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I am trying to reproduce the steps of a paper that gives the two principal complex roots of the following equation:

$$ a_{11}z^4 + \left( 2a_{12}+a_{66} \right) z^2 + a_{22}=0 \, , \tag{1}\label{1}$$

where $a_{11},\, a_{12},\, a_{22},\, a_{66} \in \Re$. After formulating it, the author states that

it has been proven that, except for the following special cases

a) $a_{22}=0$ ,

b) $a_{22}=2a_{12}+a_{66}=0$ ,

a) $a_{11}=0$ ,

d) $a_{11}=2a_{12}+a_{66}=0$ ,

the roots of \eqref{1} are either complex, or purely imaginary numbers.

He then goes on to say that the principal roots (and I think he defines them by the ones with a positive imaginary part) of the equation have the following relationships:

$$ z_1 z_2 = -\sqrt{\frac{a_{22}}{a_{11}}}\, , \tag{2}\label{2}$$

$$ z_1^2+z_2^2=-\frac{2a_{12}+a_{66}}{a_{11}}\, , \tag{3}\label{3}$$

$$ i\left( z_1+z_2 \right) = -\sqrt{\frac{2a_{12}+a_{66}}{a_{11}}+2\sqrt{\frac{a_{22}}{a_{11}}}}\, . \tag{4}\label{4}$$

I don't know how he obtains the relationship in \eqref{4}. I think I get that he obtains the first two ones from Vieta's formula applied to the biquadratic equation, and considering that, in the rest of the paper, whenever he takes a square root of these complex solutions, he takes the negative one (for physical significance). To put this last sentence into equations, let us define $x=z^2$; then the roots of

$$ a_{11}x^2 + \left( 2a_{12}+a_{66} \right) x + a_{22}=0 \, , \tag{5}$$

satisfy that

$$ x_1+x_2=-\frac{2a_{12}+a_{66}}{a_{11}} \tag{6}\, , $$

from where \eqref{3} is obtained, and

$$ x_1 x_2 = \frac{a_{22}}{a_{11}}\, ; \tag{7}\label{7}$$

applying the definition of the complex square root stated before, \eqref{7} transforms into \eqref{2}:

$$ \left(z_1z_2 \right)^2= \frac{a_{22}}{a_{11}} \Rightarrow -z_1z_2 = \sqrt\frac{a_{22}}{a_{11}}\tag{8}\, . $$

Finally, he writes:

[t]he two principal roots, $z_1$ and $z_2$, can be solved using equations \eqref{2} to \eqref{4}. The solution is:

$$ z_1=\frac{i}{2}\left[ \sqrt{ \frac{2a_{12}+a_{66}}{a_{11}} +2\sqrt{\frac{a_{22}}{a_{11}}} } + \sqrt{ \frac{2a_{12}+a_{66}}{a_{11}} -2\sqrt{\frac{a_{22}}{a_{11}}} } \right] \, ,\tag{9} \label{9}$$ $$ z_2=\frac{i}{2}\left[ \sqrt{ \frac{2a_{12}+a_{66}}{a_{11}} +2\sqrt{\frac{a_{22}}{a_{11}}} } - \sqrt{ \frac{2a_{12}+a_{66}}{a_{11}} -2\sqrt{\frac{a_{22}}{a_{11}}} } \right] \, .\tag{10} \label{10}$$

My questions are: how to get \eqref{4}? And, then, how does he get \eqref{9}, \eqref{10} from \eqref{2} to \eqref{4}?

Excuse my complex algebra illiteracy if the answer is obvious.

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    $\begingroup$ (4) follows from $(z_1+z_2)^2 = z_1^2+2z_1z_2+z_2^2$ taking (2) and (3) into consideration. (9) and (10) require similar consideration of $(z_1-z_2)^2$ which doesn't seem to appear explicitly but then follow immediately as solutions of the $2\times2$ system of equations. $\endgroup$ – sharding4 Jun 29 '17 at 3:18
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enter image description here I have solved the above question. You can see in the image below

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