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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.

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  • $\begingroup$ I think it would help if you make yourself a drawing of how the squareroots of a complex number look like (what position they are in relation to each other). If this is clear for you, you will not have any problem showing this=) $\endgroup$ – flawr Sep 13 '14 at 22:24
  • $\begingroup$ @flawr That works for me when I consider a, b and c real. It may sound silly, but my problem is mainly with this step: $((z+\frac b {2a})^ 2)^{\frac 1 2}=z+\frac b {2a}$ $\endgroup$ – Lessa121 Sep 13 '14 at 22:30
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    $\begingroup$ @LunaSage I just wanted to tell you: If $z$ is complex, you still have two possibilities $x,y$ so that $z = x^2$ and $z=y^2$. In the end you will see (if you make a drawing) that $x = -y$ so they behave the same as the square roots of real numbers. $\endgroup$ – flawr Sep 13 '14 at 22:35
  • $\begingroup$ I think I got it. Thank you so much! $\endgroup$ – Lessa121 Sep 13 '14 at 22:41
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    $\begingroup$ $w^\frac{1}{2}$ is multi-valued, and the values differ from each other by a minus sign. That's exactly the $\pm$ that appears in the quadratic formula. $\endgroup$ – Zavosh Sep 14 '14 at 0:41
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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).

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