B What is the Error in Calculating (e^(iπ))^i?

[Anonymous Vegetable]

Before I start, there are only really two pieces of information this concerns and that is the idea that 1^x = 1 and that e^i*π = -1

So it would follow that (e^i*π)ⁱ = -1ⁱ
And so that would mean that i²ⁱ = e^-π which doesn't seem to be right at all. Where is the issue here as there must be one but I am sure I don't have the knowledge required to figure it out.

 

[Bystander]

the negative number to the power of i however I'm not sure how to sort it

Re-investigate this aspect.

 

[Anonymous Vegetable]

Re-investigate this aspect.

I've edited it to make another point anyway hahaha but yeah I shall

 

[Bystander]

Point of order: please do NOT make changes to your original post. It makes very confusing reading for late arriving participants.

 

[Anonymous Vegetable]

Point of order: please do NOT make changes to your original post. It makes very confusing reading for late arriving participants.

My humbumblest apologies and it shan't happen again.

 

[fresh_42]

Maybe this could help you:
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

 

[Bystander]

it shan't happen again.

De nada. You're new to the forum.

 

[mfb]

Not all exponentiation laws work with complex numbers, and with a complex base those exponents are not unique any more.

$$i^{2i} = e^{2i \log(i)} = e^{2i (i \pi/2)} = e^{- \pi}$$ using the principal value of the logarithm, indeed.

 

[Anonymous Vegetable]

Maybe this could help you:
https://www.physicsforums.com/insights/things-can-go-wrong-complex-numbers/

Well I can't say it clears things up all that much hahaha, but it identified the problem with the real a and b exponent part so thank you.

 

[Anonymous Vegetable]

I just find it amusing that what appears to be an extremely non real value seems to equal a simple real number

 

[Anonymous Vegetable]

Not all exponentiation laws work with complex numbers, and with a complex base those exponents are not unique any more.

$$i^{2i} = e^{2i \log(i)} = e^{2i (i \pi/2)} = e^{- \pi}$$ using the principal value of the logarithm, indeed.

I assume your log refers to ln? Sorry just being picky

 

[micromass]

I assume your log refers to ln? Sorry just being picky

Outside of high school, logarithms with base $e$ are always denoted as $\log$. The notation ln is not really used anymore.

 

[Mark44]

Outside of high school, logarithms with base $e$ are always denoted as $\log$. The notation ln is not really used anymore.

I don't believe that's true. Every calculus textbook I have distinguishes between log (meaning base-10 logarithm) and ln. Granted, all of my textbooks are at least 15 to 20 years, and some are older.

 

[Battlemage!]

Outside of high school, logarithms with base $e$ are always denoted as $\log$. The notation ln is not really used anymore.

I don't believe that's true. Every calculus textbook I have distinguishes between log (meaning base-10 logarithm) and ln. Granted, all of my textbooks are at least 15 to 20 years, and some are older.

I know it adds little to the conversation, but I have to concur. Pretty much all my textbooks use ln. Maybe it's an undergrad thing?

 

[Isaac0427]

So it would follow that (ei*π)i = -1i
And so that would mean that i2i = e-π which doesn't seem to be right at all. Where is the issue here as there must be one but I am sure I don't have the knowledge required to figure it out.

What seems to be the problem? I don't see one.