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.
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.
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?
$\displaystyle z=e^{\theta i}=\cos(\theta)+i\sin(\theta)$
Now, use the 4 values we found above for $\displaystyle \theta$, i.e., $\displaystyle \theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4} ,\frac{7\pi}{4}$.
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.
Hello, Tala!
$\displaystyle \text{Solve: }\:z^4 +1 \:=\: 0$
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 . . . . . . $\displaystyle z\;=\; \cos\left(\tfrac{\pi}{4}\!+\!\tfrac{\pi}{2}n\right ) + i\sin\left(\tfrac{\pi}{4}\!+\!\tfrac{\pi}{2}n \right) \quad \text{ for }n = 0,1,2,3$