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

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The discussion centers on the calculation of the expression (e^(iπ))^i, leading to the conclusion that i^(2i) equals e^(-π). Participants highlight that exponentiation laws do not universally apply to complex numbers, making the results non-unique. The principal value of the logarithm is crucial in this context, as it clarifies the relationship between complex exponentiation and real numbers. Misunderstandings about logarithmic notation, specifically the use of "log" versus "ln," also arise, indicating a need for clarity in mathematical communication.

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

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.
 
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Anonymous Vegetable said:
the negative number to the power of i however I'm not sure how to sort it
Re-investigate this aspect.
 
Bystander said:
Re-investigate this aspect.
I've edited it to make another point anyway hahaha but yeah I shall
 
Point of order: please do NOT make changes to your original post. It makes very confusing reading for late arriving participants.
 
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Bystander said:
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.
 
Anonymous Vegetable said:
it shan't happen again.
De nada. You're new to the forum.
 
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.
 
  • #10
I just find it amusing that what appears to be an extremely non real value seems to equal a simple real number
 
  • #11
mfb said:
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
 
  • #12
Anonymous Vegetable said:
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.
 
  • #13
micromass said:
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.
 
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  • #14
Mark44 said:
micromass said:
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?
 
  • #15
Anonymous Vegetable said:
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.
 

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