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?
 A: The way people will really solve it.
If we have a root $z_0$, then $-z_0$ is also a root. Since this is a quadratic, there can't be more than $2$ roots, so we just need to guess one root $z_0$. How do we do that?
Observe that $|z^2| = \sqrt{3^2 + 4^2} = 5$, therefore $|z|=\sqrt{5}$. Now, the only nice numbers with that absolute value are $\pm 1 \pm 2i$ and $\pm 2 \pm i$. Geometric intuition says that one of the roots must be found in the 4th quadrant. A quick check shows that indeed $z_0 = 2-i$ fits the bill:
$$
(2-i)^2 = 4 - 4i + i^2 = 3 - 4i.
$$
So the answer is $2-i$ and $-2 + i$.
This is how a lazy person like me solves a problem like this. It is perfectly legitimate to guess the answer (if you have a good argument why other answers are impossible, which in this case follows from the general knowledge that a quadratic has at most two roots). Only when guessing doesn't work right away do people start testing more powerful general approaches.
A: Hint: Write $3 - 4i$ in polar form,
$$3 - 4i = r e^{i\theta}$$
for an appropriate positive $r$ and $0 \le \theta < 2\pi$. Then use the fact that the two square roots of $re^{i\theta}$ are
$$\pm \sqrt{r} e^{i\theta/2}$$
A: Denesting $\,\sqrt{a+b\sqrt{n}}\,$ can be done by a simple rule that I discovered in my youth.
Simple Denesting Rule $\rm\ \ \, \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \  \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $  
and, furthermore,  $\rm\:w\:$ has trace $\rm\: =\: w+w' =  (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\:  2\:a$

Here $\:3-4i\:$ has norm $= 25.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = -5\ $ yields $\ \  8-4i\:$
which has $\rm\ \sqrt{trace}\: =\: \sqrt{16}\ =\ 4.\ \ \ \color{brown}{Dividing\ it\ out}\ $ yields $\ \ (8-4i)/4\, =\, 2-i$ 

Or:  $\:3-4i\:$ has norm $= 25.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ \ = \ 5\ \ $ yields $\  {-}2-4i\:$
which has $\rm\ \sqrt{trace}\: =\: \sqrt{-4} = 2i.\ \ \color{brown}{Dividing\ it\ out}\ $ yields $\ (-2-4i)/(2i)\, =\, i-2$ 

One may choose the sign of the square-root so that the arithmetic is simpler, as in the first case above. For many further worked examples see my prior posts on denesting.
A: 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 $z = -2 + i$
A: HINT:
Let $\displaystyle z=a+ib$
So, we have $\displaystyle a^2-b^2+2ab i=3-4i$
Equate the real & the imaginary parts 
Use $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2$ to find $a,b$
Observe that the signs of $a,b$ are opposite
