# Euler's identity

Euler's identity: $e^{i\pi} =-1$

A different approach to derive this equation without using series

$\displaystyle \int_{0} ^{x} \frac{1}{1+x^{2}} dx =\arctan x$

$\displaystyle \frac{1}{1+x^{2}} =\frac{1}{2} \left(\frac{1}{1+xi} +\frac{1}{1-xi}\right)$

$\displaystyle \int_{0}^{x} \frac{1}{2} \left(\frac{1}{1+xi} +\frac{1}{1-xi}\right) dx = \frac{1}{2i} \left(\ln\left(1+xi\right)-\ln\left(1-xi\right)\right)$

$\frac{1}{2i} \left(\ln\left(1+xi\right)-\ln\left(1-xi\right)\right) =\arctan x$

We know $\arctan 1 =\frac{\pi}{4}$

$\frac{1}{2i} \left(\ln\left(1+i\right)-\ln\left(1-i\right)\right) =\frac{\pi}{4}$

$\left(\ln\left(1+i\right)-\ln\left(1-i\right)\right) =\frac{\pi\times i}{2}$

$\displaystyle \ln\left(\frac{1+i}{1-i}\right) =\frac{\pi\times i}{2}$

$\displaystyle\ln\left(\frac{\left(1+i\right)^{2}}{2}\right) =\frac{\pi\times i}{2}$

$\displaystyle\ln\left(\frac{1+2i -1}{2}\right) =\frac{\pi\times i}{2}$

$\displaystyle\ln\left(i\right) =\frac{\pi\times i}{2}$

$\displaystyle \exp^{\ln\left( i \right)} =\exp^{\frac{i \pi}{2}}$

$\displaystyle i=e^{\frac{i \pi}{2}}$

square both sides to obtain $-1 =e^{i\pi}$ Note by Amal Hari
1 week, 1 day ago

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