puzzle with powers of i

[pearl night]

i^123456787 = (i^4×a) = (i^4)^a
i^4 = 1, 1^anything = 1, no?

the answer is actually -i, but i don’t see where my logic went wrong,

[knotty gorge]

you can only write 123456787 = 4a if 123456787 is divisible by 4

[slender atlas]

complex exponentiation is multibranch

[pearl night]

i meant that a = 123456787/4

[pearl night]

so does this mean theres multiple solutions to my problem?

[knotty gorge]

if you only allow integer exponents, then powers of complex numbers have a single well defined value, and the exponent rules you are used to are true generally

if you start adding fractional exponents then there are multiple values that you have to choose from, and some exponent rules are no longer valid (because they switch which choice you made)

[past vapor]

i^(123456784+3)=i^(123456784) into i^(3) ,since 123456784 is divisible by 4 so i^(123456784)=1, and i^(3)=-i , which is answer

[slender atlas]

no

[lusty cloud]

! m0lc14kk
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[vivid jacinth]

Reference to a table of $i$ values, depending on the exponent:

[
\begin{array}{c|c}
\text{Exponent } n & i^n \
\hline
1 & i \
2 & -1 \
3 & -i \
4 & 1 \
\end{array}
]

The pattern starts repeating every 4 powers. To determine $i^n$ for any positive integer $n$, divide $n$ by 4 and use the remainder to select the corresponding value from the table.

[lusty cloud]

! m0lc14kk

[vivid jacinth]

[
123456787 = 4 \cdot 30864196 + 3
]

Since the remainder is $3$, we use the third pattern in the $i$ table:

[
i^{123456787} = -i
]

[lusty cloud]

! m0lc14kk