Annoying Imaginary Friends

Using Eulers formula:

$i = e^{i\frac{\pi}{2}}$

Substituting this for i:

${(e^{i\frac{\pi}{2}})} ^ {0!}$ + ${(e^{i\frac{\pi}{2}})} ^ {1!}$ + ${(e^{i\frac{\pi}{2}})} ^ {2!}$ + ${(e^{i\frac{\pi}{2}})} ^ {4!}$ + ... + ${(e^{i\frac{\pi}{2}})} ^ {100!}$

$e^{i\frac{\pi}{2}}$ + $e^{i\frac{\pi}{2}}$ + ${e^{i\pi}}$ + ${({e}^{i\pi})} ^ {3}$ + ${({({e}^{i\pi})} ^ {3})} ^ {4}$ + ....

$e^{i\frac{\pi}{2}}$ + $e^{i\frac{\pi}{2}}$ + ${e^{i\pi}}$ + ${e}^{3i\pi}$ + ${e}^{12i\pi}$ + .... + ${e}^{(99!*50)i\pi}$

Note that $e^{i\theta}$ is cyclical, repeating with multiples of $2\pi$ . If $\theta$ is an even multiple of $\pi$ , than $e^{i\theta} = e^{0} =1$ . Using this fact, the sum becomes

$i + i - 1 - 1 + 1 + 1 + ... +1 = 95 + 2i$

$i^n = 1$ if $n \equiv 0 \quad \left(\bmod \text{ 4} \right)$ . Additionally, if $x \geq 4$ then $4 | x!$ . The summation can be rewritten as

$= i^{0!} + i^{1!} + i^{2!} + i^{3!} + 1 \cdot 97 \\ = i + i -1 -1 + 97 \\ = 95 + 2i$

Using i^{4k} = i^4 = 1 where k is a positive integer, you can break the sum into i^{0!}+i^{1!}+i^{2!}+i^{3!}+(100-3) = i+i+(-1)+(-1)+97 = 95 + 2i.