B Implications of e^i*pi = -1

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1. Aug 17, 2016

Anonymous Vegetable

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.

2. Aug 17, 2016

Bystander

Re-investigate this aspect.

3. Aug 17, 2016

Anonymous Vegetable

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

4. Aug 17, 2016

Bystander

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

5. Aug 17, 2016

Anonymous Vegetable

My humbumblest apologies and it shan't happen again.

6. Aug 17, 2016

Staff: Mentor

7. Aug 17, 2016

Bystander

De nada. You're new to the forum.

8. Aug 17, 2016

Staff: Mentor

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.

9. Aug 17, 2016

Anonymous Vegetable

10. Aug 17, 2016

Anonymous Vegetable

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

11. Aug 17, 2016

Anonymous Vegetable

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

12. Aug 17, 2016

micromass

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

13. Aug 17, 2016

Staff: Mentor

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.

14. Aug 21, 2016

Battlemage!

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. Aug 22, 2016

Isaac0427

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