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The answer is $ \pm 2i \ $ and -$1 + - i$

Well another problem I am struggling with, not even sure how to start on this on, so can't even give out information on what I am trying to do.

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closed as off-topic by Shaun, Nosrati, Travis, Did, Theoretical Economist Sep 3 '18 at 0:02

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  • $\begingroup$ J.doe, was one of roots given? $\endgroup$ – user376343 Sep 2 '18 at 14:24
  • $\begingroup$ @j.doe Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details HERE $\endgroup$ – gimusi Oct 23 '18 at 20:50
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Recall that since the coefficients are reals we have that $(z^2+4)$ divides $z⁴+2z³+6z²+8z+8=0$.

Therefore performing long division we obtain

$$z⁴+2z³+6z²+8z+8=(z^2+4)(z^2+2z+2)$$

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  • $\begingroup$ Right.. I really should have known this right away. thanks $\endgroup$ – j.doe Sep 2 '18 at 14:33
  • $\begingroup$ @j.doe You are welcome! Bye $\endgroup$ – gimusi Sep 2 '18 at 14:51
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The equation in question has real coefficients. If $a+bi$ is a zero, so is $a-bi$, hence not only $-2i$ is a solution, but also $2i$. To find the other zeroes, divide the polynomial by $(z-2i)(z+2i)$ and solve the resulting quadratic.

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Alt. hint: $\, + 2i\,$ must also be a root, and let $\,a,b\,$ be the other two roots. By Vieta's relations $\,-2i +2i + a + b=-2 \implies a+b=-2\,$ and $\,(-2i)(2i) a b = 8 \implies ab=2\,$. It follows that $\,a,b\,$ are the roots of the quadratic $\,z^2+2z+2=0\,$

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