# Polynomial with real coefficients but complex roots

• Sep 17th 2012, 10:41 AM
Tala
Polynomial with real coefficients but complex roots
Hi everyone

I'm trying to solve this equation:

z4+1 = 0

I've tried every method I can think of, such as De Moivre's formula, but I only get two complex roots.

What method do I use ?

I just want to know the method/procedure.
• Sep 17th 2012, 10:49 AM
MarkFL
Re: Polynomial with real coefficients but complex roots
You could rewrite the equation as:

$\displaystyle z^4-i^2=0$
• Sep 17th 2012, 10:52 AM
Tala
Re: Polynomial with real coefficients but complex roots
Should I still use De Moivre's formula ? Or ?
• Sep 17th 2012, 11:08 AM
MarkFL
Re: Polynomial with real coefficients but complex roots
You could use Euler/de Moivre (which are easier), but another method would be to set:

$\displaystyle z^2=\pm i$

$\displaystyle (a+bi)^2=\pm i$

$\displaystyle (a^2-b^2)+(2ab)i=0\pm i$

By equating coefficients, you may determine the 4 roots.

Using Euler and de Moivre:

$\displaystyle e^{4\theta i}=\cos(\pi+2k\pi)+i\sin(\pi+2k\pi)$

$\displaystyle \theta=\frac{\pi}{4}(2k+1)$ where $\displaystyle k\in\{0,1,2,3\}$

Can you finish?
• Sep 17th 2012, 11:19 AM
Tala
Re: Polynomial with real coefficients but complex roots
I have to be honest and say that both methods seem a bit unclear to me.
• Sep 17th 2012, 11:28 AM
MarkFL
Re: Polynomial with real coefficients but complex roots
$\displaystyle z=e^{\theta i}=\cos(\theta)+i\sin(\theta)$

Now, use the 4 values we found above for $\displaystyle \theta$, i.e., $\displaystyle \theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4} ,\frac{7\pi}{4}$.
• Sep 17th 2012, 11:30 AM
Tala
Re: Polynomial with real coefficients but complex roots
Can you explain step 1 in Euler and De Moivre ?
• Sep 17th 2012, 11:44 AM
MarkFL
Re: Polynomial with real coefficients but complex roots
We have:

$\displaystyle z^4=-1=-1+0i$

We find: $\displaystyle |z|=\sqrt{(-1)^2+0^2}=1$ hence, we may state:

$\displaystyle z=e^{\theta i}\:\therefore\:z^4=e^{4\theta i}$

So, we must have:

$\displaystyle e^{4\theta i}=-1+0i=\cos(\pi+2k\pi)+i\sin(\pi+2k\pi)$ where $\displaystyle k\in\mathbb{Z}$

By Euler's formula, we know then:

$\displaystyle 4\theta=\pi(2k+1)$

$\displaystyle \theta=\frac{\pi}{4}(2k+1)$

Now, we want:

$\displaystyle 0\le \theta<2\pi$

$\displaystyle 0\le \frac{\pi}{4}(2k+1)<2\pi$

$\displaystyle 0\le 2k+1<8$

Since k is an integer, we have $\displaystyle k\in\{0,1,2,3\}$

Hence, we have:

$\displaystyle z=\cos\left(\frac{\pi}{4}(2k+1) \right)+i\sin\left(\frac{\pi}{4}(2k+1) \right)$

Now, use the values we found for k to compute the 4 roots.
• Sep 17th 2012, 11:51 AM
Tala
Re: Polynomial with real coefficients but complex roots
I understand now !

Thank you so much
• Sep 17th 2012, 02:09 PM
Soroban
Re: Polynomial with real coefficients but complex roots
Hello, Tala!

Quote:

$\displaystyle \text{Solve: }\:z^4 +1 \:=\: 0$

DeMoivre's Theorem is the approach I would use.

We have: .$\displaystyle z^4 \;=\;-1 \;=\;\cos(\pi\!+\!2\pi n) + i\sin(\pi\!+\!2\pi n)$

Hence: . .$\displaystyle z \;=\;\bigg[\cos(\pi\!+\!2\pi n) + i\sin(\pi\!+\!2\pi n)\bigg]^{\frac{1}{4}}$

n . . . . . . $\displaystyle z\;=\; \cos\left(\tfrac{\pi}{4}\!+\!\tfrac{\pi}{2}n\right ) + i\sin\left(\tfrac{\pi}{4}\!+\!\tfrac{\pi}{2}n \right) \quad \text{ for }n = 0,1,2,3$