<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 value of $a$ is calculated incorrectly when $b=1$.  From $2ab=-4$ there is a unique value of $a$ for every value of $b$. When $b=1$, $a$ can only have the value $-2$ (not $+2$ as stated in the question), and then $z = i - 2$.
The quartic equation for $b$, although derived by squaring, is correct and is not responsible for introducing any extra solutions.  It could have been derived without squaring, using $ab=−2$.  The error is in the step "subbing into $a^2=3+b^2$ ". That quadratic equation derives two values of $a$ from each value of $b$, when we know from $2ab=−4$ that there is only one value of $a$ for each value of $b$.  If $2ab=-4$ had been used for the determination of $a$ from $b$ no extra solutions would appear, and every value of $b$ would determine a correct value of $a$.

What did the book do now?

The book does not say, but I guess it meant: eliminate one of the variables using $xy = -2$, obtaining a quartic (degree 4) equation in the other variable that is a quadratic equation in $x^2$ (or $y^2$, whichever is the surviving variable).  The quartic has two real and two imaginary roots.  Where the book says there are two solutions, that means solutions with real values of $x$ and $y$.  These can be found by taking the two real solutions of the quartic and solving for the other variable through $xy=-2$; if one variable is real, the value found for the other variable will also be real.
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