When the imaginary unit, i, is raised to increasingly higher powers, a cyclic (repetitive) pattern emerges. Remember that i 2 = -1.
 Repeating Pattern of Powers of i : i0 = 1 i4 = i3 • i = (-i) • i = -i2 = 1 i8 = i 4• i4 = 1 • 1 = 1 i1 = i i5 = i 4• i = 1 • (i) = i i9 = i 4• i 4• i = 1 • 1• i = i i2 = -1 i6 = i 4• i2 = 1 • (-1) = -1 i10 = (i 4)2 • i2 = 1 • (-1) = -1 i3 = i2 • i = (-1) • i = -i i7 = i 4• i3 = 1 • (-i) = -i i11 = (i 4)2 • i3 = 1 • (-i) = -i

The powers of i repeat in a definite pattern: (i, -1, -i, 1)

 Powers of i i1 i2 i3 i4 i5 i6 i7 i8 ... Simplified Form i -1 -i 1 i -1 -i 1 ...

Simplifying powers of i:
You will need to remember (or establish) the powers of 1 through 4 of i to obtain one cycle of the pattern. From that list of values, you can easily determine any other positive integer powers of i.

 Method 1: When the exponent is greater than or equal to 5, use the fact that i 4 = 1 and the rules for working with exponents to simplify higher powers of i. Break the power down to show the factors of four.

 When raising i to any positive integer power, the answer is always i, -1, -i or 1.
 Another way to look at the simplification: Method 2: Divide the exponent by 4: • if the remainder is 0, the answer is 1 (i0). • if the remainder is 1, the answer is i (i1). • if the remainder is 2, the answer is -1 (i2). • if the remainder is 3, the answer is -i (i3).

Simplify i87

 By Method 1: Break down the power to show factors of 4. (84 is the largest multiple of 4) By Method 2: Divide the power by 4 to find the remainder. 87 ÷ 4 = 21 with remainder 3 The answer is i3 which is -i.