Finding complex roots of fourth degree polynomial $z^4 + 8z^3 + 16z^2 + 9$

Question

I have the equation: $$z^4 + 8z^3 + 16z^2 + 9 = 0$$ I need to find all the complex solutions and I've got no clue how to approach it. I've tried factoring but nothing came out of it. I'm still very new to the world of complex numbers so I'll appreciate any help.

Answer 1

If you substitute $$x=z+2$$ the equation turns into $$x^4-8x^2+25=0$$ which can be solved by a further substitution $y=x^2$

Answer 2

$$z^2(z+4)^2 =9i^2$$

so $$z^2+4z-3i=0\;\;\;\;{\rm or}\;\;\;\;z^2+4z+3i=0$$

You have a discriminant $D = 2(3+i)^2$ in the first case and $D = 2(3-i)^2$ in the second case.

Answer 3

We can use factorization $x^2+y^2 = (x+iy)(x-iy)$: \begin{align} z^4+8z^3+16z^2+9 &= z^2(z^2+8z+16) + 9 \\ &= z^2(z+4)^2+9 \\ &= (z(z+4)+3i)(z(z+4)-3i) \\ &= ((z+2)^2 - (4-3i))((z+2)^2-(4+3i))\\ &= (z+2-z_0)(z+2+z_0)(z+2-\overline {z_0})(z+2+\overline {z_0}) \end{align}

where $z_0^2 = 4-3i$.

An explanation for the last step: if $z_0$ is a root of $z^2 = w$, then $-z_0$ is a root as well, and $\pm\overline{z_0}$ are the roots of $z^2 = \overline w$. Notice how this is in agreement with the known fact that roots of a polynomial with real coefficients are either real or come in complex conjugate pairs.