>>11686757let f(z) = e^iz so that f(?) = e^i?
f'(z) = ie^iz
f'(z) = i * f(z)
So wherever f(z) is, it is headed in the direction counterclockwise around the unit circle (since f(0) trivially equals 1).
The distance to -1 on this trajectory is ?, so f(?) = -1
e^i? = -1
e^i? + 1 = 0
bada-bing bada-boom 5 different cool numbas.
