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
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?
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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!
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that makes much more sense, thank you!
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