<s>[INST] How to solve $z^4 +6z^2 +25 = 0$ into complex conjugate? How to solve $z^4 +6z^2 +25 =  0$ into complex conjugate?
I started with
$$(z^2 + 3)^2 + 16 = 0$$
$$(z^2 + 3)^2  = - 16$$
$$z^2 + 3 = \pm 4i$$
Is this the way to start solving the equation or am I completely of?
 [/INST] You have $z^2=-3\pm4i$.  You need the two square roots of $-3+4i$ and the two square roots of $-3-4i$.  I can think of two ways to do this.  One is polar coordinates:
$$
-3+4i = \sqrt{3^2+4^2}\left(\cos\theta+i\sin\theta\right) = 5\left(\cos\theta+i\sin\theta\right)
$$
where $\theta$ is an angle between $\pi/2$ and $\pi$ whose tangent is $-4/3$.  Since $\arctan(-4/3)=-\arctan(4/3)$ is between $-\pi/2$ and $0$, we add $\pi$ to it and get $\theta=\pi-\arctan(4/3)$.
Then the square roots of $-3+4i$ are
$$
\pm\sqrt{5}\left(\cos\left(\frac\theta 2\right)+i\sin\left(\frac\theta 2\right)\right).
$$
And a similar thing works for the square roots of $-3-4i$, and in fact you get
$$
\pm\sqrt{5}\left(\cos\left(\frac\theta 2\right)-i\sin\left(\frac\theta 2\right)\right)
$$
where $\theta$ is the same number as above.
The other method is this.  Say $a+bi$ is a square root of $-3+4i$, where $a$ and $b$ are real.  Then $(a+bi)^2=-3+4i$.  So $a^2-b^2 + 2abi = -3+4i$.  Then
$$
\begin{align}
a^2 - b^2 & = -3, \\
2ab & = 4.
\end{align}
$$
If you write $b=2/a$ and substitute that for $b$ in the first equation, you have and equation in $a$ that can be solved.
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