The N-th power of a (complex number) is located as shown, and all triangles are similarity. because the absolute values are multiplied and the arguments are added to yield the polar form of the product. the points are on a logarithmic spiral.
Put a=(1+i θ/N), and take a limit N→∞ then the logarithmic spiral approaches the unit circle. The argument of a is equal to arctan(θ/N). This argument is approximately equal to θ/N in case of N»1.
The N-th power of a is located as shown.
To sum up, take the limit N→∞ then a^N approaches cos(θ)+i sin(θ).
Remember the definition of e^z, We obtain Euler’s formula, e^i θ=cos(θ)+i sin(θ).