z = 1 + 2 i z=1+2i z = 1 + 2 i is also a root of equation. We obtain this by conjugating the equation. Dividing the polynomial z 3 + z + 10 z^3+z+10 z 3 + z + 10 by ( z − 1 − 2 i ) ( z − 1 + 2 i ) = z 2 − 2 z + 5 (z-1-2i)(z-1+2i)=z^2-2z+5 ( z − 1 − 2 i ) ( z − 1 + 2 i ) = z 2 − 2 z + 5
we receive z + 2 z+2 z + 2 . Thus, z = − 2 z=-2 z = − 2 is the real root.
Now we consider equation z 4 + z 3 + z − 1 = 0 z^4+z^3+z-1=0 z 4 + z 3 + z − 1 = 0 . We will check that z 1 = i z_1=i z 1 = i is a root of equation. Namely, we have: z 1 4 + z 1 3 + z 1 − 1 = 1 − i + i − 1 = 0 z_1^4+z_1^3+z_1-1=1-i+i-1=0 z 1 4 + z 1 3 + z 1 − 1 = 1 − i + i − 1 = 0
It is clear that z 2 = − i z_2=-i z 2 = − i is also a root of equation.
This can be obtained from the fact that z 2 = z ˉ 1 z_2=\bar{z}_1 z 2 = z ˉ 1
We shall divide the equation by ( z − i ) ( z + i ) = z 2 + 1 (z-i)(z+i)=z^2+1 ( z − i ) ( z + i ) = z 2 + 1 .
Then one receives z 2 − 1 + z . z^2-1+z. z 2 − 1 + z . It remains to solve z 2 + z − 1 = 0 z^2+z-1=0 z 2 + z − 1 = 0 .
The latter quadratic equation has solutions (see e.g. https://en.wikipedia.org/wiki/Quadratic_equation for detalis)
z 3 = − 1 + 5 2 , z 4 = − 1 − 5 2 z_3 = \frac{-1+\sqrt{5}}2, \,\,z_4 = \frac{-1-\sqrt{5}}2 z 3 = 2 − 1 + 5 , z 4 = 2 − 1 − 5 .
Thus, we have the roots z 1 = i , z 2 = − i , z 3 = − 1 + 5 2 , z 4 = − 1 − 5 2 , z_1=i,\quad z_2=-i,\quad z_3 = \frac{-1+\sqrt{5}}2,\quad z_4 = \frac{-1-\sqrt{5}}2, z 1 = i , z 2 = − i , z 3 = 2 − 1 + 5 , z 4 = 2 − 1 − 5 ,