Quadratic formula with complex coefficients Let $a,b$ and $c$ be complex numbers. I'm trying to prove that this version of the usual quadratic formula:
$$z=\frac{-b+(b^2-4ac)^{\frac{1}{2}}} {2a}$$
solves the quadratic equation
$$az^2+bz+c=0$$
This seemed fairly easy to do, but I came across some doubts. This is what I've done so far, please correct me if I made a mistake anywhere:
$$az^2+bz+c=0$$
$$z^2+\frac b a z+\frac c a=0$$
$$(z+\frac b {2a})^ 2=\frac {b^2} {4a^2} -\frac c a$$
Now, if we were working with real numbers, I would just take square root on both sides of the equation and that would be all. The problem is, the complex function $w^{\frac 1 2}$ is multivaluated. I don't quite know how to work around it without reducing the image set in order to work with an injective function. Please, if somebody know how to finish this proof, I would really appreciate it. 
PS: I apologize in advance for the misspelling/structure mistakes of this post. 
 A: For the reals the convention that $\sqrt5$ means the positive square root is firmly established. There is a similar convention for complex numbers. $\sqrt w$ is called the "principal square root" and is defined as the one with $-\frac{\pi}{2}<\arg w\le\frac{\pi}{2}$. So if the square root of $w$ has a real part, then $\sqrt w$ means then square root with positive real part. For a pure imaginary square root, we take the one with positive imaginary part. If you try out a few cases, that looks the most natural convention even if you are using the form $a+ib$ rather than the mod and arg.
So in your formula for the quadratic equation, it is fine, indeed preferable to carry on using the $\sqrt{b^2-4ac}$ symbol. If you think about it, the two square roots are still $\pm$ the square root. So curiously almost nothing changes. :) 
Of course, when you plug in numerical values, it is better to evaluate things and end up with $a+ib$ for some reals $a,b$ (if you can, it is not always straightforward). 
Anyone at all picky will complain that you before dividing by $a$ you need to comment that $a$ is assumed to be $\ne0$ (unless already stated in the question).
