Question

Complex Numbers Question

Answer 1

Since
`i = √(-1)`, it follows that
`i^2 = -1`,
`i^3 = -i`, and
`i^4 = 1`.

Then

`i^5 = i^4* i = i \\\\ i^6 = i^4* i^2 = -1 \\\\ i^7 = i^4* i^3 = -i \\\\ i^8 = i^4 * i^4 = 1`

and the pattern repeats. The value of
`i^n` boils down to finding the remainder upon dividing
`n` by 4.

We have

`i^(11) = i^8* i^3 = -i \\\\ i^(33) = i^(32) * i = \left(i^4\right)^8* i = i \\\\ i^(257) = i^(256) * i = \left(i^4\right)^(64)* i = i`

so that

`13 - 6i^(11) + 2i^(33) - 5i^(257) = 13 + 6i + 2i - 5i = \boxed{13 + 3i}`

and so the answer is C.