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 : 
i^{0} = 1 
i^{4} = i^{3 }• i = (i) • i = i^{2} = 1 
i^{8} = i ^{4}• i^{4} = 1 • 1 = 1 
i^{1} = i 
i^{5} = i ^{4}• i = 1 • (i) = i 
i^{9} = i ^{4}• i ^{4}• i = 1 • 1• i = i 
i^{2} = 1 
i^{6} = i ^{4}• i^{2} = 1 • (1) = 1 
i^{10} = (i ^{4})^{2} • i^{2} = 1 • (1) = 1 
i^{3} = i^{2 }• i = (1) • i = i 
i^{7} = i ^{4}• i^{3} = 1 • (i) = i 
i^{11} = (i ^{4})^{2} • i^{3} = 1 • (i) = i 
The powers of i repeat in a definite pattern: (i, 1, i, 1)
Powers of i 
i^{1} 
i^{2} 
i^{3} 
i^{4} 
i^{5} 
i^{6} 
i^{7} 
i^{8} 
... 
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 (i^{0}).
• if the remainder is 1, the answer is i (i^{1}).
• if the remainder is 2, the answer is 1 (i^{2}).
• if the remainder is 3, the answer is i (i^{3}). 

