Answer:
1+i
Step-by-step explanation:
To find the 8th roots of unity, you have to find the trigonometric form of unity.
1. Since
then

and

This gives you 
Thus,

2. The 8th roots can be calculated using following formula:
![\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bz%7D%3D%5C%7B%5Csqrt%5B8%5D%7B%7Cz%7C%7D%20%28%5Ccos%5Cdfrac%7B%5Cvarphi%2B2%5Cpi%20k%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B%5Cvarphi%2B2%5Cpi%20k%7D%7B8%7D%29%2C%20k%3D0%2C%5C%201%2C%5Cdots%2C7%5C%7D.)
Now
at k=0, ![z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;](https://tex.z-dn.net/?f=z_0%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%200%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%200%7D%7B8%7D%29%3D1%5Ccdot%20%281%2B0%5Ccdot%20i%29%3D1%3B)
at k=1, ![z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_1%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%201%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%201%7D%7B8%7D%29%3D1%5Ccdot%20%28%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=2, ![z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;](https://tex.z-dn.net/?f=z_2%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%202%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%202%7D%7B8%7D%29%3D1%5Ccdot%20%280%2B1%5Ccdot%20i%29%3Di%3B)
at k=3, ![z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_3%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%203%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%203%7D%7B8%7D%29%3D1%5Ccdot%20%28-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=4, ![z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;](https://tex.z-dn.net/?f=z_4%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%204%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%204%7D%7B8%7D%29%3D1%5Ccdot%20%28-1%2B0%5Ccdot%20i%29%3D-1%3B)
at k=5, ![z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_5%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%205%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%205%7D%7B8%7D%29%3D1%5Ccdot%20%28-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=6, ![z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;](https://tex.z-dn.net/?f=z_6%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%206%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%206%7D%7B8%7D%29%3D1%5Ccdot%20%280-1%5Ccdot%20i%29%3D-i%3B)
at k=7, ![z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_7%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%207%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%207%7D%7B8%7D%29%3D1%5Ccdot%20%28%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
The 8th roots are

Option C is icncorrect.
Answer:
use something called tiger algerbra
Step-by-step explanation:
Step-by-step explanation:
A regular n-sided polygon has n axes of symmetry. The angle between two consecutive axes is 360° / n.
In the case of an octagon, the angle is 360° / 8 = 45°.
Is the algebra, if so what type?