What's wrong in this manipulation of a square root of a negative number?

[zenterix]

Homework Statement

I am working on differential equations and in my calculations I ran into the sequence of expressions

$$\sqrt{1+c}=\sqrt{-|1+c|}=i\sqrt{|1+c|}=i\sqrt{-(1+c)}=i^2\sqrt{1+c}=-\sqrt{1+c}$$

where $c<-1$.

Relevant Equations

Is the sequence of expression above correct? It does not seem so.

Now, it is very likely that I am making a super silly mistake here. But I can't see it.

For context, I have a 2x2 system of linear first order differential equations

$$\vec{x}'=A\vec{x}$$

where

$$A=\begin{bmatrix} 0 & 1\\c&-2\end{bmatrix}$$

and the characteristic polynomial is

$$\lambda^2+2\lambda-c=0$$

The two eigenvalues are $-1\pm\sqrt{1+c}$.

Suppose $c<-1$. Then $1+c<0$.

So I was just in the middle of writing the eigenvalues given these values of $c$.

 

[Hill]

$$\sqrt{1+c}=\sqrt{-|1+c|}=i\sqrt{|1+c|}=i\sqrt{-(1+c)}=i^2\sqrt{1+c}=-\sqrt{1+c}$$

It can be simplified, e.g.:
$\sqrt 3=\sqrt{-(-3)}=i\sqrt{-3}=i^2\sqrt 3=-\sqrt 3$

 

[zenterix]

It can be simplified, e.g.:
$\sqrt 3=\sqrt{-(-3)}=i\sqrt{-3}=i^2\sqrt 3=-\sqrt 3$

How can it be correct that $\sqrt{3}=-\sqrt{3}$?

 

[Hill]

How can it be correct that $\sqrt{3}=-\sqrt{3}$?

It is not. This example should make it easier to see what is wrong.

 

[zenterix]

The issue seems to be with how we treat $i$ after the second and third equal signs in your expressions.

$\sqrt{-(-3)}=i\sqrt{-3}$ seems correct to me.

It also seems correct to do the next step where we write $\sqrt{-3}=i\sqrt{3}$.

But then we have $i\cdot i$.

Is this the same thing as $i^2$? I think it is.

I can't see the error.

We could also write

$$\sqrt{3}=\sqrt{-1}\sqrt{-1}\sqrt{3}$$

We can either then write the rhs as $i^2\sqrt{3}=-\sqrt{3}$ which gives us an incorrect result, or $\sqrt{(-1)^2}\sqrt{3}=\sqrt{3}$ which is the correct result but I don't know how to justify why this is the way to go.

 

[docnet]

Hint: For every complex number $z$, there exist two numbers $w$ such that $w^2=z$.

 

[zenterix]

Hint: For every complex number z, there exist two numbers w such that w2=z.

Given a $z\in\mathbb{C}$, $z=re^{i(\theta+2\pi k)}$, the square root is a complex $w$ such that

$$w^2=z=re^{i(\theta+2\pi k)}$$

$$w=\sqrt{r}e^{i\left (\frac{\theta}{2}+\pi k\right )},\ \ \ \ \ k=0,1$$

In the case of the number $-1$, the square root is

$$\sqrt{-1}=\sqrt{1}e^{i\left (\frac{\pi}{2}+\pi k\right )},\ \ \ \ \ k=0,1$$

So the two roots are

$$e^{i\frac{\pi}{2}}=i$$

$$e^{i\frac{3\pi}{2}}=-i$$

But how does this fit into the sequence of expressions we're looking at. I think I know the rules of how to manipulate complex numbers. However, I must be missing some rule because in the sequence of expressions I don't see a mistake. Do you claim there is an actual error in the manipulations shown in the OP or in the example with $\sqrt{3}$?

 

[Orodruin]

Assuming that by $\sqrt{\ }$ we refer to the principal branch, you cannot just break out a $-1$ of the square root to obtain an $i$. You need to be more careful than that. It will not be generally true that $\sqrt{ab} = \sqrt{a}\sqrt{b}$.

 

[docnet]

Let $a,b>0$, then it's not true that $\sqrt{-a} \sqrt{-b} = \sqrt {(-a)(-b)}$.

Given a complex number $z$ there are two solutions to the equation$w^2=z$. There is no consistent way to pick one of these roots and call it $\sqrt z$ unless it's real and positive.

 

[Orodruin]

Let $a,b>0$, then it's not true that $\sqrt a \sqrt b = \sqrt {(-a)(-b)}$.

Yes it is because $(-a)(-b)$ is still positive. What is not generally true is the inference that it equals $\sqrt{-a}\sqrt{-b}$ keeping to the principal branch.

 

[docnet]

Yes it is because $(-a)(-b)$ is still positive. What is not generally true is the inference that it equals $\sqrt{-a}\sqrt{-b}$ keeping to the principal branch.

Sorry, it was a typo. I just fixed it. Hopefully it makes more sense now.

 

[Hill]

It will not be generally true that $\sqrt{ab} = \sqrt{a}\sqrt{b}$.

then it's not true that $\sqrt{-a} \sqrt{-b} = \sqrt {(-a)(-b)}$.

To add to the above,

"Even in 1770 the situation was still sufficiently confused that it was possible for so great a mathematician as Euler to mistakenly argue that $\sqrt{-2} \sqrt{-3} = \sqrt 6$."
(Needham, Visual Complex Analysis, p. 2)

 

[Orodruin]

To add to the above,

I mean, I am pretty sure Newton never actually passed a Calculus 101 exam …

Edit: Now I was reminded of one of my favorite math jokes …

Why didn’t Newton invent group theory?
He wasn’t Abel.

 

[zenterix]

@Orodruin What is the "principal branch"?

 

[Orodruin]

@Orodruin What is the "principal branch"?

https://en.m.wikipedia.org/wiki/Principal_branch

 

[PeroK]

Homework Statement: I am working on differential equations and in my calculations I ran into the sequence of expressions

$$\sqrt{1+c}=\sqrt{-|1+c|}=i\sqrt{|1+c|}=i\sqrt{-(1+c)}=i^2\sqrt{1+c}=-\sqrt{1+c}$$

Your error boils down to assuming that for positive $a, b$, we have:
$$\sqrt{ab} = \sqrt{(-a)(-b)} = \sqrt{-a}\sqrt{-b} = (i\sqrt a)(i\sqrt b) = - \sqrt{ab}$$However, for positive $a, b$, the following is not true:
$$\sqrt{(-a)(-b)} \ne \sqrt{-a}\sqrt{-b}$$Take $a = b = 1$ for a simple counterexample.

 

[WWGD]

@Orodruin What is the "principal branch"?

The one in which points on the x-axis have argument equal to 0.

 

[Orodruin]

The one in which points on the x-axis have argument equal to 0.

If you want to be specific that’s not enough to fully specify the branch though. You would also need to specify where the branch cut is (the negative real axis).

 

[WWGD]

If you want to be specific that’s not enough to fully specify the branch though. You would also need to specify where the branch cut is (the negative real axis).

Well, you need $\mathbb \pi$ worth of angles.

 

[Orodruin]

Well, you need $\mathbb \pi$ worth of angles.

The principal branch of the square root is defined as the function which takes the positive square root of the magnitude of the argument and divides its argument (in the branch $(-\pi,\pi]$) by two. This is $2\pi$ worth of argument in the domain, which obviously becomes $\pi$ in the codomain.