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Trying to understand this derivation by Jean-Claude Arbaut, of why formula of quadratic form must hold also for complex variable with complex coefficents.


(derivation starts)

Complex resolution

Again, we start with the equation

$$ax^2+bx+c=0$$

Where $a\neq0$, and $a,b,c\in\Bbb C$.

Then, rewrite the equation

$$\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}=0$$

Now, a complex number $z$ has always a square root (two distinct ones if $z\neq0$, see for example here). We go on with a square root $\alpha$ of $b^2-4ac$, that is, $\alpha^2=b^2-4ac$, so that the equation can be written

$$\left(x+\frac{b}{2a}\right)^2-\frac{\alpha^2}{4a^2}=0$$

It's a difference of squares, and we know that $A^2-B^2=(A+B)(A-B)$, so

$$\left(x+\frac{b}{2a}+\frac{\alpha}{2a}\right)\left(x+\frac{b}{2a}-\frac{\alpha}{2a}\right)=0$$

So the roots are

$$x=-\frac{b}{2a}\pm\frac{\alpha}{2a}$$

Or if you prefer

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Important note: there is no canonical way to write that $\sqrt{t}$ is a specific square root of $t\in\Bbb C$, hence this symbol is meaningless. However, since it appears with a $\pm$, we will always work with both roots, so using the symbol here is harmless. Nevertheless, it should be avoided.

(derivation ends)


First of all to get into rewritten form

$$ax^2+bx+c=0 \Rightarrow \left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}=0$$ $$\Rightarrow \left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2}$$

But now to me there seems to be 2 options. Either this step is not true or $a^2$ means here $aa$ instead of $aa^*$. Where $a^*$ is it's complex conjugate.

As $$\frac{a}{a^2}=\frac{a}{a^*a}=\frac{a}{a^*}\frac{a^*}{a^2}=\frac{1}{a^*}\ne \frac{1}{a}$$

What is the actual justification for this step?

As now if $a^2$ would mean $aa$ following step seems not to make any sense, if we assume the $b^2$ means $bb^*$. (I've before used this formula in physics, and correct answers have required $b^2=bb^*$, so that's why I assume this.)

$$\frac{b^2}{4a^2}=\frac{bb^*}{4aa}=(\frac{b}{2a})^2$$

Thus both of the ways I can think this with, seems to fail, so I assume there's something fundamental I've understood wrong now.

And also as last one the step of opening the difference of squares would require

$$\frac{\alpha}{2a}\frac{\alpha}{2a}=\frac{\alpha ^2}{4a^2}$$

Which to my knowledge implies $a^2=aa$ (again), but which seemed unconsistent with the step before.

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  • $\begingroup$ $a^2$ always means $a\cdot a$. $a\cdot a^{\ast}$ (or more conventionally $a\cdot \overline{a}$) is $\lvert a\rvert^2$ - which only coincides with $a^2$ when $a$ is real. $\endgroup$ – Daniel Fischer Sep 6 '16 at 14:01
  • $\begingroup$ You say "Trying to understand (this) derivation" but there is nothing "behind" this "this"... $\endgroup$ – Jean Marie Sep 6 '16 at 14:07
  • $\begingroup$ @JeanMarie Whole derivation is in the post, separated by lines, could add link too though. $\endgroup$ – PyHop Sep 6 '16 at 14:28
  • $\begingroup$ @DanielFischer So the $b^2$ in the formula should actually also be treated as $bb$. Alright thanks, makes this consistent, and pushes problem elsewhere. $\endgroup$ – PyHop Sep 6 '16 at 14:36
  • $\begingroup$ All right, if nothing more can be found in the initial post... $\endgroup$ – Jean Marie Sep 6 '16 at 14:36
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Seems unnecessarily complicated. For $z\in \mathbb C$ with $z\ne 0$ there are exactly 2 solutions $y_1,y_2$ to $y^2=z,$ with $y_1+y_2=0.$ And for $a\ne 0$ we have $u^2=z/4a^2 \iff u=y/2a$ where $y$ is a solution of $y^2=z.$... Also $y^2=0\iff y=0.$

So just take the real-quadratic-formula method verbatim: For $a\ne 0,$ $$ax^2+bx+c=0\iff (x+b/2a)^2=(b^2-4ac)/4a^2\iff x\in \{-b/2a+ y/2a: y\in S\}$$ where $S=\{y: y^2=b^2-4ac\}.$

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