# Solving Quartic

• Apr 2nd 2006, 11:18 AM
c_323_h
Solving Quartic
Find all roots of x^4 + i =0. Write answers in trigonometric form.

Don't know how to solve quartics.
Any help would be appreciated.
• Apr 2nd 2006, 01:05 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
Find all roots of x^4 + i =0. Write answers in trigonometric form.

Don't know how to solve quartics.
Any help would be appreciated.

You need to use a theorem know as De' Moivre's
.
You need to express
$\displaystyle x^4+i=0$
as,
$\displaystyle x^4=-i$
Now, express $\displaystyle i$ in trigonomteric form,
$\displaystyle i=\cos\frac{-\pi}{2}+i\sin\frac{-\pi}{2}$
Thus,
$\displaystyle x^4=\cos\frac{-\pi}{2}+i\sin\frac{-\pi}{2}$
Thus,
$\displaystyle x=\left(\cos\frac{-\pi}{2}+i\sin\frac{-\pi}{2}\right)^{1/4}$
By De Moivre's Theorem,
$\displaystyle x=\cos\left(-\frac{\pi}{8}\right)+i\sin\left(-\frac{\pi}{8}\right)$
But, this is only one solution there are 3 more (because of the degree). In this case, we divide, $\displaystyle 2\pi$ by the degree, (4) to get, $\displaystyle \pi/2$ and each other solution is added by $\displaystyle \pi/2$
Thus,
$\displaystyle x=\left(\cos\frac{-\pi}{8}+i\sin\frac{-\pi}{8}\right)$
$\displaystyle x=\left(\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8} \right)$
$\displaystyle x=\left(\cos\frac{7\pi}{8}+i\sin\frac{7\pi}{8} \right)$
$\displaystyle x=\left(\cos\frac{11\pi}{8}+i\sin\frac{11\pi}{8} \right)$
If you want you can add $\displaystyle 2\pi$ to your first solution to get,

$\displaystyle x=\left(\cos\frac{3\pi}{8}+i\sin\frac{3\pi}{8} \right)$
$\displaystyle x=\left(\cos\frac{7\pi}{8}+i\sin\frac{7\pi}{8} \right)$
$\displaystyle x=\left(\cos\frac{11\pi}{8}+i\sin\frac{11\pi}{8} \right)$
$\displaystyle x=\left(\cos\frac{15\pi}{8}+i\sin\frac{15\pi}{8} \right)$

The beauty of the second result is that it is very easy to see, that each one is divided by 8 and the numerator is increasing by 4.
• Apr 2nd 2006, 02:17 PM
c_323_h
Thanks!

when you solve for roots do you always solve for x?
• Apr 2nd 2006, 03:11 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
Thanks!

when you solve for roots do you always solve for x?

Not sure I understand your question
• Apr 2nd 2006, 03:44 PM
c_323_h
$\displaystyle x^2 + i = 0$

You subracted $\displaystyle i$ from both sides to get $\displaystyle x^4=-i$

Why did you subtract $\displaystyle i$?
• Apr 2nd 2006, 03:51 PM
ThePerfectHacker
Quote:

Originally Posted by c_323_h
$\displaystyle x^2 + i = 0$

You subracted $\displaystyle i$ from both sides to derive $\displaystyle x^4=-i$

Why did you subtract $\displaystyle i$?

You mean, in a regular poynomial you factor one side so you can set its factors equal to zero but over here you do not do this?

Because a special equation of polynomial is,
$\displaystyle x^n=y,y\not =0$.
We can show that they are EXACTLY $\displaystyle n$ solutions and there are called Roots of Unity which are of quite importance in Abstract Algebra/Field Theory.

Whenever, you see and equation like this you bring the constant term to the other side and then use De Moivre's Theorem to find its roots.
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It is interesting to note, that the soltuions that you get, are in trigonomteric form. Some of them can be simplified in algebraic terms (only using +,-,x,/ and roots) while some of them cannot (impossible) and thus you need to leave it in the form it is in.