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)

A better way to solve  $ b^4+ 3b^2-4=0 $ is to write it as $(b^2 +4)(b^2     -1)=0$.
This gives $b=1,-1,2i,-2i$.  (Note, you missed $-2i$ in your write up.)
Then find the $a$ values as you did...
As for why $2+i$ wasn't in the answer key:
When you squared both sides of $ab=-2$ to get $a^2b^2=4$, you introduced new solutions. So in the end, you need to check which solutions you obtained work. This is why $2+i$ wasn't a solution to the original problem. In fact:  $(2+i)^2=3+4i$. 
In your last step, when solving for $a$ when $b=2i$, you obtain $a^2=-1$, so $a=i$ or $a=-i$ (you missed $-i$).
For $b=-2i$, you get again $a=i$ or $a=-i$.
This gives you four solutions that need checking: $-2+i, -2-i, 2+i, 2-i$. 

Your update is off...
You are correct up to the point when you obtain $b^4+3b^2-4=0$.
As I mentioned above, to solve this, write
$$
   b^4+ 3b^2-4=0 \iff (b^2 +4)(b^2     -1)=0\iff b=1,-1,2i,-2i.
$$
For each of these $b$ values, we find the corresponding $a$ value(s):
For $b=1$: $a^2=3+1^2\Rightarrow a=2,-2$.
This gives $a+bi=2+i$ 
or $a+bi=-2+i$.
 
For $b=-1$: $a^2=3+(-1)^2\Rightarrow a=2,-2$.
This gives $a+bi=2-i$ 
or $a+bi=-2-i$. 
 
For $b=2i$: $a^2=3+(2i)^2\Rightarrow a^2=-1\Rightarrow a=i,-i$.
This gives $a+bi=i+(2i)i=-2+i$ 
or $a+bi=-i+(2i)i=-2-i$. 
 
For $b=-2i$: $a^2=3+(-2i)^2\Rightarrow a^2=-1\Rightarrow a=i,-i$.
This gives $a+bi=i+(-2i)i=2+i$ 
or $a+bi=-i+(-2i)i=2-i$. 
 
So we have four possible solutions:
$$
2+i, 2-i, -2+i , -2-i
$$
Now we check which of these is a solution to the original equation $(a+bi)^2=3-4i$.
Taking the squares of the possible solutions: 
$$(−2+i)^2=3−4i, (−2−i)^2=3+4i, (2+i)^2=3+4i, (2−i)^2=3−4i.$$
So, there are two solutions: −2+i and 2−i.
