Euler's identity

Euler's identity: eiπ=1e^{i\pi} =-1

A different approach to derive this equation without using series

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

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

0x12(11+xi+11xi)dx=12i(ln(1+xi)ln(1xi))\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(1xi))=arctanx\frac{1}{2i} \left(\ln\left(1+xi\right)-\ln\left(1-xi\right)\right) =\arctan x

We know arctan1=π4\arctan 1 =\frac{\pi}{4}

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

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

ln(1+i1i)=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+2i12)=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}

expln(i)=expiπ2\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}

Note by Amal Hari
1 week, 1 day ago

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