Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Exponent problems with i and Euler's formula?
e^(2πi) = 1, so (e^(πi))² = 1. When x²=1, x=±1. Shouldn't e^(πi)=±1? What prevents e^(πi) from equaling 1?
Also,
e^(2πi) = 1, so one of the 2πi roots of 1 must be e. But e is not on the unit circle in a complex plane. How is this possible?
1 Answer
- Anonymous8 years agoFavorite Answer
For your first question, it is true that e^{πi} = -1. The thing to remember is that for a complex number z = x+iy, we have e^z = e^x(cos(y) + i*sin(y)). So e^{πi} = (e^0)(cos(π) + i*sin(π)) = 1*(-1 + i*0) = -1.
For the second question, you're again right that e is one of the 2πi-th roots of 1, but it's not true that a complex root of 1 should have modulus 1, as you've just observed.
Source(s): Mathematics student.