Question

Which of the following is not one of the 8th roots of unity?

-i
i
1 + i
√2/2 + √2/2 i

Answer 1

1+i

Explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1. Since
`z=1=1+0\cdot i,` then

`Rez=1,\\ \\Im z=0`

and

`|z|=√(1^2+0^2)=1,\\ \\\\\cos\varphi =(Rez)/(|z|)=(1)/(1)=1,\\ \\\sin\varphi =(Imz)/(|z|)=(0)/(1)=0.`

This gives you
`\varphi=0.`

Thus,

`z=1\cdot(\cos 0+i\sin 0).`

2. The 8th roots can be calculated using following formula:

`\sqrt[8]{z}=\{\sqrt[8]z (\cos(\varphi+2\pi k)/(8)+i\sin (\varphi+2\pi k)/(8)), k=0,\ 1,\dots,7\}.`

Now

at k=0,
`z_0=\sqrt[8]{1} (\cos(0+2\pi \cdot 0)/(8)+i\sin (0+2\pi \cdot 0)/(8))=1\cdot (1+0\cdot i)=1;`

at k=1,
`z_1=\sqrt[8]{1} (\cos(0+2\pi \cdot 1)/(8)+i\sin (0+2\pi \cdot 1)/(8))=1\cdot ((√(2))/(2)+i(√(2))/(2))=(√(2))/(2)+i(√(2))/(2);`

at k=2,
`z_2=\sqrt[8]{1} (\cos(0+2\pi \cdot 2)/(8)+i\sin (0+2\pi \cdot 2)/(8))=1\cdot (0+1\cdot i)=i;`

at k=3,
`z_3=\sqrt[8]{1} (\cos(0+2\pi \cdot 3)/(8)+i\sin (0+2\pi \cdot 3)/(8))=1\cdot (-(√(2))/(2)+i(√(2))/(2))=-(√(2))/(2)+i(√(2))/(2);`

at k=4,
`z_4=\sqrt[8]{1} (\cos(0+2\pi \cdot 4)/(8)+i\sin (0+2\pi \cdot 4)/(8))=1\cdot (-1+0\cdot i)=-1;`

at k=5,
`z_5=\sqrt[8]{1} (\cos(0+2\pi \cdot 5)/(8)+i\sin (0+2\pi \cdot 5)/(8))=1\cdot (-(√(2))/(2)-i(√(2))/(2))=-(√(2))/(2)-i(√(2))/(2);`

at k=6,
`z_6=\sqrt[8]{1} (\cos(0+2\pi \cdot 6)/(8)+i\sin (0+2\pi \cdot 6)/(8))=1\cdot (0-1\cdot i)=-i;`

at k=7,
`z_7=\sqrt[8]{1} (\cos(0+2\pi \cdot 7)/(8)+i\sin (0+2\pi \cdot 7)/(8))=1\cdot ((√(2))/(2)-i(√(2))/(2))=(√(2))/(2)-i(√(2))/(2);`

The 8th roots are

`\{1,\ (√(2))/(2)+i(√(2))/(2),\ i, -(√(2))/(2)+i(√(2))/(2),\ -1, -(√(2))/(2)-i(√(2))/(2),\ -i,\ (√(2))/(2)-i(√(2))/(2)\}.`

Option C is icncorrect.