<s>[INST] Square Roots of Complex Number $3-4i$ 
What I did
$z^2=3-4i$
$(a+bi)^2 = 3-4i$
$a^2-b^2+2abi = 3-4i$
Then got 2 simultaneous equations
$a^2-b^2=3$ and $2ab=-4$
Solve for $a^2$ in 1st equation: $a^2=3+b^2$
Subbed into 2nd equation to power of 2
$(3+b^2)b^2=4$
$b^4+3b^2-4=0$
Is there a better way than below? Solving power 4 equation then cubic?
Then solved solved power 4 equation ... 
$(b-1)(Ab^2+Bb^2+Cb+D)=b^4+3b^2-4$
$(b-1)(b^3+b^2+4b+4)=b^4+3b^2-4$
then  solved for cubic equation, getting ...
$(b-1)(b+1)(b^2+4)=0$
So
$b=\pm 1 \text{ or } 2i$
What did the book do now? 
I did: (subbing into $a^2=3+b^2$)
When $b=1, a^2=3+1^2 \Longrightarrow a = 2$
When $b=-1, a^2=3+(-1)^2 \Longrightarrow a=2$
When $b=2i, a^2=3+(2i)^2=3+4(-1)=-1, a=i$
So I will have 3 equations


*

*$2+i$ // why is this not in the book?

*$2-i$

*$i-2$


UPDATE (In response to @David Mitra)
Ok. I let $x=a^2$ and $b=b^2$ giving the quadratic equation: $x=3+y$ and $xy=4$. Then (after subbing) $(y-1)(y+4)=0$
Then when $y=1, x=4$. $y=-4, x=-1$
Then $b=\pm 1 or \pm 2i$ and $a=\pm 2i or \pm i$
Finally testing equations: 
$(2+i)^2 = 3+4i$ (rej)
$(2-i)^2=5$ what to do?
$(i-2)^2=3$
$(-2-i)^2=3$
... 8 equations (but 4 unique)
 [/INST] The high school way:
If $(x+iy)^2=3-4i$, expanding the l.h.s.  and identifying the real and imaginary parts, you obtain the system of equations
$\;\begin{cases}
x^2-y^2=3,\\2xy=-4.
\end{cases}$
Now, you can also write that $|x+iy|^2=|3-4i|$, i.e. $\;x^2+y^2=5$. Therefore you obtain the linear system in $x^2$ and $y^2$:
$$\begin{cases}
x^2-y^2=3,\\x^2+y^2=5
\end{cases}\iff \begin{cases}
x^2=4,\\y^2=1 
\end{cases}$$
Thus, $x=\pm 2,\;y=\pm 1$. Furthermore, the unused equation $xy=-2$ says that $x$ and $y$ have opposite signs. Consequently the solutions are
$$x+iy=\pm(2-i).$$
</s>