<s>[INST] Find all roots of $z^2=3-4i$. Find all roots of $z^2=3-4i$.
$z^2=3-4i$
$z^2+4i-3=0$
But how do I go on from here?
 [/INST] Solve: $z^2 = 3 - 4i$
Let
$$z= x + yi$$
Rewrite as
$$(x + yi)^2 = 3 - 4i$$
Expand
$$x^2 +2xyi + y^2i^2 = 3 - 4i$$
Simplify
$$x^2 - y^2 + 2xyi = 3 - 4i$$
real/imaginary parts
$$x^2 - y^2 = 3$$
and
$$2xy = -4$$
$$xy = -2$$
Therefore $y = -\frac{2}{x}$.
Substitute $y = -\frac 2x$ into $x^2 - y^2 = 3$, giving
$$x^2 - \frac{4}{x^2} = 3$$
Multiply by $x^2$:
$x^4 - 4 = 3x^2$
Factor
$$x^4 - 3x^2 - 4 = 0$$
$$(x^2 + 1) (x^2 - 4) = 0$$
$$(x^2+ 1) (x + 2) (x - 2) = 0$$
Given $xy = -2$
for $x = 2$ then $y = -1$ therefore $z = 2 - i$
for $x = -2$ then $y = 1$ therefore $z = -2 + i$
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