Squaring and Square roots

[CraigD]

… this question is still bugging the hell out of me.

 

Does the equation using -ve roots have any solutions at all, real or imaginary?

Though I’ve not had a chance to work on it to my satisfaction, I believe the answer is, for the complex numbers at least, “no”.

 

Throwing out unnecessary constants, the question is: [math]\sqrt{z} \overset?= -1[/math] for some complex number z.

 

Because [math]e^{ix} = \cos x + i\sin x[/math] (Euler’s formula), this amounts to solving for x and y the equation

[math]-1 = \sqrt{z} = \sqrt{ye^{ix}} = \sqrt{y} e^{\frac{ix}2} = \sqrt{y} (\cos\frac{x}2 +i\sin\frac{x}2)[/math],

Which has a unique solution [math]y=1[/math], [math]x=2\pi[/math].

 

However, solutions with [math]x \ge 2\pi[/math] don’t uniquely map to the complex numbers – in particular, [math]x=2\pi \rightarrow z=1 \leftarrow x=0[/math].

 

So, the more general equation

[math]z^a = -1[/math] appears to have solutions only for real numbers [math]a \gt \frac12[/math].

 

I don’t feel like I intuitive understand this, however. I’m just poking my head into this thread while breaking from some high-deadline-pressure coding, so won’t be able to work on my comfort level with this for some days, yet.

[sebbysteiny]

I think that deserves a QP on it's own CraigD.

 

I think you have come to the same conclusion as me.

 

It seems the -ve root has no solutions. So if you take a linear equation with one solution, and you take the square root, you should lose a solution. But of course, you shouldn't as if one side equals another, the roots should also equal.

 

Similarly, if you square that equation back again, you will not get another solution from the possibility of the negative root because, all you have done is end up with the original equation that had only one solution.

 

So the only way those two observations can be put together is if the negative root possibility has no solutions. You can't create new solutions simply by sqare rooting both sides and then squaring them again.

 

From this, it follows that the flaw in my knowledge was the assumption that when you square an equation, you get new solutions. This is only true where the negative root contains solutions. Instead, you should conclude that new solutions MAY be created and you can only ignore this effect once you have shown that any new solutions would have to come from an equation that has no solutions.

 

Well, I don't know about you, but I think I learnt something here.

 

Good job gang.

[Qfwfq]

Does the equation using -ve roots have any solutions at all, real or imaginary?

Any equation of the type:

 

[math]\sqrt{f(x)}=-a[/math]

 

has the same solutions as:

 

[math]\sqrt{f(x)}=a[/math].

 

so it isn't that the former has no solution, it just doesn't add any more of them to the set of solutions to the squared equation.

 

More in general, all the equations:

 

[math]\sqrt[r]{f(x)}=a\,e^{\frac{2k\pi}{r}i}[/math]

 

for each integer k, all have the same solutions. I think this ought to help Craig to grasp what he has trouble in understanding intuitively.

:cup: