Find all solutions of the equation $z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$ Find all solutions of the equation:
$z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$
I don't have any  other idea than to guess them.
 A: HINT:
You can write the same equation as:
$$(z^2-2z+5)(z^2+2i\sqrt2 z-2)=0$$
Then you can find the roots of each brackets separately...
SOLUTION:
$$z=1-2i$$
$$z = 1+2i$$
$$z = -i\sqrt2$$
A: If you are interested in a systematic approach to factorising this polynomial, start by assuming that it can be factorised into two quadratic polynomials such as $$(z^2+Az+B)(z^2+Cz+D)$$
Thus the given polynomial equation is equivalent to $$z^4+z^3(A+C)+z^2(B+AC+D)+z(AD+BC)+BD=0$$
We can further speculate that $B$ and $D$ might be integers whose product is $-10$ so we can choose either $(B,D)=(5,-2)\text{or}(-5,2)$. Note that I am ignoring other possibilities for the time being.
So this gives you two sets of calculations (at most) to attempt. 
If you choose the pair $(-5,2)$ you will find the resulting equations in $A$ and $C$ are inconsistent, whereas the first choice works.
I will leave the working to you, but actually it's quite quick.
A: The peculiar form of the polynomial, with the same $\,\sqrt{2}\,$ present in the imaginary parts of the coefficients, only, suggests splitting it into two terms for closer inspection:
$$
\begin{align}
P(z) &= z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10 \\
&= (z^4-2z^3+3z^2+4z-10) + i \,\sqrt{2}\,(2z^3 - 4 z^2+10z) \\
&= (z^4-2z^3+3z^2+4z-10) + i \,2\sqrt{2}\,z(z^2 - 2 z+5) \\
\end{align}
$$
The quadratic factor $\,z^2 - 2 z+5\,$ in the second term further suggests checking whether the first term maybe has it as a factor as well, and it turns out that it does, indeed:
$$
z^4-2z^3+3z^2+4z-10 = (z^2-2)(z^2 - 2 z+5)
$$
From here on, the polynomial factors, and the rest follows by routine calculations:
$$
P(z) = (z^2 + i\,2\,\sqrt{2}z- 2)(z^2-2z+5)
$$
Disclaimer: the shortcut above only works because the polynomial was specifically set up the way it was. Such "luck" virtually never happens in "real life" outside homework/contest problems.
