# 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?

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What does "into complex conjugate" mean? – Henning Makholm Nov 4 '11 at 13:34
Subtract 3 from both sides, take square root. Then, figure out the two square roots of $\sqrt{-3 + 4i}$ and the two of $\sqrt{-3 - 4i}$. – Graphth Nov 4 '11 at 13:34
This is a good start. Now you get 2 quadratic equations (one with plus, the other with minus), solve each one of them. – Joel Cohen Nov 4 '11 at 13:35
+1 for showing your work. I think it got a more useful answer than just posting the problem. – Ross Millikan Nov 4 '11 at 17:50

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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I finally solved it=) Thank you – David Nov 4 '11 at 15:42
@David I fixed a typo: One needs $\sqrt{5}$, not just $5$, as the coefficient at the beginning. – Michael Hardy Nov 4 '11 at 17:45

The aim is to find a good representation of the square roots of $-3 \pm 4i$. But these are Gaussian integers, members of a unique factorization domain, so you should try to factor them into primes. Voilà: $-3+4i = (1+2i)^2$ and $-3-4i=(1-2i)^2$. So the solutions of your equation are $\pm1\pm2i$.

The method? The Gaussian primes are $1+i$ (dividing $2$), the ordinary primes congruent to $3$ modulo $4$, and for each prime $p\equiv 1\bmod 4$, a pair of primes $\pi_1$ and $\pi_2 = \overline{\pi_1}$ with $\pi_1\pi_2=p$. This splitting of $p$ arises out of the theorem that such $p$ is the sum of two squares. In particular, $5=(1+2i)(1-2i)=1+4$. So you look at $-3+4i$ and notice that $(-3+4i)(-3-4i)=25=(1+2i)^2(1-2i)^2$ and use Unique Factorization, but judiciously, 'cause you have to take into account the fact that there are four units, $\pm1$ and $\pm i$.

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