Euler's Formula
e^{i theta}=cos theta + i sin thetaeiθ=cosθ+isinθ
Let us first review some useful power series.
e^x=1/{0!}+x/{1!}+x^2/{2!}+cdotsex=10!+x1!+x22!+⋯
cos x=1/{0!}-x^2/{2!}+x^4/{4!}-cdotscosx=10!−x22!+x44!−⋯
sin x=x/{1!}-x^3/{3!}+x^5/{5!}-cdotssinx=x1!−x33!+x55!−⋯
Now, we are ready to prove Euler's Formula.
Proof
By rewriting as a power series,
e^{i theta}=1/{0!}+(i theta)/{1!}+(itheta)^2/{2!}+(i theta)^3/{3!}+(i theta)^4/{4!}+(i theta)^5/{5!}+cdotseiθ=10!+iθ1!+(iθ)22!+(iθ)33!+(iθ)44!+(iθ)55!+⋯
by distributing the powers,
=1/{0!}+i theta/{1!}+i^2 theta^2/{2!}+i^3 theta^3/{3!}+i^4 theta^4/{4!}+i^5 theta^5/{5!}+cdots=10!+iθ1!+i2θ22!+i3θ33!+i4θ44!+i5θ55!+⋯
by i^2=-1i2=−1
=1/{0!}+i theta/{1!}-theta^2/{2!}-i theta^3/{3!}+theta^4/{4!}+i theta^5/{5!}-cdots=10!+iθ1!−θ22!−iθ33!+θ44!+iθ55!−⋯
by separating the real part and the imaginary part,
=(1/{0!}-theta^2/{2!}+theta^4/{4!}-cdots)+i(theta/{1!}-theta^3/{3!}+theta^5/{5!}-cdots)=(10!−θ22!+θ44!−⋯)+i(θ1!−θ33!+θ55!−⋯)
by identifying the power series,
=cos theta + i sin theta=cosθ+isinθ
Hence, we have Euler's Formula
e^{i theta}=cos theta+i sin thetaeiθ=cosθ+isinθ.
I hope that this was helpful.
