How do you simplify [tex]i^{29}[/tex]?
(Note: [tex]i[/tex] is the imaginary unit, where [tex]i^2 = -1[/tex].)
Answer (2)
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 '.
To simplify i 29 , we first recognize that the powers of i repeat every 4 terms. By dividing 29 by 4, we find it corresponds to the first position in the cycle, which is i . Thus, i 29 = i .
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