47 ms·
I think Euler’s identity becomes more elegant when using 2π instead of π: e^(2πi) = 1
by niklasbuschmann 7y ago
I think Euler’s identity becomes more elegant when using 2π instead of π:
e^(2πi) = 1
- epage 7y agoThe form I was taught: e^jpi + 1 = 0 The reason for this form is that it doesn't just contain an imaginary unit and the two most popular constants but the additive and multiplicative identities.