Question

Complex Numbers Question

Answer 1

Since $i = \sqrt{-1}$, it follows that $i^2 = -1$, $i^3 = -i$, and $i^4 = 1$.

Then

$i^5 = i^4\times i = i \\\\ i^6 = i^4\times i^2 = -1 \\\\ i^7 = i^4\times i^3 = -i \\\\ i^8 = i^4 \times i^4 = 1$

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

We have

$i^{11} = i^8\times i^3 = -i \\\\ i^{33} = i^{32} \times i = \left(i^4\right)^8\times i = i \\\\ i^{257} = i^{256} \times i = \left(i^4\right)^{64}\times i = i$

so that

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

and so the answer is C.