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

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Discussion Overview

The discussion revolves around the calculation of the expression \((e^{i\pi})^i\) and the potential errors in this calculation. Participants explore the implications of exponentiation with complex numbers, particularly focusing on the properties of logarithms and the uniqueness of complex exponentiation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that \((e^{i\pi})^i\) leads to the expression \(i^{2i} = e^{-\pi}\), questioning the validity of this result.
  • Another participant mentions the complexity of exponentiation laws with complex numbers, indicating that these laws may not apply as expected.
  • There are discussions about the notation of logarithms, with some participants asserting that \(\log\) refers to the natural logarithm, while others argue that it can also refer to base-10 logarithms.
  • One participant expresses amusement at the result of a complex exponent yielding a simple real number, highlighting the unexpected nature of the outcome.
  • Several participants emphasize the importance of the principal value of the logarithm in these calculations, suggesting that it may lead to different interpretations of the results.

Areas of Agreement / Disagreement

Participants express differing views on the properties of logarithms and exponentiation with complex numbers. There is no consensus on the correct interpretation of the logarithmic notation, and the discussion remains unresolved regarding the calculation of \((e^{i\pi})^i\).

Contextual Notes

Participants acknowledge that not all exponentiation laws apply to complex numbers, and there are discussions about the uniqueness of complex exponentiation, which may influence the interpretation of results.

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 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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