

Originally Posted by Unknown008
Well, I've not done complex no.s yet, but consulting my book, I think you can get it that way;

1)
$z^4 + 4z^2 + 16 =0$

$z^2 = \frac{-4 \pm \sqrt{4^2 - 4(1)(16)}}{2(1)}$

$z^2 = \frac{-4 \pm \sqrt{16- 64}}{2}$

$z^2 = \frac{-4 \pm \sqrt{-48}}{2}$

$z^2 =-2 \pm 2\sqrt{3}i$

So,

$z = -\sqrt{-2 \pm 2\sqrt{3}i}\ or\ +\sqrt{-2 \pm 2\sqrt{3}i}$

Then you simplify. Note that I'm not sure if that's how it is done, I'm trying.
 Vi Nguyen Posts: 48, Reputation: 2 Junior Member #5 Jul 4, 2009, 11:11 AM
roots of complex no.s
Can someone show the workings to finding the roots of:

(Z+i)^5 - (Z-i)^5=0

Apparently it can be done using complex exponentials and roots of unity (whatever that means), anyway the answer is Z=±√((5±2√5)/5)
 ScottGem Posts: 64,966, Reputation: 6056 Computer Expert and Renaissance Man #6 Jul 4, 2009, 12:02 PM

Please review the guidelines on asking for help with homework that can be found here:

Ask Me Help Desk - Announcements in Forum : Arts & Literature
 galactus Posts: 2,271, Reputation: 282 Ultra Member #7 Jul 4, 2009, 03:18 PM
Originally Posted by Vi Nguyen
Can someone show me how to find the roots of:
(Z^4)+(Z^3)+(Z^2)+Z+1

Please show how you worked this out. Thanks

Rewrite as:

$\left(x^{2}-\frac{(\sqrt{5}-1)x}{2}+1\right)\left(x^{2}+\frac{(\sqrt{5}+1)x}{2 }+1\right)$

Now, use the quadratic formula on each of those and find the 4 roots. All 4 are complex.
 Vi Nguyen Posts: 48, Reputation: 2 Junior Member #8 Jul 5, 2009, 01:13 AM
If you haven't noticed it's the end of semester break, so this isn't for homework, just asking these questions to tie up loose ends.

Originally Posted by ScottGem
Please review the guidelines on asking for help with homework that can be found here:

Ask Me Help Desk - Announcements in Forum : Arts & Literature
 ScottGem Posts: 64,966, Reputation: 6056 Computer Expert and Renaissance Man #9 Jul 5, 2009, 05:20 AM

Originally Posted by Vi Nguyen
If you haven't noticed it's the end of semester break, so this isn't for homework, just asking these questions to tie up loose ends.
If you haven't noticed, schools have different semesters all over the world. Since you don't list your location, I have no idea what your school schedule is.

But that is not the issue. I've reviewed several of your threads, in all those threads, you post a problem and ask someone to solve it for you. You show no effort to do the work yourself. And that violates the rules of this site. Any further attempts on your part to get people to do your work for you will be removed.
 galactus Posts: 2,271, Reputation: 282 Ultra Member #10 Jul 5, 2009, 08:09 AM
Originally Posted by Vi Nguyen
Can someone show the workings to finding the roots of:

(Z+i)^5 - (Z-i)^5=0

Apparently it can be done using complex exponentials and roots of unity (whatever that means), anyway the answer is Z=±√((5±2√5)/5)
Did you try factoring?

$x^{5}-y^{5}=(x-y)(x^{4}+x^{3}y+x^{2}y^{2}+xy^{3}+y^{4})$

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