# Roots of Negative Numbers (Complex Analysis)

1. Jul 13, 2015

### MossEE

1. The problem statement, all variables and given/known data
Express (-1)1/10 in exponential form

(My first time posting - I hope I got the syntax right!)

2. Relevant equations

3. The attempt at a solution

I got the solution, it's ejπ/10, but I'm not sure why. Here's my work:

(-1)1/10 = (cos(π) + jsin(π))1/10 = cos(pi/10) + jsin(pi/10) = ejπ/10,

I'm wondering why, when moving from step 2 to 3 using De Moivre's Theorem, the jsin(pi) term doesn't drop out. The above math checks out if we would refer to pi as the generic angle 'theta', but shouldn't the complex part fall out once we know that theta is just pi?

2. Jul 13, 2015

### SammyS

Staff Emeritus
Hello MossEE. Welcome to PF !

Your question is not clear. I don't see θ anywhere in your post.

3. Jul 13, 2015

### andrewkirk

What do you mean by 'fall out'? That's not a well-defined mathematical concept.

It sounds like you mean 'replace j sin (π) by 0'. Well we can do that, but nothing is compelling us to do it. Since doing that prevents us from taking the next step, we don't do it. We keep it in the form (cos(π) + jsin(π))1/10, and De Moivre then allows us to state that that is equal to cos(pi/10) + jsin(pi/10).

Maybe a lesson to learn from this is that simplifying is not always the best option. In this case simplifying, by substituting in the zero, prevents us from getting to the solution.

Another example of unsimplifying creating an opportunity for a solution is when we add and subtract the same thing to a formula - thereby making it longer - because it will allow subsequent cancellations that are more helpful. John Bell does this a number of times in his famous entanglement paper.

4. Jul 13, 2015

### MossEE

Thanks for the warm welcome. I was referring to theta in the context of not simplifying the equation (by plugging in pi for theta), sorry for the ambiguity.

Haha that's my engineering background showing through. You are absolutely correct though, precision of language is important, especially in STEM.

This makes sense. I guess I'm only left with a tangential question then - I can understand why simplifying early (by plugging in) won't give us a complete solution, but why does it produce a solution that is mathematically incorrect?

5. Jul 13, 2015

### SammyS

Staff Emeritus
What are you plugging in and into what is it plugged?

6. Jul 13, 2015

### andrewkirk

$(-1)^{1/10} = (\cos(\pi) + j \sin(\pi))^{1/10} = \cos(\pi/10) + j \sin(\pi/10) = e^{j\pi/10}$

If you substitute 0 for $j \sin(\pi)$ in the second stage you are left with $(\cos(\pi) + 0)^{1/10}= (\cos\pi) ^{1/10}$, which no longer has an imaginary part and hence you cannot apply de Moivre's theorem.

I think you're thinking that you'll apply de Moivre and get $\cos(\pi/10)$, which would indeed be wrong. But that would be misusing de Moivre because it only applies when you have appropriately related real and imaginary parts and the simplified expression doesn't.

7. Jul 13, 2015

### MossEE

Right on. Thanks for your help!