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Vi Nguyen · Jul 3, 2009, 04:43 AM

Need to find the roots of :

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

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

thanks :)

Unknown008 · 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 · Jul 4, 2009, 10:57 AM

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 · 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 · Jul 4, 2009, 11:11 AM

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 · 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 · 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 · 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 · 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 · 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?

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