Find all solutions of the equation $z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$

Question

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.

Answer 1

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: $$z=1-2i$$ $$z = 1+2i$$ $$z = -i\sqrt2$$

Answer 2

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.

Answer 3

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.