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Complex Numbers — “i”

Q:  Compute i^45 + i^160 – i^74 – i^103 – i^51.


A:  First, we must remember that i is the square-root of -1.  In other words:

i² = -1

So, what is i³?

i³ = i * i * i = i² * i = -1 * i = -1

And,

i^4 = i² * i² = 1

i^5 = i^4 * i = 1 * i = i

See the pattern?

i = i

i² = -1

i³ = -i

i^4 = 1

i^5 = i

i^6 = -1

i^7 = -i

i^8 = 1

….. continued …..

So,

i^45 = i

i^160 = 1

i^74 = -1

i^103 = -i

i^51 = -i

Therefore:

i^45 + i^160 – i^74 – i^103 – i^51 = i + 1 – (-1) -(-i) – (-i) = 2 + 3i

3 thoughts on “Complex Numbers — “i”

  1. the explanation helped a lot but I’m getting a little confused on i^74 and i^51. Shouldn’t i^51= i and 74= 1?

  2. Josefino, there was a mistake with i^51, which is now corrected.

    i^51 = -i

    and

    i^74 = -1

    To verify, all exponents divisible by 4 are equal to “1”. 72 is divisible by 4, so:
    i^72 = 1
    i^73 = i
    i^74 = -1

    Sorry for my hasty error. As mentioned, I had a screaming baby and was unable to edit before posting and logging off!

    1. that makes much more sense, thank you!

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