Hi everyone

I'm trying to solve this equation:

z^{4}+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.

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- Sep 17th 2012, 10:41 AMTalaPolynomial with real coefficients but complex roots
Hi everyone

I'm trying to solve this equation:

z^{4}+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 AMMarkFLRe: Polynomial with real coefficients but complex roots
You could rewrite the equation as:

$\displaystyle z^4-i^2=0$ - Sep 17th 2012, 10:52 AMTalaRe: Polynomial with real coefficients but complex roots
Should I still use De Moivre's formula ? Or ?

- Sep 17th 2012, 11:08 AMMarkFLRe: 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 AMTalaRe: 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 AMMarkFLRe: 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 AMTalaRe: Polynomial with real coefficients but complex roots
Can you explain step 1 in Euler and De Moivre ?

- Sep 17th 2012, 11:44 AMMarkFLRe: 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 AMTalaRe: Polynomial with real coefficients but complex roots
I understand now !

Thank you so much - Sep 17th 2012, 02:09 PMSorobanRe: 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 . . . . . . $\displaystyle z\;=\; \cos\left(\tfrac{\pi}{4}\!+\!\tfrac{\pi}{2}n\right ) + i\sin\left(\tfrac{\pi}{4}\!+\!\tfrac{\pi}{2}n \right) \quad \text{ for }n = 0,1,2,3$