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 '.