I got as far as taking the square root of both sides, and I'm ashamed to say that I'm already stuck. Any pointers?
In regards to the comment, I got as far as $q^4+3q^2-4=0$ by equating the parts that had $i$ and the parts that didn't, aka $p^2-q^2=3$ and $2pqi=-4i.$
If you know about De Moivre's theorem then the easiest solution would be to put the right-hand side into polar coordinates, take the square root using the theorem and then put it back into the form $a + bi$.
A link to the Wikipedia page that describes it, look under "roots of complex numbers"
Post-expansion of the left hand side, you have a system of two nonlinear equations with two unknowns $$\begin{cases} p^2-q^2=3\\[1ex] 2pq=-4 \end{cases}$$ which can be solved via substitution of variables. Rewrite the first second equation as, say, $p=-\frac{2}{q}$ (with $q\neq0$). Then in the first equation, you have $$\left(-\frac{2}{q}\right)^2-q^2=\frac{4}{q^2}-q^2=3\implies q^4+3q^2-4=0$$ which is quadratic in $q^2$. That is, you can set $r=q^2$ and rewrite your equation as $$r^2+3r-4=0,$$ find the solution for $r$, undo the substitution, solve for $q$, back-substitute into the first equation of the system, and finally find $p$.
We can take these equations you have derived and solve them for $p$ and $q$. Starting from scratch,
$(p+qi)^2 = p^2 + 2pqi + (qi)^2 = p^2 - q^2 + 2pqi$
Hence we have $p^2 - q^2 = 3$ and $2pq = -4$. Setting $q = \frac{-2}{p}$ and plugging this into the first equation, $p^2 - \left( \frac{-2}{p} \right)^2 = 3$, or $p^4 - 3p^2 - 4 = 0$.
This can be factored to give $(p^2 - 4)(p^2 + 1) = 0$. There are no real solutions to $p^2 = -1$, so we must have $p^2 = 4$, or $p = \pm 2$. Then $q = -2/p = \mp 1$.
Hence $p + qi \in \{ 2 - i, -2 + i \}$.
This can be computed by a very simple $ $ Square Root Denesting Rule:
Simple Denesting Rule $\rm\ \ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $
$\ 3-4 i\ $ has norm $= 25.\:$ $\rm\ \color{blue}{subtracting\ out}\,\ \sqrt{norm}\ = -5\,\ $ yields $\,\ 8-4i\:$
with $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{16}\, =\, 4.\ \ \ \rm \color{brown}{Dividing\ it\ out}\ $ of the above yields $\,\ 2- i$
Remark $\ $ Many more worked examples are in prior posts on this denesting rule.
Taking squared absolute values on both sides of $(p+qi)^2=3-4i$, we get $p^2+q^2=|p+qi|^2 = |(p+qi)^2| = 5$. So $p=\pm 1$ and $q=\pm 2$ may yield a solution. By inspection, $1-2i$ is a solution.
An interesting way to think about this problem:
If you square the imaginary number $re^{i\theta}$, you get the imaginary number $r^2e^{i2\theta}$. Using the tangent half-angle formula, we see that $\tan\left(\frac{\tan^{-1}\left(-\frac{4}{3}\right)}{2}\right)=-\frac{1}{2}$. So, $p=-2q$. Since $|3-4i|=5$, $|p+qi|=\sqrt{5}$. We conclude that the only two possible solutions are $2-i$ and $-2+i$.
