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Roots of Negative Numbers (Complex Analysis)

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Homework Statement


Express (-1)1/10 in exponential form

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

Homework Equations



The Attempt at a Solution


[/B]
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?
 

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  • #2
SammyS
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Homework Statement


Express (-1)1/10 in exponential form

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

Homework Equations



The Attempt at a Solution


[/B]
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?
Hello MossEE. Welcome to PF !

Your question is not clear. I don't see θ anywhere in your post.
 
  • #3
andrewkirk
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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.
 
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Hello MossEE. Welcome to PF !

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

What do you mean by 'fall out'? That's not a well-defined mathematical concept.
Haha that's my engineering background showing through. You are absolutely correct though, precision of language is important, especially in STEM.

It sounds like you mean 'replace j sin (π) by 0'. Well we can do that, but nothing is compelling us to do it.

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?
 
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SammyS
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What are you plugging in and into what is it plugged?
 
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andrewkirk
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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?
It doesn't. Your sequence is:

##(-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.
 
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It doesn't. Your sequence is:

##(-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.
Right on. Thanks for your help!
 

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