How do you simplify i²⁹? (if you can't tell, that's the imaginary number i^29).

The imaginary number ' i ' is the square root of -1.

i = √-1
i² = -1
i³ = -√-1
i to the 4th power = +1
i to the 5th power = √-1
etc.

and all the higher powers keep going in the same 4-step cycle.

What's 29 divided by 4 ?
It's 7 with a remainder of 1.

So 29 with all the 4s thrown away is 1, and ' i ' to the 29th power is
the same thing as ' i ' to the 1st power = √-1  or just  plain ' i '.