It appears that the Latex commands have been lost during the transition? So my latex math isn't showing up as it use to. Don't know if you can read my equations above.

It looks like the blackslash command doesn't work anymore - at least when I try to edit. Aha, you have to double the backslash like \\

I found a general solution, but it isn't a single formula -- I can't figure out how to piece the data together into one.

I started with the idea that, well, -1 is just i2:

\[ \large \frac{i^m + i^2}{i^n + i^2} \]

I then did a lot of unnecessary work, and eventually realized that you can just modulus m or n by four because i5 is just i*i4, or put another way, it's 1*i, which is just i.

This means that m and n collapse to either 1 or 3.

I noticed that you can reduce i1 and i3 in this way:

\[ \large \frac{i^3 + i^2}{i^1 + i^2} \]

\[ \large \frac{i^2 ( i + 1 )}{i ( i + 1 ) } \]

The i+i cancels out, and you're left with some power of i.

I then looked all cases, changing 1 to i4 where appropriate:

If m is 1 and n is 3, the answer is i-1

If m is 3 and n is 1, the answer is i1

If m is 1 and n is 1, the answer is i4

If m is 3 and n is 3, the answer is i4

Given more time, I could probably reduce this into a single formula (since that's kinda been my specialty lately), but I couldn't quite figure out how to put it together.

If n is 0 or a multiple of 4 the denominator is 0 so the fraction ((i^m)-1)/0 is meaningless; not a power of i.
Otherwise, if m is 0 or a multiple of 4 the numerator is 0 so the fraction 0/((i^n)-1) is 0. This is not a power of i. Otherwise if m and n are congruent modulo 4 then the numerator and denominator are equal so the fraction has value 1. This is a power of i.
Otherwise you have either (i-1)/(-i-1) or (-i-1)/(i-1)
-i = i^3 so (-i-1)/(i-1)=((i^3)-1)/(i-1)=(i^2)+i+1 = -1 + i +1 = i.
(i-1)/(-i-1) is 1/i is -i. These last two values are both powers of i.
The fraction never evaluates to -1. That’s the only power of i to which the expression does not evaluate for any m or n.

Can you calculate i^i ?
Can you prove it’s a real number?