An algebra problem by Achal Jain

-1/I = -I/(I^2) = -I/-1 = +I, so e^(-pi/I) + 1 = e^(pi*I) + 1 = 0 Ed Gray

e^iπ+1=0 ,

e^iπ×i÷i + 1=0 ,

e^-π÷1 + 1=0

The Euler's formula says that $e^{i \theta}=\cos\theta + i \sin \theta$ which means that in The Euler's formula says that in $e^{\frac{-\pi}{i}}$ we have an angle $\theta$ to find. Multiplying the numerator and the denominator of the fraction in the exponent by $i$ we get $e^{\frac{- i \pi}{i^2}}=e^{i\frac{-\pi}{i^2}}=e^{\frac{-\pi}{-1}}=e^{i \pi }$ . Then, $e^{\frac{-\pi}{i}}+1=0$