Euler's identity

Euler's identity: eiπ=−1e^{i\pi} =-1

A different approach to derive this equation without using series

∫0x11+x2dx=arctan⁡x\displaystyle \int_{0} ^{x} \frac{1}{1+x^{2}} dx =\arctan x

11+x2=12(11+xi+11−xi)\displaystyle \frac{1}{1+x^{2}} =\frac{1}{2} \left(\frac{1}{1+xi} +\frac{1}{1-xi}\right)

∫0x12(11+xi+11−xi)dx=12i(ln⁡(1+xi)−ln⁡(1−xi))\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)

12i(ln⁡(1+xi)−ln⁡(1−xi))=arctan⁡x\frac{1}{2i} \left(\ln\left(1+xi\right)-\ln\left(1-xi\right)\right) =\arctan x

We know arctan⁡1=π4\arctan 1 =\frac{\pi}{4}

12i(ln⁡(1+i)−ln⁡(1−i))=π4\frac{1}{2i} \left(\ln\left(1+i\right)-\ln\left(1-i\right)\right) =\frac{\pi}{4}

(ln⁡(1+i)−ln⁡(1−i))=π×i2\left(\ln\left(1+i\right)-\ln\left(1-i\right)\right) =\frac{\pi\times i}{2}

ln⁡(1+i1−i)=π×i2\displaystyle \ln\left(\frac{1+i}{1-i}\right) =\frac{\pi\times i}{2}

ln⁡((1+i)22)=π×i2\displaystyle\ln\left(\frac{\left(1+i\right)^{2}}{2}\right) =\frac{\pi\times i}{2}

ln⁡(1+2i−12)=π×i2\displaystyle\ln\left(\frac{1+2i -1}{2}\right) =\frac{\pi\times i}{2}

ln⁡(i)=π×i2\displaystyle\ln\left(i\right) =\frac{\pi\times i}{2}

exp⁡ln⁡(i)=exp⁡iπ2\displaystyle \exp^{\ln\left( i \right)} =\exp^{\frac{i \pi}{2}}

i=eiπ2\displaystyle i=e^{\frac{i \pi}{2}}

square both sides to obtain −1=eiπ-1 =e^{i\pi}

#Calculus

Note by Amal Hari
1 year, 6 months ago

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