Basically, the question is to solve $z^8= 1$. I have factored this down to $$(z+1)(z-1)(z^2+1)(z^4+1)=0$$
I have simplified $(z^4+1)$ to $z =\pm \sqrt{i}$ and online I know that this can be simplified to the following four solutions:
z= 0.7071 + 0.7071 i
z= -0.7071 + 0.7071 i
z= -0.7071 - 0.7071 i
z= 0.7071 - 0.7071 i
How would I get this without a calculator?
Thank you!
We can have$$\begin{align}z^4+1&=z^4+2z^2+1-2z^2\\&=(z^2+1)^2-(\sqrt 2\ z)^2\\&=(z^2-\sqrt 2\ z+1)(z^2+\sqrt 2\ z+1)\end{align}$$
use the Euler's Identity $$e^{\pi i}=-1$$ and $$e^{ix}=\cos x+i\sin x$$
$$z^4=-1$$ $$z^4=e^{\pi i}$$ $$z=e^{\frac{\pi }{4}+\frac{n\pi}{2}}$$ $$z_1=\cos(\frac{\pi }{4}+\frac{\pi}{2})+i\sin(\frac{\pi }{4}+\frac{\pi}{2})$$ $$z_2=\cos(\frac{\pi }{4}+\frac{2\pi}{2})+i\sin(\frac{\pi }{4}+\frac{2\pi}{2})$$ $$z_3=\cos(\frac{\pi }{4}+\frac{3\pi}{2})+i\sin(\frac{\pi }{4}+\frac{3\pi}{2})$$ $$z_4=\cos(\frac{\pi }{4}+\frac{4\pi}{2})+i\sin(\frac{\pi }{4}+\frac{4\pi}{2})$$
Use Euler's formula, $\mathrm{e}^{x + \mathrm{i} y} = \mathrm{e}^x (\cos y + \mathrm{i} \sin y)$ to compute the root.
$$z^4=-1\Longleftrightarrow$$ $$z^4=|-1|\cdot e^{\arg(-1) i}\Longleftrightarrow$$ $$z^4=1\cdot e^{\pi i}\Longleftrightarrow$$ $$z=\left(1\cdot e^{\left(\pi+2\pi k\right) i}\right)^{\frac{1}{4}}\Longleftrightarrow$$ $$z=\left(1\cdot e^{\left(\pi+2\pi k\right) i}\right)^{\frac{1}{4}}\Longleftrightarrow$$ $$z=1^{\frac{1}{4}}\cdot e^{{\frac{1}{4}}\left(\pi+2\pi k\right) i}\Longleftrightarrow$$ $$z=1\cdot e^{{\frac{1}{4}}\left(\pi+2\pi k\right) i}\Longleftrightarrow$$ $$z=e^{{\frac{1}{4}}\left(\pi+2\pi k\right) i}$$
With $k\in\mathbb{Z}$ and $k:0-3$
