Hyperfinity
Answered

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Complex Numbers Question

Complex Numbers Question class=

Sagot :

LammettHash

Since [tex]i = \sqrt{-1}[/tex], it follows that [tex]i^2 = -1[/tex], [tex]i^3 = -i[/tex], and [tex]i^4 = 1[/tex].

Then

[tex]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[/tex]

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

We have

[tex]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[/tex]

so that

[tex]13 - 6i^{11} + 2i^{33} - 5i^{257} = 13 + 6i + 2i - 5i = \boxed{13 + 3i}[/tex]

and so the answer is C.

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