I think because z is a variable and having multiple values for 3i on the right hand side provides all the possible solutions. I have a feeling only certain n values will work. There is always an exponent z that makes this equation possible without considering the multiple representations of i at the base. Using n breaks that association. This is my guess btw
If n=1 you have 5pi at the bottom.
Cool!
z = (4m+1)/(4k+1) - i[2.ln3/π(4k+1)]
For principle values m=k=0
z = 1 - i*2ln3/π
Dividing i^(4m + z) = 3•i^(4n + 1) by i^(4n + 1)
i^(4m + z - 4n - 1) = 3
=> i^(4(m - n) + z - 1) = 3
Or i^(4k+ z - 1) = 3, where k = m - n
Maybe do an exponent that multiplies? Like, can you do: W^Z =nW? or at least i^Z = ni? I am a fan of general cases, you see?
Great idea! 😍
Did you do i to the power of z equals negative i? e to the power of z equals negative e?
sounds like great ideas!
Ooooh. That should be fun!!!
Why wouldn't you need n? Isn’t it integral to the substitution for i?
I think because z is a variable and having multiple values for 3i on the right hand side provides all the possible solutions. I have a feeling only certain n values will work. There is always an exponent z that makes this equation possible without considering the multiple representations of i at the base. Using n breaks that association. This is my guess btw