Euler's Formula
e^{i theta}=cos theta + i sin theta
Let us first review some useful power series.
e^x=1/{0!}+x/{1!}+x^2/{2!}+cdots
cos x=1/{0!}-x^2/{2!}+x^4/{4!}-cdots
sin x=x/{1!}-x^3/{3!}+x^5/{5!}-cdots
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!}+cdots
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
by i^2=-1
=1/{0!}+i theta/{1!}-theta^2/{2!}-i theta^3/{3!}+theta^4/{4!}+i theta^5/{5!}-cdots
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)
by identifying the power series,
=cos theta + i sin theta
Hence, we have Euler's Formula
e^{i theta}=cos theta+i sin theta.
I hope that this was helpful.
