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How Do You Find Higher Powers of i?

How do you find higher powers of i?

Summary

  1. 'i' is the imaginary unit and equal to the square root of -1
  2. We can break higher powers of 'i' into i4 terms multiplied together
  3. Then we can simplify each expression by replacing i4 with 1
  4. Divide the exponent by 4 to figure out how many times we can group 4 'i's together
  5. If the exponent on your 'i' term is divisible by 4, the whole thing will simplify to 1
  6. If you get a remainder of 2 when you divide the exponent by 4, your 'i' term will simplify to -1
  7. If you get a remainder of 3 when you divide the exponent by 4, your 'i' term will simplify to '-i'
  8. If you get a remainder of 1 when you divide the exponent by 4, your 'i' term will simplify to just 'i'

Notes

    1. 'i' is equal to the square root of -1, so squaring both sides of i = √(-1) will give us i2 = -1
    2. i4 is the same as i2 times i2
    3. Since we know i2 equals -1, that means i4 equals -1 times -1
    4. -1 times -1 is just 1, so i4 = 1
    1. Let's take a look at how we can use i4=1 to find higher powers of 'i'
    1. Putting the 'i' terms into groups of four will allow us to use i4 = 1
    2. If we divide the exponent 12 by 4, we can see how many times we have i4 being multiplied
    3. Remember, when we multiply terms with the same base, we can just add their exponents
    4. Since 12÷4=3, we can rewrite 'i12' as three 'i4' terms multiplied together
    1. i12 is the same as multiplying three i4 terms together
    2. Since i4=1, multiplying three of them together will still give you 1
    3. When you have 'i' raised to a power that is divisible by 4, it will always be equal to 1
    1. In this example, we start with i22
    2. Putting the 'i' terms into groups of four will allow us to use i4 = 1
    3. If we divide the exponent 22 by 4, we can see how many times we have i4 being multiplied
    4. Since 22÷4=5 with a remainder of 2, that means we have five i4 terms with two 'i's left over
    5. We can write those two leftover 'i's as i2
    6. 'R2' just means a 'remainder of 2'
    1. i22 is the same as multiplying five i4 terms and one i2 together
    2. Since i4=1, multiplying five of them together will still give you 1
    3. Since i2 = -1, multiplying it by the 1 from the i4 terms will give you -1
    4. Any time you have a remainder of 2 when you divide the exponent by 4, your 'i' term will simplify to -1
    1. Putting the 'i' terms into groups of four will allow us to use i4 = 1
    2. If we divide the exponent 31 by 4, we can see how many times we have i4 being multiplied
    3. Since 31÷4=7 with a remainder of 3, that means we have seven i4 terms with three 'i's left over
    4. Multiplying i4 together 7 times is the same as taking (i4)7, so we can rewrite this as i28
    5. 28 is divisible by 4, which means that i28 is equal to 1
    6. 'R3' just means a 'remainder of 3'
    1. i31 is the same as i28 times i3, which can be rewritten as i28•i2•i
    2. Since 28 is evenly divisible by 4, we know i28 will give us 1
    3. Knowing that i2 = -1, we're left with 1•-1•i, which is just '-i'
    4. Any time you have a remainder of 3 when you divide the exponent by 4, your 'i' term will simplify to '-i'
    1. In this example, we start with i45
    2. Putting the 'i' terms into groups of four will allow us to use i4 = 1
    3. Since 45÷4=11 with a remainder of 1, that means we have eleven i4 terms with one 'i' left over
    4. Multiplying i4 together 11 times is the same as taking (i4)11, so we can rewrite this as i44
    5. 'R1' just means a 'remainder of 1'
    1. i45 is the same as i44•i
    2. Since 44 is evenly divisible by 4, we know it will give us 1
    3. 'i' multiplied by 1 is just 'i'!
    4. Any time you have a remainder of 1 when you divide the exponent by 4, your 'i' term will simplify to just 'i'