"z=1+2i" is also a root of equation. We obtain this by conjugating the equation. Dividing the polynomial "z^3+z+10" by "(z-1-2i)(z-1+2i)=z^2-2z+5"
we receive "z+2" . Thus, "z=-2" is the real root.
Now we consider equation "z^4+z^3+z-1=0" . We will check that "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"
It is clear that "z_2=-i" is also a root of equation.
This can be obtained from the fact that "z_2=\\bar{z}_1"
We shall divide the equation by "(z-i)(z+i)=z^2+1" .
Then one receives "z^2-1+z." It remains to solve "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 = \\frac{-1+\\sqrt{5}}2, \\,\\,z_4 = \\frac{-1-\\sqrt{5}}2" .
Thus, we have the roots "z_1=i,\\quad z_2=-i,\\quad z_3 = \\frac{-1+\\sqrt{5}}2,\\quad z_4 = \\frac{-1-\\sqrt{5}}2,"
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